mike mcgarry - 加拿大vs摩洛哥欧赔

one month gre study schedule for advanced students

thu, 12 sep 2024 15:00:37 +0000

the content in this post applies in 2024 to the new, shorter gre!

for those who are focused and ambitious, a 30-day gre study schedule can work really well, as long as you have the time it will demand.

however, if you feel that life will intrude in one way or another, or if you need to improve your score by hundreds of points, this schedule may not be for you. if an honest look at life’s daily demands reveals that dedicating hour upon hour to gre study will be unlikely, ask yourself: is pushing back your test date possible?

if you are able to register or reschedule your exam for a later date, that might be in your best interest, and then you can check out some of magoosh’s other, longer schedule options.

however, if you are all-in on preparing for the gre in thirty days, we’ve got you! just remember, this is an aggressive plan, and you will need to be able to dedicate, on average, 1.5-2.5 hours a day at least six days per week.

click here to download a pdf of the study schedule!

for advanced gre students: an emphasis on reviewing & drilling

this schedule was designed with advanced students in mind, which means that most of the assigned lessons skew toward refining the test-taking skills you already have. before you dive in, ask yourself whether you currently have a working knowledge of the fundamentals–the basic topics, techniques, and strategies necessary to successfully navigate the gre exam. if that’s not you or you’d benefit from more introductory material, we do have a 30-day option for beginners. also, because this schedule assumes that you won’t need to cover much (or any) of the introductory lessons, there is also a greater emphasis on practice questions and review.

now, if this schedule is mostly a good fit, but at certain points you find the higher-level content a bit too challenging, you can always make modifications to better suit your own prep needs. once you have a sense of your personal strengths and weaknesses, you can explore all the available lessons on the magoosh dashboard, swap out lesson material that you aren’t quite ready to tackle for introductory or mid-tier concepts, and set up custom practice sessions that focus on your desired topic(s) of study.

essential material

magoosh gre prep
ets’s official guide to the gre book
ets powerprep free tests 1 & 2: if you would prefer to take the practice test on paper, you can print out a copy of ets’s paper-delivered test. however, taking the paper-based gre is pretty rare these days. if you are like most test takers and will be taking the exam online, we highly recommend that your practice and prep are also online–mimic the medium you’ll be using on exam day as much as possible.
magoosh’s gre vocabulary flashcards app and magoosh’s gre math flashcards app: they’re free and you can use them on the web, iphone/ipad or android!
magoosh’s complete guide to the gre pdf: this comprehensive, web-based guide to the gre gives you the quick but very helpful overview you need to understand this test. you’ll see how the gre is designed and scored, what skills it tests, how to find and use the best gre prep, and how to study for each test section.
reading materials and journal (error log): more detail on each of these below, in the “about the material” section.

optional material

top 5 free gre practice resources: this page includes instructions on where to find good full-length gre practice tests, and how to take practice tests and incorporate them into your studies. this page also has links to magoosh’s free gre diagnostic quizzes.
quizlet.com: gives you online access to flashcards, making it easier, say, to quiz yourself on your mobile device
vocabulary.com: provides conversational example sentences and a flood of example sentences. if you are really trying to achieve a high verbal score, you can also check out lists of crazy hard gre words that other users have compiled.

about the material

about vocabulary

coming back to math after years away

tue, 25 jun 2024 15:52:26 +0000

suppose you have been away from math for a while, perhaps since algebra two and the sat in high school. after these necessarily evils, you bid it a hasty farewell and gave it the bum’s rush from your entire life. now, several years later, you have completed a baccalaureate degree in something, and wish to go on to advanced degree — all highly commendable — but alas! graduate schools require the gre, and the gre involves math. how does one get back into math after years away?

math anxiety

some folks who have not dealt with math for a wrestle with “math anxiety”, and may even be heard to utter such things as “i’ve never been good at math.” it actually says something quite profound about the american math educational system that such a substantial proportion of intelligent and highly educated people claim to be “bad at math.”

study after study has shown that “math anxiety” involves self-fulfilling prophecies galore: folks who don’t like math avoid it, don’t practice it, and thus get rusty at it and are more likely to make mistakes because they are rusty — which reinforces the anxiety. in most cultures in which “math anxiety” is not a huge issue, proficiency at math is regarded as the result of simple hard work and practice, not of some kind of genetic windfall that a nerdy few have and most people don’t.

all of this is to say: dive in. you’re an intelligent and well-educated person: you can do this! do math every day between now and your gre. math is not a spectator sport: the only way you learn is by doing it. yes, you will make mistakes in the beginning. that’s fine. that’s exactly what is supposed to happen in the math-learning process. just learn from your mistakes and keep working.

practice, practice, practice

before you even begin your formal gre practice, keep a few gre math tips in mind!

start warming up with mental math. you should be able, in your head, to add and subtract random two digit numbers, and multiply all single digit numbers. make sure you know your “times table” inside out, and drill it until you know it. start reviewing arithmetic with fractions and decimals, with percents and ratios. if you are a magoosh member, you will find our introductory math video lessons helpful. this mcgraw-hill book is also a very gentle refresher.

one of the biggest traps on the gre quantitative section is calculator overdependence. math-phobes sometimes get lazy and think: i don’t really need to know what 5 times 7 is, because i’ll always have a calculator on the gre. that is a stupendously bad approach! one gre quant question after another is written specifically to punish all those people who instinctually reach for the calculator all the time. the more you can rely on your own mental math, and not touch the calculator, the more you will avoid these traps, and the more successful you will be. there is no substitute for strong mental math skills. ideally, you should use the calculator no more than two or three times at most in an entire gre quant section.

get comfortable with estimation. for example, suppose however the question is framed, it involves multiplying 298 times 5.2. a benighted approach would use the calculator to calculate that exact answer. a considerably more sophisticated approach would recognize immediately —- 298 is almost 300, and 5.2 is very close to 5, so (298)*(5.2) must be very close to 300*5, and your mental math has to be good enough that you can do that multiplication in your head: 300*5 = 1,500. quite often, when ugly numbers are given in the question, answer choices are spaced apart enough that estimation easily will isolate a single answer.

learn the divisibility tricks. learn the doubling & halving shortcut. part of success on gre math is developing something called “number sense”, which comes only from “playing” with numbers (see chris’s description of the “99 game” in this post.) the more you can develop a sense of “play” in games such as these, the more it will build your confidence in math.

math and your brain

i would also recommend getting curious about your own brain and how it interacts with mathematical ideas. this post talks a little about left-brain vs. right-brain approaches to math. many of the folks who really struggle with math are right-brain, big-picture types, perhaps with aptitude for the arts, and typically not good with details, organization, and precision. these people typically get the big ideas very quickly, but then can’t apply them in the rough and tumble of individual problems.

if this is you, i have some hard left-brain medicine for you. unrelated to anything specifically mathematical, you need to practice (a) detail management; (b) organization; and (c) precision, in any areas in which it is possible. organize that messy drawer of your dresser, or your kitchen cabinet, or your medicine cabinet, in the most hyper-rational anal-retentive way possible. pay attention to subtle precise differences in colors, in fonts, in leaf shapes, in window designs, in bird calls, etc. etc. volunteer to plan and organize things for people, forcing yourself to manage all the relevant details. in all the environments in which you are the most familiar (your house, your place of work, store where you frequently shop, etc.), practice noticing a new detail, something you have never seen before, each and every time you are there. in all of these, you are trying to exercise your left-brain capacities: just a little strength more on that side of your brain will have huge payoffs in your ability to handle math problems.

summary

you are an intelligent person. therefore, you are eminently capable of learning enough math to perform well on the gre quant section. don’t let your own habitual fears psyche you out. practice feeling confidence in math, and practice math questions with a sense of curiosity and adventure. believe in your ability to develop the necessarily skills and perspectives, and don’t let yourself down!

the post coming back to math after years away appeared first on magoosh blog — gre® test.

]]>

how to round to the nearest integer | gre math

fri, 05 apr 2024 12:40:43 +0000

what does it mean to round a number? rounding means to make a number shorter or simpler, but keeping it as close in value as possible to the original number. let’s take a closer look in this edition of gre math tips.

rounding to the nearest integer

the most common type of rounding is to round to the nearest integer. the rule for rounding is simple: look at the digits in the tenth’s place (the first digit to the right of the decimal point). if the digit in the tenths place is less than 5, then round down, which means the units digit remains the same; if the digit in the tenths place is 5 or greater, then round up, which means you should increase the unit digit by one.

here are a couple of other things to know:

observation #1: under most circumstances, rounding changes the decimal to whatever integer is closer. for example, 4.3 is rounded to 4, and 4.9 is rounded to 5. the exception is when the decimal is smack dab between two integers: 4.5 is exactly equidistant to both 4 and 5, but because of the “tie-breaker” rule of rounding, anything with a 5 in the tenths digit is rounded up. this is the only case in which the “go to the closer integer” interpretation will fail.

observation #2: do not double-round. some people look at a number like, say, 7.49, and they erroneously think — well, that 9 would round the 4 up to 5, and then a 5 gets round up, to this number would round to 8. wrong! never round a number “in stages.” rounding is a one-shot deal, a one-step process. when the number we need to round is 7.49, we only need notice that the tenth’s digit is a 4, which means the number is rounded down to 7. one step, case closed. in fact, all of the following numbers get rounded to 7:

7.49
7.499999
7.4999999999999999999

—> all get rounded to 7

here’s the truly mind-boggling part:

how many numbers would there be larger than this last number, but still lower than 7.5? infinity! no matter how many additional 9’s we slap on to the end of that number, there’s still a continuous infinity of decimals larger than that number and below 7.5 no matter how finely we chop up the real number line, each tiny fragment of the line, no matter how small, still contains a continuous infinite of numbers.

observation #3: the “tie-breaker” rule can be tricky with negative values. for example, +2.5 gets rounded up to 3, but –2.5 gets rounded down … to –3. as with positive numbers, the negative number ending in .5 is rounded to the higher absolute value integer, but with negatives, that’s rounding down. (this is not the only way to formulate this rule, but this is the convention that ets follows.)

rounding to any other decimal place

rounding to the nearest integer is really rounding to the nearest units place. sometimes, you will be asked to round to the nearest hundreds, or to the nearest hundredths — to some decimal place other than the units place. the rule is just a more generalized version of the previous rounding rule.

suppose we are asked to round to some specific decimal place — call this the “target place.” you always look at only one digit, the digit immediately to the right of the target place. if this digit immediately to the right is {0, 1, 2, 3, 4}, then you “round down”, and the digit in the target place remains unchanged. if this digit immediately to the right is {5, 6, 7, 8, 9}, then you “round up”, and the digit in the target place increases by 1.

what do you have to know about how to round to the nearest integer for the gre?

not many gre questions will say, “here’s a number: round it to the nearest such-and-such.” by contrast, many questions, in the course of asking something else, could ask you to round your answer to the nearest such-and-such. in this way, rounding is one math skill you need to know for the gre. there are a few tricky issues, which i will address here.

other cases of rounding

very occasionally, a gre math question may ask you not to round to a particular decimal place, but rather to the nearest multiple of something. for example, suppose you are asked to round, say, to the nearest 0.05 — how do you do that?

well, let’s think about the results first. the result of rounding to the nearest 0.05 would be something divisible by 0.05 — that is to say, a decimal with either a 0 or a 5 in the hundredth place, no digits to the right of that, and any digits to the left of that. the following are examples of numbers which could be the result of rounding to the nearest 0.05:

0.35
1.40
3.15
5.2
8

notice: the second, (b) is the square root of 2 (sqrt{2} = 1.414213562….) rounded to the nearest 0.05, and the third, (c), is pi rounded to the nearest 0.05.

let’s demonstrate the rounding by means of an example. what numbers, when rounded to the nearest 0.05, would be rounded to 2.35? well, for starters, 2.35 and other “tenths” around it would be rounded to 2.35

2.32 — rounded down to 2.30

2.33 — rounded up to 2.35

2.34 — rounded up to 2.35

2.35 — stays at value

2.36 — rounded down to 2.35

2.37 — rounded down to 2.35

2.38 — rounded up to 2.40

now, the tricky regions are those between the values that are rounded in different directions. for example, 2.32 is rounded down and 2.33 is rounded up, so something fishy is happening between those two. let’s think about the hundredths between 2.32 and 2.33 — exactly between them is 2.325, the midpoint between 2.30 and 2.35, and like all midpoints, according to the “tiebreaker” rule, it gets rounded up. thus:

2.320 —- rounded down to 2.30

2.321 —- rounded down to 2.30

2.322 —- rounded down to 2.30

2.323 —- rounded down to 2.30

2.324 —- rounded down to 2.30

2.325 —- rounded up to 2.35 (the “tie-breaker” rule)

2.326 —- rounded up to 2.35

2.327 —- rounded up to 2.35

2.328 —- rounded up to 2.35

2.329 —- rounded up to 2.35

2.330 —- rounded up to 2.35

this is all probably far more detail than you will need to know for gre math, but this does demonstrate the steps you would take to round any decimal to the nearest 0.05. by analogy, you could round any decimal to any specified multiple.

got rounding to the nearest integer mastered? test your skills on magoosh’s gre math practice questions with answers and explanations, or check out the rounding question below!

need more help?

improve your gre scores with magoosh gre, you can choose between a live cohorted class with an instructor (which includes all our lessons and practice questions) or access to the self-study option by itself.

the post how to round to the nearest integer | gre math appeared first on magoosh blog — gre® test.

]]>

90 day gre study plan for advanced students

fri, 23 feb 2024 16:00:54 +0000

the content in this post applies in 2024 to the new, shorter gre!

if you’re aiming for a top score on the gre, and you’re starting out with a strong background in the content areas that the gre tests, then this 90 day study plan can help get you where you want to be! here, you’ll find the guidance and materials you’ll need from a 3-month gre study plan to master the material.

click here to download a pdf of the study schedule!

this 90-day plan is designed to help you refine the test-taking skills you already have in order to achieve your scoring goals. the schedule requires twelve weeks, assuming 1.5-2.5 hours for each of the five weekdays, and one longer stint on the weekend (“day six”). if you would prefer to work on both weekend days, and free up some time during the week, feel free to make those changes.

for advanced gre students: an emphasis on reviewing & drilling

this schedule is best suited for advanced students. most of the assigned lessons skew towards covering topics that require a working knowledge of the fundamentals–the basic topics, techniques, and strategies that establish a solid foundation for how to approach the gre exam. there is also a greater emphasis on working and reviewing practice questions.

as you work through this schedule, if you find that some of the higher-level content is a bit too challenging, you can always make modifications to better suit your own prep needs. once you have a sense of your personal strengths and weaknesses, you can explore all the available lessons on the magoosh dashboard, swap out lesson material that you aren’t quite ready to tackle for introductory or mid-tier concepts, and set up custom practice sessions that focus on your desired topic(s) of study.

curious about other study plan options or how to best modify a schedule depending on your situation? check out this post: gre study plans and guides.

essential material
optional material
about the material
week 1
week 2
week 3
week 4
week 5
week 6
week 7
week 8
week 9
week 10
week 11
week 12
week 12.5
takeaways

essential material

magoosh gre prep
ets’s official guide to the gre book
ets powerprep free tests 1 & 2: if you would prefer to take the practice test on paper, you can print out a copy of ets’s paper-delivered test. however, taking the paper-based gre is pretty rare these days. if you are like most test takers and will be taking the exam online, we highly recommend that your practice and prep are also online–mimic the medium you’ll be using on exam day as much as possible.
magoosh’s gre vocabulary flashcards app and magoosh’s gre math flashcards app: they’re free and you can use them on the web, iphone/ipad or android!
magoosh’s complete guide to the gre pdf: this comprehensive, web-based guide to the gre gives you the quick but very helpful overview you need to understand this test. you’ll see how the gre is designed and scored, what skills it tests, how to find and use the best gre prep, and how to study for each test section.
reading materials and journal (error log): more detail on each of these below, in the “about the material” section.

optional material

top 5 free gre practice resources: this page includes instructions on where to find good full-length gre practice tests, and how to take practice tests and incorporate them into your studies. this page also has links to magoosh’s free gre diagnostic quizzes.
quizlet.com: gives you online access to flashcards, making it easier, say, to quiz yourself on your mobile device
vocabulary.com: provides conversational example sentences and a flood of example sentences. if you are really trying to achieve a high verbal score, you can also check out lists of crazy hard gre words that other users have compiled.

about the material

about vocabulary

90 day gre study plan for beginners

fri, 23 feb 2024 16:00:32 +0000

the content in this post applies in 2024 to the new, shorter gre!

if you’re starting more or less from scratch, you may be wondering if 90 days (~3 months) of gre prep is even close to enough time. it is! but there’s a key component to preparing for the gre: you need to be strategic about how you organize your study. luckily, magoosh has an expert-written plan that can help you figure out how to prepare to meet and exceed your gre goals in the next three months!

click here to download a pdf of the study schedule!

this 90-day plan is designed to have you improve as much as you can in a three-month period. we have designed twelve weeks, assuming 1.5-2.5 hours for each of the five weekdays, and one longer stint on the weekend (“day six”). if you would prefer to work on both weekend days, and free up some time during the week, feel free to make those changes.

for gre beginners: starting with the basics

this schedule was designed with beginners in mind, which means that most of the assigned lessons skew towards covering the fundamentals–the basic topics, techniques, and strategies that will give you a solid foundation for how to approach the gre exam.

if you find that you are already comfortable with a lot of the introductory material, you can always modify this schedule to suit your own prep needs. once you have a sense of your personal strengths and weaknesses, you can dig into all the available lessons on the magoosh dashboard, swap out lesson material that you’re well-versed in for more advanced concepts, and set up custom practice sessions that focus on your desired topic of study. this schedule was designed to be easily modified based on the needs of individual users.

curious about other study plan options or how to best modify a schedule depending on your situation? check out this post: gre study plans and guides.

essential material
optional material
about the material
week 1
week 2
week 3
week 4
week 5
week 6
week 7
week 8
week 9
week 10
week 11
week 12
week 12.5
takeaways

essential material

magoosh gre prep
ets’s official guide to the gre book
ets powerprep free tests 1 & 2: if you would prefer to take the practice test on paper, you can print out a copy of ets’s paper-delivered test. however, taking the paper-based gre is pretty rare these days. if you are like most test takers and will be taking the exam online, we highly recommend that your practice and prep are also online–mimic the medium you’ll be using on exam day as much as possible.
magoosh’s gre vocabulary flashcards app and magoosh’s gre math flashcards app: they’re free and you can use them on the web, iphone/ipad or android!
magoosh’s complete guide to the gre pdf: this comprehensive, web-based guide to the gre gives you the quick but very helpful overview you need to understand this test. you’ll see how the gre is designed and scored, what skills it tests, how to find and use the best gre prep, and how to study for each test section.
reading materials and journal (error log): more detail on each of these below, in the “about the material” section.

optional material

top 5 free gre practice resources: this page includes instructions on where to find good full-length gre practice tests, and how to take practice tests and incorporate them into your studies. this page also has links to magoosh’s free gre diagnostic quizzes.
quizlet.com: gives you online access to flashcards, making it easier, say, to quiz yourself on your mobile device
vocabulary.com: provides conversational example sentences and a flood of example sentences. if you are really trying to achieve a high verbal score, you can also check out lists of crazy hard gre words that other users have compiled.

about the material

about vocabulary

gre geometry formulas

tue, 08 dec 2020 03:00:50 +0000

perhaps you recall with more than a little dread your high school geometry class: fifteen-step proofs and the law of cosines were two things you were certain you’d never see again. well, if you are taking the gre, you were by no means mistaken—proofs and indeed all of trig are nowhere to be found on gre math.

that doesn’t mean that the gre is void of geometry. below are the gre geometry rules and geometry formulas you have to know. this list is by no means exhaustive, but for must-know concepts you’ve come to the right place.

gre geometry formulas

if you’re interested in more than geometry, try our more general gre math formula ebook!

gre geometry formulas at a glance

need to see the key geometry formulas in one place? you’re in luck! here’s a list of geometry formulas you’ll need on test day. bookmark this page, and read on for more in-depth explanations of each geometry formula.

category	rule/formula
angles	click here for all rules 6 month gre study schedule sun, 18 oct 2020 21:00:32 +0000 the content in this post applies in 2024 to the new, shorter gre! the magoosh 6 month study plan moves at a relaxed pace that should still be followed diligently, lest procrastination leads to things getting out of hand. it’s designed to expose you to everything that the gre might throw at you on test day as well as equip you with strategies and techniques to help you achieve your dream score. click here to download a pdf of the study schedule! time commitment students should assume each session will take somewhere between 1.5-2.5 hours. earlier in the schedule, most of the work is assigned across 4 days, with days 5-6 used in whatever way will best suit your prep needs. day 7 is a rest day. later in the schedule, day 5 is allocated to verbal practice. why does verbal get its own day? well, based on the number of concepts tested, math topics far outweigh verbal topics. therefore, to ensure that verbal doesn’t slip by the wayside once its lessons are complete, starting in the second half of the schedule, you’ll see dedicated time for verbal practice. designed to be flexible now, how you approach the schedule is really up to you. you can opt for a monday-saturday routine. you can combine days if you want more study-free days during the week. you can work two, long weekend days to free up time on the weekdays. feel free to make any changes that best suit the unique demands of your daily life. one of our goals was to make this schedule easy to modify because we know that balancing responsibilities is tough and a study schedule is rarely “one-size fits all.” another thing to note, this schedule was designed for those who have the time to become gre experts in mind. this means that the lessons cover everything from test-taking fundamentals to the most advanced, rarely tested content. if you find that you are already comfortable with a lot of the introductory material, you can always make changes to suit your unique prep needs. once you have a sense of your personal strengths and weaknesses, you can swap out lesson material that you’re well-versed in for more advanced concepts, and set up custom practice sessions that focus on your individual priorities. curious about other study plan options or how to best modify a schedule depending on your situation? check out this post: gre study plans and guides. feeling like just getting started is a challenge? check out how other students were able to balance the demands of work, life, and gre preparation in our post how to study for the gre. table of contents essential material optional material about the material month 1 month 2 month 3 month 4 month 5 month 6 takeaways essential material magoosh gre prep ets’s official guide to the gre book ets powerprep free tests 1 & 2: if you would prefer to take the practice test on paper, you can print out a copy of ets’s paper-delivered test. however, taking the paper-based gre is pretty rare these days. if you are like most test takers and will be taking the exam online, we highly recommend that your practice and prep are also online–mimic the medium you’ll be using on exam day as much as possible. magoosh’s gre vocabulary flashcards app and magoosh’s gre math flashcards app: they’re free and you can use them on the web, iphone/ipad or android! magoosh’s complete guide to the gre pdf: this comprehensive, web-based guide to the gre gives you the quick but very helpful overview you need to understand this test. you’ll see how the gre is designed and scored, what skills it tests, how to find and use the best gre prep, and how to study for each test section. reading materials and journal (error log): more detail on each of these below, in the “about the material” section. optional material top 5 free gre practice resources: this page includes instructions on where to find good full-length gre practice tests, and how to take practice tests and incorporate them into your studies. this page also has links to magoosh’s free gre diagnostic quizzes. quizlet.com: gives you online access to flashcards, making it easier, say, to quiz yourself on your mobile device vocabulary.com: provides conversational example sentences and a flood of example sentences. if you are really trying to achieve a high verbal score, you can also check out lists of crazy hard gre words that other users have compiled. manhattan’s 5lb book of gre practice problems is not absolutely necessary, but if you are in need of even more practice material, this is a quality source. don’t feel like you need the latest version. the content of the gre has not changed significantly, so an older edition will suit your practice needs just fine. full disclosure: we find the math content to be very test-like. the verbal content, on the other hand, tends to veer a bit harder than what you’d see on the gre, especially for text completions and sentence equivalence. about the material about vocabulary positive and negative square roots on the gre wed, 07 sep 2016 22:52:50 +0000 some questions to start: can a square root be negative? yes, there is such a thing as a negative square root. how would you deal with negative roots on the gre? this question is actually fairly complicated, because it depends on how the gre words the question. let’s start with some gre math practice problems before i explain. negative square root practice problems first of all, consider these two similar, but not identical, quantitative comparison questions. these are a bit easier than you would see on the gre, but they illustrate an important distinction about possible negative square roots. 1) 2) all i will say right now is: despite apparent similarities, those two questions about whether roots are positive or negative, have *two completely different answers. the distinction between them is the subject of this article. explanations to these practice problems will appear at the end of this blog article. jump ahead by clicking here. do i include the negative roots? often, students are confused about this question. for example, in the questions above, we know +4 is a square root of 16, but can’t -4 be one as well? or can it? can a square root be negative? do we include the negative square root as part of column b or not? does it matter how the question is framed? all of these questions about possible negative square roots are resolved by understanding the following two cases. case i: the symbol appears in case one, the test-maker, in writing the question, uses this symbol. this symbol appears printed on the page in the question itself.* what is this symbol? well, the most folks call this simply a “square-root” symbol, but the proper name is the “principal square root” symbol. here, “principal” (in the sense of “main” or “most important”) means: you take one and only one root, the most important, or principal, one — the positive root only. that is the deep meaning of this symbol. thus, in all cases in which this symbol appears as part of the question itself, you never consider the negative square root, and only take the positive square root. case ii: the symbol doesn’t appear in this case, that special symbol does not appear as part of the problem. what does appear is, for instance, a variable squared, or some other combination of algebra that leads to a variable squared, and you yourself, in your process of solving the problem, have to take the square root of something in order to solve it. the act of “square rooting” is not initiated by the test maker in the act of writing the question; rather, it is you who initiated the square-rooting. in this case, 100% of the time, you always have to consider both the positive and negative square roots. summary: can a square root be negative? can a square root be negative? well, the answer is: it depends on what was printed in the problem. the principal square root symbol never has a negative output, so if the test maker printed that symbol, it’s restrictions have to be respected: all square roots then are positive. on the other hand, if the problem contains a variable squared, or some other algebra that leads to a variable squared, and you yourself take a square root as part of the act of solving, then you always have to consider all possible solutions, both the positive square roots and the negative square roots. if you master that distinction, you will always understand when to consider both positive and negative roots vs. when to consider only the positive root. you may want to go back to those two qcs at the beginning and think them through again before reading the solutions below. additional resources if you’re having trouble with quantitative comparison questions on the gre, i would recommend taking a look at some of these additional resources. they may help clear up some of the concepts that you’re struggling with, and can provide some extra practice. quantitative comparison tips qc tip #1: dealing with variables qc tip #2: striving for equality qc tip #3: logic over algebra qc tip #4: comparing in parts qc tip #5: estimation with a twist quantitative comparison strategies qc: the devil is in the details qc: “the relationship cannot be determined from the information given” answer choice qc and manipulation gre math: solving quantitative comparisons quantitative comparison practice gre qc practice problems gre qc geometry practice problems gre math help gre math formula ebook practice problem solutions 1) here, the principal square root symbol appears as part of the problem itself. we are in case i. of course, that symbol means: take the positive square root only. so column b can only equal +4. of course, that’s always bigger than 3. answer = (b). 2) here, there’s no square root symbol printed as part of the problem itself. we are in case ii. for any square roots we take as part of our solution, we are liable to account for both the positive and negative roots. sure enough, the very first thing we encounter in the prompt is a variable squared, and when we solve for x, we have to account for both roots: x = ±4. the variable x could equal well have either one of those values. now, when we proceed to the qc, we see that the different values of x would give different answers. if x = +4, then column b is greater, but if x = -4, then column a is greater. different values lead to different conclusions, and this situation means we don’t have enough information to establish a definitive relationship. answer = (d) special note: to find out where roots sit in the “big picture” of gre quant, and what other quant concepts you should study, check out our post entitled: what kind of math is on the gre? breakdown of quant concepts by frequency editor’s note: this post was originally published in september, 2012, and has been updated for freshness, accuracy, and comprehensiveness. the post positive and negative square roots on the gre appeared first on magoosh blog — gre® test. ]]> diagonals of a polygon in gre geometry wed, 07 sep 2016 22:32:36 +0000 students love to skip over the basics, asking questions like: how many vertices does an octagon have? how many diagonals does an octagon have? what’s the difference between a regular octagon and, well, an octagon? and gre geometry does delve into some complex polygon math. but before we get to that, i will begin with two challenging gre math problems. gre geometry: polygon problems 1) regular pentagon p has all five diagonals drawn. what is the angle between two of these diagonals where they meet at a vertex of the pentagon? (a) 12° (b) 36° (c) 54° (d) 60° (e) 72° 2) how many diagonals does a regular 20-sided polygon have? (a) 60 (b) 120 (c) 170 (d) 240 (e) 400 explanations to these practice problems will appear at the end of this blog article. jump ahead by clicking here. polygons first, some basic terminology to begin this discussion. a polygon is any geometric shape all of whose sides are straight line-segments. any triangle is a polygon. any quadrilateral (including trapezoids, parallelograms, rhombuses, rectangles and squares) are polygons. a pentagon is a 5-sided polygon. a hexagon is a 6-sided polygon. an octagon is an eight-sided polygon. a circle or parabola or anything with a curved side is not a polygon. a point where two of the sides of a polygon meet is called a vertex. the number of vertices a polygon has is always equal to the number of sides it has. another important polygon fact concerns the sum of angles. you may know that the sum of the three angles in any triangle is 180°. you may even know that the sum of the four angles in any quadrilateral is 360°. this pattern generalizes. the sum of all n angle in any n-sided polygon is: *sum of angles = (n – 2)180°** thus, any pentagon (n = 5) would have angles that add up to 3180 = 540°. any hexagon (n = 6) would have angles that add up to 4180 = 720°. any octagon (n = 8) would have angles that add up to 6180 = 1080°. (see the blog on gre geometric formulas) finally, there is this paradoxical word “regular.” in everyday language, “regular” means “ordinary, unexceptional, commonplace.” in geometry, it connotes the exact opposite! a shape is regular if and only if it is both equilateral and equiangular—that is, if and only if all the sides have the same length and all the angles are equal. the “regular” version of any polygon is the most elite, most symmetrical version possible of that polygon. the “regular triangle” would be what we known as an equilateral triangle. the “regular quadrilateral” is the square. for higher polygons, you are most likely to see the regular version on the gre, because the test (like all mathematicians) loves symmetry. diagonals of a polygon now, we can talk about diagonals. a diagonal* is any line through the interior of a polygon that connects two non-adjacent vertices. what does this mean? well, first of all, starting from any vertex, an adjacent vertex is either vertex connected to the starting vertex by one side of the polygon. consider an irregular quadrilateral: let’s start at vertex a. starting from vertex a, we are connected by sides of this quadrilateral to both b and d; vertices b & d are the ones that are adjacent to vertex a. the only vertex not connected to a by a side of the quadrilateral is c. c is a’s only non-adjacent vertex, and a is c’s. thus, one diagonal goes from a to c. it’s not hard to see that the other goes from b to d. any quadrilateral has just two diagonals. notice that triangles never have diagonals: if we start from any vertex in a triangle, the other two vertices are adjacent. there simply are no non-adjacent vertices in a triangle, so diagonals are not possible. among quadrilaterals, there are special rules for the diagonals of a parallelogram and the categories within parallelograms: the diagonals of a parallelogram bisect each other: that is to say, the intersection point of the two diagonals is the midpoint of each one. a rhombus is a parallelogram with four equal sides. the diagonals of a rhombus bisect each other and are perpendicular. a rectangle is a parallelogram with four 90° angles. the rectangle of a rhombus bisect each other and have equal length. this is related to an old trick among carpenters. when a carpenter cuts two pairs of equal lengths to make the sides of a door or window frame, he knows he has a parallelogram because of the equal lengths, but how does he know whether he has a rectangle? without precise equipment, it’s very hard to measure the difference between, say, an 89° or 90° angle. well, all the carpenter has to do is measure the two diagonals: if these two easy-to-measure lengths are equal, then it’s guaranteed that he has four right angles! a square is a parallelogram, a rectangle, and a rhombus. it is a regular quadrilateral with four equal sides and four 90° angles. the diagonals of a square bisect each other, have equal length, and are perpendicular. diagonals of a regular pentagon a pentagon is any five-sided polygon, and the sum of its angles is 540°, as we saw above. the only pentagon you are likely to meet on the gre is the most symmetrical, the regular pentagon. since the angles are equal, we can divide the sum of the angles by five. 540°/5 = 108° that’s the angle of each of the five angles in the pentagon. here’s a regular pentagon with the five diagonals drawn. how many diagonals does a pentagon have? any pentagon has exactly five diagonals. these diagonals trace out the shape of a classic five-pointed star, such as that those on the flag of the united states of america. the lengths and divisions of this star are intimately related to that magical and mystical number, the golden ratio; sacred geometry is purported to give insight into the meaning of life, but you don’t need to know any of that for the gre! how would we find any angles in this shape? well, we know each big angle of the pentagon is 108°. look, for example, triangle abc. this triangle is an isosceles triangle, because ab = bc, and we know that angle abc = 108°. the other two angles must be equal: call them x. 108° + x + x = 180* 2x = 180° – 108° = 72° x = 36° this means that angle bac = angle bca = 36°, and so do many other symmetrically related angles around the shape. we could subtract (angle bac) from (angle bae) to get (angle cae) angle cae = (angle bae) – (angle bac) = 108° – 36° = 72° from that, we could find many other angles inside the shape. we could use analogous means to find angles involving the diagonals of any higher polygon. diagonals of a regular hexagon a hexagon is any six-sided polygon, and the sum of its angles is 720°, as we saw above. in a regular hexagon, each angle = 720°/6 = 120° how many diagonals does a hexagon have? starting from one vertex, two other vertices are adjacent, so 3 vertices are non-adjacent, making possible three diagonals from one vertex. from a, we can draw diagonals to c, d, and e. from each vertex, there are three diagonals. since there are six vertices, you might think there would be a total of 36 = 18 diagonals, but that counting method double-counts everything. you see, the diagonal from a to c would get counted once as a diagonal from a and again as a diagonal from c to a. thus, the number of diagonals in a hexagon is 18/2 = 9. these can be groups into two kinds. the six shorter diagonals together make a six-sided star, the magen david. the three longer diagonals form just three symmetrically criss-crossing segments, what is called in mathematics a “degenerate six-pointed star.” these two diagrams show the nine diagonals of a regular hexagon. of course, the six-pointed star simply consists of two overlapping equilateral triangles, pointing in opposite directions. (that geometric fact led to extensive mystical speculation in about the star of david in the kabbalah, but again, you don’t need to understand any mysticism for the gre!) diagonals of a regular heptagon a heptagon* is any seven-sided polygon (n = 7). sometimes it is called a “septagon,” but “heptagon” is the preferred mathematical name. the sum of its angles would be (n – 2)180° = 5180° = 900° this means that each of the seven angles in a regular heptagon would have a measure of each angle = 900°/7 = 128.5714286…° the angle measures are not integers! this is why the gre is exceptionally unlikely to ask you anything about the regular heptagon, and it’s also why you probably never talked much about regular heptagons in high school geometry. it’s why most people aren’t even clear on the proper name for this beast! their non-integer angle measure makes them the first “black sheep” in the regular polygon family! i won’t say anything else about them, because they almost never make an appearance on the gre, but i will show you the two possible seven pointed stars from their diagonals: these stars are hauntingly beautiful, because of their idiosyncratic symmetry. diagonals of a regular octagon an octagon is any eight-sided polygon, and the sum of its angles is 1080°, as we saw above. in a regular octagon, each angle = 1080°/8 = 135° that angle is the supplement of a 45° angle. the regular octagon is the typical stop sign shape in many parts of the world. how many diagonals does an octagon have? how many vertices does an octagon have? starting from one vertex, two other vertices are adjacent, so five vertices are non-adjacent, making possible five diagonals from one vertex. from a, b & h are the symmetrical vertices, so we can draw diagonals to c, d, e, f, and g. similar logic as with the hexagon: five at each vertex, eight vertices, but that counts each diagonal twice, so the total number is 58/2 = 20. ac and ag are what we might call “3 vertex diagonals”: there are eight of these, which form a star. af and ad, each parallel to two sides, are what we might call “4 vertex diagonals”: eight of these form another star. finally, ae is like a diameter of the whole octagon, cutting across its center: there are four such lines, and they form a degenerate eight-pointed star. just as the six-pointed star consisted of two overlapping equilateral triangles, the first eight-pointed star on the left consists of two separate overlapping squares: square aceg and square bdfg. (the names of these squares are reminiscent of the lines at w. 4th street!) between these three stars (counting the degenerate thing on the right as a “star”) we have all 20 of the regular octagon’s diagonals. summary onward and upward! the methods discussed in this blog can be extended to apply to a nonagon (n = 9), a decagon (n = 10), or any higher polygon. armed with this information, you should be able to answer anything the gre asks you about a diagonal of a polygon! if you had any “aha” moments while reading this blog, you might want to give the practice problems on the top another peek before reading the explanations below. practice problem explanations 1) let’s take a look at the pentagon with its five diagonals. an example of the angle between two diagonals at a vertex would be angle ebd, where diagonals bd and be meet at vertex b. we will follow the logic outlined above. triangle bcd is isosceles with bc = cd, and angle bcd = 108°. the other two angles are equal: call them each x. 108° + x + x = 180 2x = 180° – 108° = 72° x = 36° so, angle cbd = 36°. well, triangle abe is in every way equal to triangle bcd, so angle abe must also equal 36°. thus, we can subtract from the big angle at vertex b. (angle ebd) = (angle abc) – (angle cbd) – (angle abe) (angle ebd) = 108° – 36° – 36° = 36° answer = (b) 2) if we start at one vertex of the 20-sided polygon, then there’s an adjacent vertex on each side. not counting these three vertices, there would be 17 non-adjacent vertices, so 17 possible diagonals could be drawn from any vertex. twenty vertices, 17 diagonals from each vertex, but this method double-counts the diagonals, as pointed out above. # of diagonals = (1720)/2 = 1710 = 170 answer = (c) editor’s note: this post was originally published in january 2014 but has been updated for freshness, accuracy, and comprehensiveness. lastly, if you are looking for more gre practice material, here is a full-length, free practice test from magoosh that includes a detailed score report with a topic-by-topic breakdown of your performance. you can choose to do just the quant section or a full length exam. happy studying! the post diagonals of a polygon in gre geometry appeared first on magoosh blog — gre® test. ]]> challenging gre qc problems on number properties mon, 11 may 2015 16:00:01 +0000 get your gre math practice in with these nine gre quantitative comparison questions, ranging in difficulty from moderate to quite hard (arguably as hard as, or perhaps harder than, anything you would see on the gre quant section). solutions will come at the end of the article. remember: don’t touch a calculator! 1) 2) 3) 4) k is the set of perfect squares between 1 and 10000. 5) when 900 is divided by positive integer d, the remainder is 1. if n > 5000, then when n is divided by 23, the remainder is r. 6) f is the prime factorization of 17! 7) 8) n is a large integer, not divisible by 7. when n is divided by 35, the quotient is q₁ and the remainder is r₁. when n is divided by 7, the quotient is q₂ and the remainder is r₂, and q₂ > 5q₁. 9) the number n is any non-negative integer, and integer properties and fractions on the gre qc it’s important to remember that the gre is not particularly interested in your ability to perform long detailed calculations. in fact, if you find yourself getting sucked into a long detailed calculation anywhere on the gre quant section, you are probably doing something the hard way. the gre doesn’t care about long calculations, but it does care about your ability to use logic and number sense to simplify calculations. it is interested in the test taker’s ability to re-frame the question in a way that leads to significant simplification. nowhere is this truer than it is on the quantitative comparisons. your job on the gre qcs is to compare, not to calculate. as a general rule, a gre qc question is designed for a quite straightforward comparison or simplification. it may appear to demand a long tedious calculation, but don’t be fooled: that is not the real issue. the real issue is finding the way to look at the problem, the way to interpret it or reframe it, so that the pertinent comparison becomes much easier to see. with fractions, think about simplifying comparisons: for example, bigger numerator makes the fraction bigger, but bigger denominator makes the fraction smaller. with integer properties, there are a number of important distinctions that could come into play. you should know what is what isn’t a prime number, as well as everything it means to be a prime number. you should understand factors, multiples, prime factorizations, and remainders. for difficult problems, as some of these are, it’s very important to read the solutions carefully. notice how the problem is framed in the solution, and notice what important observations were used to simplify it. practice problem solutions 1) this one isn’t too hard, but it can be simplified. notice that 6 is a multiple of 3 and 14 is a multiple of 7. it’s much easier to find common denominators if one denominator is a multiple of the other, so group the fractions together accordingly. answer = (c) 2) for this one, do not find common denominators. it would be a nightmare to try to perform this exact calculation. instead, consider this perspective. look at the two fractions in quantity a. first, look at 10/23. we know 10/20 = 1/2, so 10/23 is less than 1/2. we also know that 19/38 = 1/2, so 19/37 is more than 1/2. in quantity a, we have (something less than 1/2) minus (something greater than 1/2) that has to be negative! quantity a is negative. meanwhile, we are adding two positive things in quantity b, so that must be positive. any positive is greater than any negative. answer = (b) 3) these numbers are probably a little bit bigger than what you would be likely to see on the gre, but notice that even for this, we don’t need a calculator! notice that both denominators are multiples of 25. 900 = 9100 = 9(425) = 3625 500 = 5100 = 5(425) = 2025 and, of course, 525 is just one more factor of 25, so 525 = 2125 we can express both denominators in that factored form, and multiply both quantities by 25. remember that cross-multiplying is a perfectly legitimate move in gre qcs, as long as all the concerned numbers are positive. answer = (b)* 4) think about the numbers in k. the first 10 perfect squares go from 1 to 100. we know that 20 squared is 400 and 30 squared is 900, so the second set of ten perfect squares goes from 100 to 400, and the third set, from 400 to 900. the spread is getting bigger. we know that 90 squared is 8100 and 100 squared is 10,000, so the last set of ten is spread out between 8100 and 10000, a spread of almost 2000. in other words, this set is clumped together at the low end and very spread out at the upper end. the outliers are toward the high end. the median is not affected by outliers, but the mean is pulled in the direction of outliers. therefore, the mean must be higher than the median. answer = (a) that’s all you need to do to answer the question. if you are interested in additional information: the median is the average of those two numbers median = 2550.5 i calculated the mean: mean = 3383.5 there’s no particularly easy way to get that number without a calculator or computer. 5) let’s think about integer d first. d must be a factor of 900 – 1 = 899. what are the factors of that number. well, we can think about this using the advanced numerical factoring, discussed in this gmat blog. thus, 899 is the product of two prime numbers and, as such, has four factors: {1, 29, 31, 899}. well, if d = 1, then dividing any positive integer by it would never yield a remainder, so d can’t be 1. d could be either 29 or 31 or 899. now, think about r. when we divide by 23, the remainder must be less than 23. well, any number less than 23 will be less than any of the possibilities for d, so d is always bigger. answer = (a) 6) consider the prime factorization of 17!—it must have many factors of 2 and 3. it has a factor of 2 for every even number from 1 to 17, plus a second factor of 2 for every number divisible by 4, plus another for every number divisible by 8, etc. by contrast, there are only three factors in 17! that contain factors of 5: {5, 10, 15}. each one contains exactly one factor of 5, so there are just three factors of 5 in 17! quantity a = 3 the factors less than 5 have many more factors that 5 does. the next prime number up is 7, which has two factors, from its multiples {7, 14}. the next three prime factors are {11, 13, 17}, and each one of those is big enough no other multiples would be part of 17! thus, these three prime number, {11, 13, 17}, each have an exponent of one in f. quantity b = 3 answer = (c) 7) to get started, think about smaller numbers. how many multiples of 5 between 45 and 200? well, 45 = 59 and 200 = 540. the number of multiples of 5 between these two is equivalent to the number of integers between 9 and 40. if we subtract, 40 – 9 = 31, that excludes 9 but includes 40. if we truly want the number of multiples between the two end points, i have to subtract one. there are 30 multiples of 5 between 45 and 200. notice that, to find the number of multiples of 5, we simply divided the top and bottom number by 5 and subtracted then, then subtracted 1. follow this pattern. we will add 1 to both quantities, because it’s the same in both and it’s in the way. in each quantity, we will factor out the lower power. now, we can compare these part-wise. first of all, of course 7 to the power of 10 must be larger than 3 to the power of 10, and subtracting 1 from both will not make a difference, so the expression in the parentheses in quantity b is larger than the corresponding expression in quantity a. then, look at the factors in front: so, every piece of quantity b is bigger than the corresponding piece of quantity a. answer = (b) 8) this is a tricky one about remainders. let’s think about this. suppose the first remainder is something less than 7, say 3. then now, if we divide this n by 7, it will go evenly into the first term, and also have a remainder of 3. in this case, the quotient when dividing by 7 would be because that’s exactly how many times 7 would go into that first term above. this doesn’t obey the condition in our problem, because the second quotient is supposed to be more than five times the first quotient. in order to do that, we would need the first remainder to be bigger than 7, say, 23. i’ve written it like that to show what happens when we divide this same number by 7. when we divide by 7, it will go five times the first quotient into that first terms, but then there are three more sevens. thus: in order for the second quotient to be more than five times the first quotient, the first quotient must have a remainder bigger than 7, so that 7 divides into it at least once, thus increasing the second quotient. thus, the first remainder is always bigger than 7, and the second remainder, the remainder when dividing by 7, absolutely has to be less than 7. answer = (a) that was an explanation that involving thinking about the numbers. here, also, is a more algebraic approach. for more detailed information, see this gmat blog on remainders. we will use the “rebuilding the dividend equation” from that blog; this can be a handy equation to know. this is the equation: n = dq + r n is the original number divided, known as the “dividend.” d is the divisor, the number by which we divided. q is the integer quotient, how many times the whole of d goes into n. r is the remainder. here save that equation. now, go to the inequality notice that the left side of this inequality is equal to the left side of the equation above. substitute the right side of the equation into the inequality: answer = (a)* 9) this is a hard problem, probably harder than anything that conceivably could appear on the gre. first of all, you should appreciate the significance of quantity a. if j is a prime number, it would have two factors. if j is the square of a prime number, it would have exactly three factors. a product of two primes has four factors, and products of non-primes have many more factors. we notice that for n = 0, j = 41, a prime number, and for n = 1, j = 43, another prime number. clearly, in those two cases, the number of factors of j is two, less than 4, and quantity b is bigger. the question is: will j always be prime? well, if you rely only on plugging numbers, this problem is particular devious. you see, this particular polynomial expression was discovered by the great mathematician leonard euler in 1772, and for all integer values from n = 0 to n = 39, this gives prime values of j. you could plug in a lot of numbers, and get a prime number every time. but, of course, plug-in alone does not establish a pattern. here’s the big idea that could help you on the gre: it is mathematically impossible to produce any algebraic rule that consistently outputs nothing but prime numbers. the pattern of prime numbers is the hardest pattern in mathematics to map. no single formula will do the job. no single formula will always, always, always give us prime numbers without exception: that is absolutely impossible. this particular formula is famous because it happens to give a lot of prime numbers at the beginning, but some of those values will not be prime numbers. for example, when n = 41, j = 4141 + 41 + 41 = 4341, which has four factors. there is not a consistent pattern. answer = (d) more information for the mathematically curious: there is no algebraic pattern for the prime numbers. the pattern of prime numbers is part and parcel of the hardest unsolved problem in modern mathematics, known as the riemann hypothesis. bernard riemann proposed this hypothesis in 1859, and since then, it has been a tantalizing problem that the finest mathematical minds in the world have been unable to solve. there’s a $1m prize on solving it, and it would have major implication in both theoretical and practical mathematics, possibly including public-key cryptography. all you need to know is that the pattern is prime numbers is not easy!! no simple algebraic formula is going to cut the mustard! the post challenging gre qc problems on number properties appeared first on magoosh blog — gre® test. ]]>

6 month gre study schedule

sun, 18 oct 2020 21:00:32 +0000

the content in this post applies in 2024 to the new, shorter gre!

the magoosh 6 month study plan moves at a relaxed pace that should still be followed diligently, lest procrastination leads to things getting out of hand. it’s designed to expose you to everything that the gre might throw at you on test day as well as equip you with strategies and techniques to help you achieve your dream score.

click here to download a pdf of the study schedule!

time commitment

students should assume each session will take somewhere between 1.5-2.5 hours. earlier in the schedule, most of the work is assigned across 4 days, with days 5-6 used in whatever way will best suit your prep needs. day 7 is a rest day. later in the schedule, day 5 is allocated to verbal practice. why does verbal get its own day? well, based on the number of concepts tested, math topics far outweigh verbal topics. therefore, to ensure that verbal doesn’t slip by the wayside once its lessons are complete, starting in the second half of the schedule, you’ll see dedicated time for verbal practice.

designed to be flexible

now, how you approach the schedule is really up to you. you can opt for a monday-saturday routine. you can combine days if you want more study-free days during the week. you can work two, long weekend days to free up time on the weekdays. feel free to make any changes that best suit the unique demands of your daily life. one of our goals was to make this schedule easy to modify because we know that balancing responsibilities is tough and a study schedule is rarely “one-size fits all.”

another thing to note, this schedule was designed for those who have the time to become gre experts in mind. this means that the lessons cover everything from test-taking fundamentals to the most advanced, rarely tested content.

if you find that you are already comfortable with a lot of the introductory material, you can always make changes to suit your unique prep needs. once you have a sense of your personal strengths and weaknesses, you can swap out lesson material that you’re well-versed in for more advanced concepts, and set up custom practice sessions that focus on your individual priorities.

curious about other study plan options or how to best modify a schedule depending on your situation? check out this post: gre study plans and guides.

feeling like just getting started is a challenge? check out how other students were able to balance the demands of work, life, and gre preparation in our post how to study for the gre.

essential material
optional material
about the material
month 1
month 2
month 3
month 4
month 5
month 6
takeaways

essential material

magoosh gre prep
ets’s official guide to the gre book
ets powerprep free tests 1 & 2: if you would prefer to take the practice test on paper, you can print out a copy of ets’s paper-delivered test. however, taking the paper-based gre is pretty rare these days. if you are like most test takers and will be taking the exam online, we highly recommend that your practice and prep are also online–mimic the medium you’ll be using on exam day as much as possible.
magoosh’s gre vocabulary flashcards app and magoosh’s gre math flashcards app: they’re free and you can use them on the web, iphone/ipad or android!
magoosh’s complete guide to the gre pdf: this comprehensive, web-based guide to the gre gives you the quick but very helpful overview you need to understand this test. you’ll see how the gre is designed and scored, what skills it tests, how to find and use the best gre prep, and how to study for each test section.
reading materials and journal (error log): more detail on each of these below, in the “about the material” section.

optional material

top 5 free gre practice resources: this page includes instructions on where to find good full-length gre practice tests, and how to take practice tests and incorporate them into your studies. this page also has links to magoosh’s free gre diagnostic quizzes.
quizlet.com: gives you online access to flashcards, making it easier, say, to quiz yourself on your mobile device
vocabulary.com: provides conversational example sentences and a flood of example sentences. if you are really trying to achieve a high verbal score, you can also check out lists of crazy hard gre words that other users have compiled.
manhattan’s 5lb book of gre practice problems is not absolutely necessary, but if you are in need of even more practice material, this is a quality source. don’t feel like you need the latest version. the content of the gre has not changed significantly, so an older edition will suit your practice needs just fine. full disclosure: we find the math content to be very test-like. the verbal content, on the other hand, tends to veer a bit harder than what you’d see on the gre, especially for text completions and sentence equivalence.

about the material

about vocabulary

positive and negative square roots on the gre

wed, 07 sep 2016 22:52:50 +0000

some questions to start: can a square root be negative? yes, there is such a thing as a negative square root. how would you deal with negative roots on the gre? this question is actually fairly complicated, because it depends on how the gre words the question. let’s start with some gre math practice problems before i explain.

negative square root practice problems

first of all, consider these two similar, but not identical, quantitative comparison questions. these are a bit easier than you would see on the gre, but they illustrate an important distinction about possible negative square roots.

all i will say right now is: despite apparent similarities, those two questions about whether roots are positive or negative, have two completely different answers. the distinction between them is the subject of this article.

explanations to these practice problems will appear at the end of this blog article. jump ahead by clicking here.

do i include the negative roots?

often, students are confused about this question. for example, in the questions above, we know +4 is a square root of 16, but can’t -4 be one as well? or can it? can a square root be negative? do we include the negative square root as part of column b or not? does it matter how the question is framed? all of these questions about possible negative square roots are resolved by understanding the following two cases.

case i: the symbol appears

in case one, the test-maker, in writing the question, uses this symbol. this symbol appears printed on the page in the question itself.

what is this symbol? well, the most folks call this simply a “square-root” symbol, but the proper name is the “principal square root” symbol. here, “principal” (in the sense of “main” or “most important”) means: you take one and only one root, the most important, or principal, one — the positive root only. that is the deep meaning of this symbol.

thus, in all cases in which this symbol appears as part of the question itself, you never consider the negative square root, and only take the positive square root.

case ii: the symbol doesn’t appear

in this case, that special symbol does not appear as part of the problem. what does appear is, for instance, a variable squared, or some other combination of algebra that leads to a variable squared, and you yourself, in your process of solving the problem, have to take the square root of something in order to solve it. the act of “square rooting” is not initiated by the test maker in the act of writing the question; rather, it is you who initiated the square-rooting.

in this case, 100% of the time, you always have to consider both the positive and negative square roots.

summary: can a square root be negative?

can a square root be negative? well, the answer is: it depends on what was printed in the problem. the principal square root symbol never has a negative output, so if the test maker printed that symbol, it’s restrictions have to be respected: all square roots then are positive. on the other hand, if the problem contains a variable squared, or some other algebra that leads to a variable squared, and you yourself take a square root as part of the act of solving, then you always have to consider all possible solutions, both the positive square roots and the negative square roots.

if you master that distinction, you will always understand when to consider both positive and negative roots vs. when to consider only the positive root. you may want to go back to those two qcs at the beginning and think them through again before reading the solutions below.

additional resources

if you’re having trouble with quantitative comparison questions on the gre, i would recommend taking a look at some of these additional resources. they may help clear up some of the concepts that you’re struggling with, and can provide some extra practice.

quantitative comparison tips

quantitative comparison strategies

quantitative comparison practice

gre math help

gre math formula ebook

practice problem solutions

1) here, the principal square root symbol appears as part of the problem itself. we are in case i. of course, that symbol means: take the positive square root only. so column b can only equal +4. of course, that’s always bigger than 3. answer = (b).

2) here, there’s no square root symbol printed as part of the problem itself. we are in case ii. for any square roots we take as part of our solution, we are liable to account for both the positive and negative roots.

sure enough, the very first thing we encounter in the prompt is a variable squared, and when we solve for x, we have to account for both roots: x = ±4. the variable x could equal well have either one of those values.

now, when we proceed to the qc, we see that the different values of x would give different answers. if x = +4, then column b is greater, but if x = -4, then column a is greater. different values lead to different conclusions, and this situation means we don’t have enough information to establish a definitive relationship. answer = (d)

special note:

to find out where roots sit in the “big picture” of gre quant, and what other quant concepts you should study, check out our post entitled:

what kind of math is on the gre? breakdown of quant concepts by frequency

editor’s note: this post was originally published in september, 2012, and has been updated for freshness, accuracy, and comprehensiveness.

the post positive and negative square roots on the gre appeared first on magoosh blog — gre® test.

]]>

diagonals of a polygon in gre geometry

wed, 07 sep 2016 22:32:36 +0000

students love to skip over the basics, asking questions like: how many vertices does an octagon have? how many diagonals does an octagon have? what’s the difference between a regular octagon and, well, an octagon? and gre geometry does delve into some complex polygon math.

but before we get to that, i will begin with two challenging gre math problems.

gre geometry: polygon problems

1) regular pentagon p has all five diagonals drawn. what is the angle between two of these diagonals where they meet at a vertex of the pentagon?

(a) 12°

(b) 36°

(d) 60°

(e) 72°

2) how many diagonals does a regular 20-sided polygon have?

(a) 60

(b) 120

(d) 240

(e) 400

explanations to these practice problems will appear at the end of this blog article. jump ahead by clicking here.

polygons

first, some basic terminology to begin this discussion. a polygon is any geometric shape all of whose sides are straight line-segments. any triangle is a polygon. any quadrilateral (including trapezoids, parallelograms, rhombuses, rectangles and squares) are polygons. a pentagon is a 5-sided polygon. a hexagon is a 6-sided polygon. an octagon is an eight-sided polygon. a circle or parabola or anything with a curved side is not a polygon.

a point where two of the sides of a polygon meet is called a vertex. the number of vertices a polygon has is always equal to the number of sides it has.

another important polygon fact concerns the sum of angles. you may know that the sum of the three angles in any triangle is 180°. you may even know that the sum of the four angles in any quadrilateral is 360°. this pattern generalizes. the sum of all n angle in any n-sided polygon is:

sum of angles = (n – 2)*180°

thus, any pentagon (n = 5) would have angles that add up to 3*180 = 540°. any hexagon (n = 6) would have angles that add up to 4*180 = 720°. any octagon (n = 8) would have angles that add up to 6*180 = 1080°. (see the blog on gre geometric formulas)

finally, there is this paradoxical word “regular.” in everyday language, “regular” means “ordinary, unexceptional, commonplace.” in geometry, it connotes the exact opposite! a shape is regular if and only if it is both equilateral and equiangular—that is, if and only if all the sides have the same length and all the angles are equal. the “regular” version of any polygon is the most elite, most symmetrical version possible of that polygon. the “regular triangle” would be what we known as an equilateral triangle. the “regular quadrilateral” is the square. for higher polygons, you are most likely to see the regular version on the gre, because the test (like all mathematicians) loves symmetry.

diagonals of a polygon

now, we can talk about diagonals. a diagonal is any line through the interior of a polygon that connects two non-adjacent vertices. what does this mean? well, first of all, starting from any vertex, an adjacent vertex is either vertex connected to the starting vertex by one side of the polygon.

consider an irregular quadrilateral:

let’s start at vertex a. starting from vertex a, we are connected by sides of this quadrilateral to both b and d; vertices b & d are the ones that are adjacent to vertex a. the only vertex not connected to a by a side of the quadrilateral is c. c is a’s only non-adjacent vertex, and a is c’s. thus, one diagonal goes from a to c. it’s not hard to see that the other goes from b to d. any quadrilateral has just two diagonals.

notice that triangles never have diagonals: if we start from any vertex in a triangle, the other two vertices are adjacent. there simply are no non-adjacent vertices in a triangle, so diagonals are not possible. among quadrilaterals, there are special rules for the diagonals of a parallelogram and the categories within parallelograms:

the diagonals of a parallelogram bisect each other: that is to say, the intersection point of the two diagonals is the midpoint of each one.

a rhombus is a parallelogram with four equal sides. the diagonals of a rhombus bisect each other and are perpendicular.

a rectangle is a parallelogram with four 90° angles. the rectangle of a rhombus bisect each other and have equal length. this is related to an old trick among carpenters. when a carpenter cuts two pairs of equal lengths to make the sides of a door or window frame, he knows he has a parallelogram because of the equal lengths, but how does he know whether he has a rectangle? without precise equipment, it’s very hard to measure the difference between, say, an 89° or 90° angle. well, all the carpenter has to do is measure the two diagonals: if these two easy-to-measure lengths are equal, then it’s guaranteed that he has four right angles!

a square is a parallelogram, a rectangle, and a rhombus. it is a regular quadrilateral with four equal sides and four 90° angles. the diagonals of a square bisect each other, have equal length, and are perpendicular.

diagonals of a regular pentagon

a pentagon is any five-sided polygon, and the sum of its angles is 540°, as we saw above. the only pentagon you are likely to meet on the gre is the most symmetrical, the regular pentagon. since the angles are equal, we can divide the sum of the angles by five.

540°/5 = 108°

that’s the angle of each of the five angles in the pentagon.

here’s a regular pentagon with the five diagonals drawn.

how many diagonals does a pentagon have? any pentagon has exactly five diagonals. these diagonals trace out the shape of a classic five-pointed star, such as that those on the flag of the united states of america. the lengths and divisions of this star are intimately related to that magical and mystical number, the golden ratio; sacred geometry is purported to give insight into the meaning of life, but you don’t need to know any of that for the gre!

how would we find any angles in this shape? well, we know each big angle of the pentagon is 108°. look, for example, triangle abc. this triangle is an isosceles triangle, because ab = bc, and we know that angle abc = 108°. the other two angles must be equal: call them x.

108° + x + x = 180*

2x = 180° – 108° = 72°

x = 36°

this means that angle bac = angle bca = 36°, and so do many other symmetrically related angles around the shape. we could subtract (angle bac) from (angle bae) to get (angle cae)

angle cae = (angle bae) – (angle bac) = 108° – 36° = 72°

from that, we could find many other angles inside the shape. we could use analogous means to find angles involving the diagonals of any higher polygon.

diagonals of a regular hexagon

a hexagon is any six-sided polygon, and the sum of its angles is 720°, as we saw above. in a regular hexagon,

each angle = 720°/6 = 120°

how many diagonals does a hexagon have? starting from one vertex, two other vertices are adjacent, so 3 vertices are non-adjacent, making possible three diagonals from one vertex. from a, we can draw diagonals to c, d, and e.

from each vertex, there are three diagonals. since there are six vertices, you might think there would be a total of 3*6 = 18 diagonals, but that counting method double-counts everything. you see, the diagonal from a to c would get counted once as a diagonal from a and again as a diagonal from c to a. thus, the number of diagonals in a hexagon is 18/2 = 9. these can be groups into two kinds. the six shorter diagonals together make a six-sided star, the magen david. the three longer diagonals form just three symmetrically criss-crossing segments, what is called in mathematics a “degenerate six-pointed star.”

these two diagrams show the nine diagonals of a regular hexagon. of course, the six-pointed star simply consists of two overlapping equilateral triangles, pointing in opposite directions. (that geometric fact led to extensive mystical speculation in about the star of david in the kabbalah, but again, you don’t need to understand any mysticism for the gre!)

diagonals of a regular heptagon

a heptagon is any seven-sided polygon (n = 7). sometimes it is called a “septagon,” but “heptagon” is the preferred mathematical name. the sum of its angles would be

(n – 2)*180° = 5*180° = 900°

this means that each of the seven angles in a regular heptagon would have a measure of

each angle = 900°/7 = 128.5714286…°

the angle measures are not integers! this is why the gre is exceptionally unlikely to ask you anything about the regular heptagon, and it’s also why you probably never talked much about regular heptagons in high school geometry. it’s why most people aren’t even clear on the proper name for this beast! their non-integer angle measure makes them the first “black sheep” in the regular polygon family! i won’t say anything else about them, because they almost never make an appearance on the gre, but i will show you the two possible seven pointed stars from their diagonals: these stars are hauntingly beautiful, because of their idiosyncratic symmetry.

diagonals of a regular octagon

an octagon is any eight-sided polygon, and the sum of its angles is 1080°, as we saw above. in a regular octagon,

each angle = 1080°/8 = 135°

that angle is the supplement of a 45° angle. the regular octagon is the typical stop sign shape in many parts of the world.

how many diagonals does an octagon have? how many vertices does an octagon have?

starting from one vertex, two other vertices are adjacent, so five vertices are non-adjacent, making possible five diagonals from one vertex. from a, b & h are the symmetrical vertices, so we can draw diagonals to c, d, e, f, and g.

similar logic as with the hexagon: five at each vertex, eight vertices, but that counts each diagonal twice, so the total number is 5*8/2 = 20. ac and ag are what we might call “3 vertex diagonals”: there are eight of these, which form a star. af and ad, each parallel to two sides, are what we might call “4 vertex diagonals”: eight of these form another star. finally, ae is like a diameter of the whole octagon, cutting across its center: there are four such lines, and they form a degenerate eight-pointed star.

just as the six-pointed star consisted of two overlapping equilateral triangles, the first eight-pointed star on the left consists of two separate overlapping squares: square aceg and square bdfg. (the names of these squares are reminiscent of the lines at w. 4th street!) between these three stars (counting the degenerate thing on the right as a “star”) we have all 20 of the regular octagon’s diagonals.

summary

onward and upward! the methods discussed in this blog can be extended to apply to a nonagon (n = 9), a decagon (n = 10), or any higher polygon. armed with this information, you should be able to answer anything the gre asks you about a diagonal of a polygon! if you had any “aha” moments while reading this blog, you might want to give the practice problems on the top another peek before reading the explanations below.

practice problem explanations

1) let’s take a look at the pentagon with its five diagonals.

an example of the angle between two diagonals at a vertex would be angle ebd, where diagonals bd and be meet at vertex b.

we will follow the logic outlined above.

triangle bcd is isosceles with bc = cd, and angle bcd = 108°. the other two angles are equal: call them each x.

108° + x + x = 180*

2x = 180° – 108° = 72°

x = 36°

so, angle cbd = 36°. well, triangle abe is in every way equal to triangle bcd, so angle abe must also equal 36°. thus, we can subtract from the big angle at vertex b.

(angle ebd) = (angle abc) – (angle cbd) – (angle abe)

(angle ebd) = 108° – 36° – 36° = 36°

answer = (b)

2) if we start at one vertex of the 20-sided polygon, then there’s an adjacent vertex on each side. not counting these three vertices, there would be 17 non-adjacent vertices, so 17 possible diagonals could be drawn from any vertex. twenty vertices, 17 diagonals from each vertex, but this method double-counts the diagonals, as pointed out above.

# of diagonals = (17*20)/2 = 17*10 = 170

answer = (c)

editor’s note: this post was originally published in january 2014 but has been updated for freshness, accuracy, and comprehensiveness.

lastly, if you are looking for more gre practice material, here is a full-length, free practice test from magoosh that includes a detailed score report with a topic-by-topic breakdown of your performance. you can choose to do just the quant section or a full length exam. happy studying!

the post diagonals of a polygon in gre geometry appeared first on magoosh blog — gre® test.

]]>

challenging gre qc problems on number properties

mon, 11 may 2015 16:00:01 +0000

get your gre math practice in with these nine gre quantitative comparison questions, ranging in difficulty from moderate to quite hard (arguably as hard as, or perhaps harder than, anything you would see on the gre quant section). solutions will come at the end of the article. remember: don’t touch a calculator!

4) k is the set of perfect squares between 1 and 10000.

5) when 900 is divided by positive integer d, the remainder is 1. if n > 5000, then when n is divided by 23, the remainder is r.

6) f is the prime factorization of 17!

8) n is a large integer, not divisible by 7. when n is divided by 35, the quotient is q₁ and the remainder is r₁. when n is divided by 7, the quotient is q₂ and the remainder is r₂, and q₂ > 5q₁.

9) the number n is any non-negative integer, and

integer properties and fractions on the gre qc

it’s important to remember that the gre is not particularly interested in your ability to perform long detailed calculations. in fact, if you find yourself getting sucked into a long detailed calculation anywhere on the gre quant section, you are probably doing something the hard way. the gre doesn’t care about long calculations, but it does care about your ability to use logic and number sense to simplify calculations. it is interested in the test taker’s ability to re-frame the question in a way that leads to significant simplification.

nowhere is this truer than it is on the quantitative comparisons. your job on the gre qcs is to compare, not to calculate. as a general rule, a gre qc question is designed for a quite straightforward comparison or simplification. it may appear to demand a long tedious calculation, but don’t be fooled: that is not the real issue. the real issue is finding the way to look at the problem, the way to interpret it or reframe it, so that the pertinent comparison becomes much easier to see.

with fractions, think about simplifying comparisons: for example, bigger numerator makes the fraction bigger, but bigger denominator makes the fraction smaller.

with integer properties, there are a number of important distinctions that could come into play. you should know what is what isn’t a prime number, as well as everything it means to be a prime number. you should understand factors, multiples, prime factorizations, and remainders.

for difficult problems, as some of these are, it’s very important to read the solutions carefully. notice how the problem is framed in the solution, and notice what important observations were used to simplify it.

practice problem solutions

1) this one isn’t too hard, but it can be simplified. notice that 6 is a multiple of 3 and 14 is a multiple of 7. it’s much easier to find common denominators if one denominator is a multiple of the other, so group the fractions together accordingly.

answer = (c)

2) for this one, do not find common denominators. it would be a nightmare to try to perform this exact calculation. instead, consider this perspective. look at the two fractions in quantity a. first, look at 10/23. we know 10/20 = 1/2, so 10/23 is less than 1/2. we also know that 19/38 = 1/2, so 19/37 is more than 1/2. in quantity a, we have
(something less than 1/2) minus (something greater than 1/2)

that has to be negative! quantity a is negative. meanwhile, we are adding two positive things in quantity b, so that must be positive. any positive is greater than any negative.

answer = (b)

3) these numbers are probably a little bit bigger than what you would be likely to see on the gre, but notice that even for this, we don’t need a calculator! notice that both denominators are multiples of 25.

900 = 9*100 = 9*(4*25) = 36*25

500 = 5*100 = 5*(4*25) = 20*25

and, of course, 525 is just one more factor of 25, so

525 = 21*25

we can express both denominators in that factored form, and multiply both quantities by 25.

remember that cross-multiplying is a perfectly legitimate move in gre qcs, as long as all the concerned numbers are positive.

answer = (b)

4) think about the numbers in k. the first 10 perfect squares go from 1 to 100. we know that 20 squared is 400 and 30 squared is 900, so the second set of ten perfect squares goes from 100 to 400, and the third set, from 400 to 900. the spread is getting bigger. we know that 90 squared is 8100 and 100 squared is 10,000, so the last set of ten is spread out between 8100 and 10000, a spread of almost 2000.

in other words, this set is clumped together at the low end and very spread out at the upper end. the outliers are toward the high end. the median is not affected by outliers, but the mean is pulled in the direction of outliers. therefore, the mean must be higher than the median.

answer = (a)

that’s all you need to do to answer the question. if you are interested in additional information:

the median is the average of those two numbers

median = 2550.5

i calculated the mean:

mean = 3383.5

there’s no particularly easy way to get that number without a calculator or computer.

5) let’s think about integer d first. d must be a factor of 900 – 1 = 899. what are the factors of that number. well, we can think about this using the advanced numerical factoring, discussed in this gmat blog.

thus, 899 is the product of two prime numbers and, as such, has four factors: {1, 29, 31, 899}. well, if d = 1, then dividing any positive integer by it would never yield a remainder, so d can’t be 1. d could be either 29 or 31 or 899.

now, think about r. when we divide by 23, the remainder must be less than 23. well, any number less than 23 will be less than any of the possibilities for d, so d is always bigger.

answer = (a)

6) consider the prime factorization of 17!—it must have many factors of 2 and 3. it has a factor of 2 for every even number from 1 to 17, plus a second factor of 2 for every number divisible by 4, plus another for every number divisible by 8, etc.

by contrast, there are only three factors in 17! that contain factors of 5: {5, 10, 15}. each one contains exactly one factor of 5, so there are just three factors of 5 in 17! quantity a = 3

the factors less than 5 have many more factors that 5 does. the next prime number up is 7, which has two factors, from its multiples {7, 14}. the next three prime factors are {11, 13, 17}, and each one of those is big enough no other multiples would be part of 17! thus, these three prime number, {11, 13, 17}, each have an exponent of one in f. quantity b = 3

answer = (c)

7) to get started, think about smaller numbers. how many multiples of 5 between 45 and 200? well, 45 = 5*9 and 200 = 5*40. the number of multiples of 5 between these two is equivalent to the number of integers between 9 and 40. if we subtract, 40 – 9 = 31, that excludes 9 but includes 40. if we truly want the number of multiples between the two end points, i have to subtract one. there are 30 multiples of 5 between 45 and 200. notice that, to find the number of multiples of 5, we simply divided the top and bottom number by 5 and subtracted then, then subtracted 1. follow this pattern.

we will add 1 to both quantities, because it’s the same in both and it’s in the way. in each quantity, we will factor out the lower power.

now, we can compare these part-wise. first of all, of course 7 to the power of 10 must be larger than 3 to the power of 10, and subtracting 1 from both will not make a difference, so the expression in the parentheses in quantity b is larger than the corresponding expression in quantity a.

then, look at the factors in front:

so, every piece of quantity b is bigger than the corresponding piece of quantity a.

answer = (b)

8) this is a tricky one about remainders. let’s think about this. suppose the first remainder is something less than 7, say 3. then

now, if we divide this n by 7, it will go evenly into the first term, and also have a remainder of 3. in this case, the quotient when dividing by 7 would be

because that’s exactly how many times 7 would go into that first term above. this doesn’t obey the condition in our problem, because the second quotient is supposed to be more than five times the first quotient. in order to do that, we would need the first remainder to be bigger than 7, say, 23.

i’ve written it like that to show what happens when we divide this same number by 7. when we divide by 7, it will go five times the first quotient into that first terms, but then there are three more sevens. thus:

in order for the second quotient to be more than five times the first quotient, the first quotient must have a remainder bigger than 7, so that 7 divides into it at least once, thus increasing the second quotient. thus, the first remainder is always bigger than 7, and the second remainder, the remainder when dividing by 7, absolutely has to be less than 7.

answer = (a)

that was an explanation that involving thinking about the numbers. here, also, is a more algebraic approach. for more detailed information, see this gmat blog on remainders. we will use the “rebuilding the dividend equation” from that blog; this can be a handy equation to know. this is the equation:

n = d*q + r

n is the original number divided, known as the “dividend.” d is the divisor, the number by which we divided. q is the integer quotient, how many times the whole of d goes into n. r is the remainder. here

save that equation. now, go to the inequality

notice that the left side of this inequality is equal to the left side of the equation above. substitute the right side of the equation into the inequality:

answer = (a)

9) this is a hard problem, probably harder than anything that conceivably could appear on the gre. first of all, you should appreciate the significance of quantity a. if j is a prime number, it would have two factors. if j is the square of a prime number, it would have exactly three factors. a product of two primes has four factors, and products of non-primes have many more factors.

we notice that for n = 0, j = 41, a prime number, and for n = 1, j = 43, another prime number. clearly, in those two cases, the number of factors of j is two, less than 4, and quantity b is bigger. the question is: will j always be prime?

well, if you rely only on plugging numbers, this problem is particular devious. you see, this particular polynomial expression was discovered by the great mathematician leonard euler in 1772, and for all integer values from n = 0 to n = 39, this gives prime values of j. you could plug in a lot of numbers, and get a prime number every time. but, of course, plug-in alone does not establish a pattern.

here’s the big idea that could help you on the gre: it is mathematically impossible to produce any algebraic rule that consistently outputs nothing but prime numbers. the pattern of prime numbers is the hardest pattern in mathematics to map. no single formula will do the job. no single formula will always, always, always give us prime numbers without exception: that is absolutely impossible. this particular formula is famous because it happens to give a lot of prime numbers at the beginning, but some of those values will not be prime numbers. for example, when n = 41, j = 41*41 + 41 + 41 = 43*41, which has four factors. there is not a consistent pattern.

answer = (d)

more information for the mathematically curious: there is no algebraic pattern for the prime numbers. the pattern of prime numbers is part and parcel of the hardest unsolved problem in modern mathematics, known as the riemann hypothesis. bernard riemann proposed this hypothesis in 1859, and since then, it has been a tantalizing problem that the finest mathematical minds in the world have been unable to solve. there’s a $1m prize on solving it, and it would have major implication in both theoretical and practical mathematics, possibly including public-key cryptography. all you need to know is that the pattern is prime numbers is not easy!! no simple algebraic formula is going to cut the mustard!

the post challenging gre qc problems on number properties appeared first on magoosh blog — gre® test.

]]>

mike mcgarry - 加拿大vs摩洛哥欧赔

one month gre study schedule for advanced students

for advanced gre students: an emphasis on reviewing & drilling

table of contents

essential material

optional material

about the material

coming back to math after years away

math anxiety

practice, practice, practice

math and your brain

summary

how to round to the nearest integer | gre math

rounding to the nearest integer

rounding to any other decimal place

what do you have to know about how to round to the nearest integer for the gre?

other cases of rounding

need more help?

90 day gre study plan for advanced students

for advanced gre students: an emphasis on reviewing & drilling

table of contents

essential material

optional material

about the material

90 day gre study plan for beginners

for gre beginners: starting with the basics

table of contents

essential material

optional material

about the material

gre geometry formulas

gre geometry formulas

gre geometry formulas at a glance

6 month gre study schedule

time commitment

designed to be flexible

table of contents

essential material

optional material

about the material

positive and negative square roots on the gre

negative square root practice problems

do i include the negative roots?

case i: the symbol appears

case ii: the symbol doesn’t appear

summary: can a square root be negative?

additional resources

quantitative comparison tips

quantitative comparison strategies

quantitative comparison practice

gre math help

practice problem solutions

special note:

diagonals of a polygon in gre geometry

gre geometry: polygon problems

polygons

diagonals of a polygon

diagonals of a regular pentagon

diagonals of a regular hexagon

diagonals of a regular heptagon

diagonals of a regular octagon

how many diagonals does an octagon have? how many vertices does an octagon have?

summary

practice problem explanations

challenging gre qc problems on number properties

integer properties and fractions on the gre qc

practice problem solutions