Quarter Wit, Quarter Wisdom: Evaluating Nasty GMAT Answer Choices

Quarter Wit, Quarter WisdomIn some Quant questions, we are given big nasty numbers in the answer choices and little else in the question stem. Often in such cases, the starting point is difficult for the test-taker to find, so today, we will discuss how to handle such questions.

The first and only rule with these types of problems is that familiarity helps. Evaluate the answer choices that make sense to you first.

Let’s look at a few questions to understand how to do that:

Which of the following is NOT prime?

(A) 1,556,551
(B) 2,442,113
(C) 3,893,257
(D) 3,999,991
(E) 9,999,991

The first thing that comes to mind when we consider how to find prime numbers should be to “check the number N for divisibility by all prime factors until we get to the √N.” But note that here, we have four numbers that are prime and one number that is not. Also, the numbers are absolutely enormous and, hence, will be very difficult to work with. So, let’s slide down to a number that seems a bit more sane: 3,999,991 (it is very close to a number with lots of 0’s).

3,999,991 = 4,000,000 – 9
= (2000)^2 – 3^2

This is something we recognise! It’s a difference of squares, which can be written as:

= (2000 + 3) * (2000 – 3)
= 2003 * 1997

Hence, we see that 3,999,991 is a product of two factors other than 1 and itself, so it is not a prime number. We have our answer! The answer is D.

Let’s try another problem:

Which of the following is a perfect square?

 (A) 649
 (B) 961
 (C) 1,664
 (D) 2,509
 (E) 100,000

Here, start by looking at the answer choices. The first one that should stand out is option E, 100,000, since multiples of 10 are always easy to handle. However, we have an odd number of zeroes here, so we know this cannot be a perfect square.

Next, we look at the answer choices that are close to the perfect squares that we intuitively know, such as 30^2 = 900, 40^2 = 1600, 50^2 = 2500. The only possible number whose perfect square could be 961 is 31 – 31^2 will end with a 1 and will be a bit greater than 900 (32^2 will end with a 4, so that cannot be the square root of 961, and the perfect squares of other greater numbers will be much greater than 900).

31^2 = (30 + 1)^2 = 900 + 1 + 2*30*1 = 961

So, we found that 961 is a perfect square and is, hence, the answer!

In case 961 were not a perfect square, we would have tried 1,664 since it is just 64 greater than 1,600. It could be the perfect square of 42, as the perfect square of 42 will end in a 4.

If 1,664 were also not a perfect square (it is not), we would have looked at 2,509. We would have known immediately that 2,509 cannot be a perfect square because it is too close to 2,500. 2,509 ends in a 9, so we may have considered 53 to be its square root, but the difference between consecutive perfect squares increases as we get to greater numbers.

(4^2 is 16 while 5^2 is 25 – the difference between them is 9. The difference between 5^2 and 6^2 will be greater than 9, and so will the difference between the perfect squares of any pair of consecutive integers greater than 6. Hence, the difference between the squares of 50 and 53 certainly cannot be 9.)

Therefore, our answer is B. Let’s try one more question:

When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?

(A) 1,296
(B) 1,369
(C) 1,681
(D) 1,764
(E) 2,500

This question is, again, on perfect squares. We can use the same concepts here, too.

30^2 = 900
31^2 = 961 (=(30+1)^2 = 900 + 1 + 2*30)

40^2= 1,600
41^2 = 1,681 (=(40+1)^2 = 1,600 + 1 + 2*40)

50^2 = 2,500
51^2 = 2,601 (=(50+1)^2 = 2,500 + 1 + 2*50)

We know that the difference between consecutive squares increases as we go to greater numbers: going from 30^2 to 31^2 is a difference of 61, while jumping from 40^2 to 41^2 is a difference of 81.

All the answer choices lie in the range from 900 to 2500. In this range, the difference between consecutive squares is between 60 and 100. So, when you add 148 to a perfect square to get another perfect square in this range, we can say that the numbers must be 2 apart, such as 33 and 35 or 42 and 44, etc. Also, the numbers must lie between 30 and 40 because twice 61 is 122 and twice 81 is 162 – 148 lies somewhere in between 122 and 162.

A and B are the only two possible options.

Consider option A – it ends in a 6, so the square root must end in a 6, too. If you add 148, then it will end with a 4 (the perfect square of a number ending in 8 will end in 4). So this answer choice works.

Consider option B – it ends in a 9, so the square root must end in a 3 or a 7. When you add 148, it ends in 7. No perfect square ends in 7, so this option is out. Our answer is, therefore, A.

We hope you see how a close evaluation of the answer choices can help you solve questions of this type. Go get ’em!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit Quarter Wisdom: What is Your Favorite Number?

Quarter Wit, Quarter WisdomFans of The Big Bang Theory will remember Sheldon Cooper’s quote from an old episode on his favorite number:

The best number is 73. Why? 73 is the 21st prime number. Its mirror, 37, is the 12th and its mirror, 21, is the product of multiplying 7 and 3… and in binary 73 is a palindrome, 1001001, which backwards is 1001001.”

Though Sheldon’s logic is infallible, my favorite number is 1001 because it has a special role in standardized tests.

1001 is 1 more than 1000 and hence, is sometimes split as (1000 + 1). It sometimes appears in the a^2 – b^2 format such as 1001^2 – 1, and its factors are 7, 11 and 13 (not the factors we usually work with).

Due to its unusual factors and its convenient location (right next to 1000), it could be a part of some tough-looking GMAT questions and should be remembered as a “special” number. Let’s look at a question to understand how to work with this  number.

Which of the following is a factor of 1001^(32) – 1 ?

(A) 768
(B) 819
(C) 826
(D) 858
(E) 924

Note that 1001 is raised to the power 32. This is not an exponent we can easily handle. If  we try to use a binomial here and split 1001 into (1000 + 1), all we will achieve is that upon expanding the given expression, 1 will be cancelled out by -1 and all other terms will have 1000 in common. None of the answer choices are factors of 1000, however, so we must look for some other factor of 1001^(32) – 1.

Without a calculator, it is not possible for us to find the factors of 1001^(32) – 1, but we do know the prime factors of 1001 and hence, the prime factors of 1001^32. We may not be able to say which numbers are factors of 1001^(32) – 1, but we will be able to say which numbers are certainly not factors of this!

Let me explain:

1001 = 7 * 11 * 13 (Try dividing 1001 by 7 and you’ll get 143. 143 is divisible by 11, giving you 13.)

1001^32 = 7^32 * 11^32 * 13^32

Now, what can we say about the prime factors of 1001^(32) – 1? Whatever they are, they are certainly not 7, 11 or 13 – two consecutive integers cannot have any common prime factor (discussed here and continued here).

Now look at the answer choices and try dividing each by 7:

(A) 768 – Not divisible by 7

(B) 819 – Divisible by 7

(C) 826 – Divisible by 7

(D) 858 – Not divisible by 7

(E) 924 – Divisible by 7

Options B, C and E are eliminated. They certainly cannot be factors of 1001^(32) – 1 since they have 7 as a prime factor, and we know 1001^(32) – 1 cannot have 7 as a prime factor.

Now try dividing the remaining options by 11:

(A) 768 – Not divisible by 11

(D) 858 – Divisible by 11

D can also be eliminated now because it has 11 as a factor. By process of elimination, the answer is A; it must be a factor of 1001^(32) – 1.

I hope you see how easily we used the factors of 1001 to help us solve this difficult-looking question. And yes, another attractive feature of 1001 – it is a palindrome in the decimal representation itself!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: How to Find Composite Numbers on the GMAT

Quarter Wit, Quarter WisdomWe love to talk about prime numbers and their various properties for GMAT preparation, but composite numbers usually aren’t mentioned. Composite numbers are often viewed as whatever is leftover after prime numbers are removed from a set of positive integers (except 1 because 1 is neither prime, nor composite), but it is important to understand how these numbers are made, what makes them special and what should come to mind when we read “composite numbers.”

Principle: Every composite number is made up of 2 or more prime numbers. The prime numbers could be the same or they could be distinct.

For example:

2*2 = 4 (Composite number)

2*3*11 = 66 (Composite number)

5*23 = 115 (Composite number)

and so on…

Look at any composite number. You will always be able to split it into 2 or more prime numbers (not necessarily distinct). For example:

72 = 2*2*2*3*3

140 = 2*2*5*7

166 = 2*83

and so on…

This principle does look quite simple and intuitive at first, but when tested, we could face problems because we don’t think much about it. Let’s look at it with the help of one of our 700+ level GMAT questions:

x is the smallest integer greater than 1000 that is not prime and that has only one factor in common with 30!. What is x?

(A) 1009

(B) 1021

(C) 1147

(D) 1273

(E) 50! + 1

If we start with the answer choices, the way we often do when dealing with prime/composite numbers, we will get stuck. If we were looking for a prime number, we would use the method of elimination – we would find factors of all other numbers and the number that was left over would be the prime number.

But in this question, we are instead looking for a composite number – a specific composite number – and some of the answer choices are probably prime. Try as we might, we will not find a factor for them, and by the time we realize that it is prime, we will have wasted a lot of precious time. Let’s start from the question stem, instead.

We need a composite number that has only one factor in common with 30!. Every positive integer will have 1 as a factor, as will 30!, hence the only factor our answer and 30! will have in common is 1.

30! = 1*2*3*…*28*29*30

30! is the product of all integers from 1 to 30, so all prime numbers less than 30 are factors of 30!.

To make a composite number which has no prime factor in common with 30!, we must use prime numbers greater than 30. The first prime number greater than 30 is 31.

(As an aside, note that if we were looking for the smallest number with no factor other than 1 in common with 31!, we would skip to 37. All integers between 31 and 37 are composite and hence, would have factors lying between 1 and 31. Similarly, if we were looking for the smallest number with no factor other than 1 in common with 50!, 53 would be the answer.)

Let’s get back to our question. If we want to make a composite number without using any primes until 30, we must use two or more prime numbers greater than 30, and the smallest prime greater than 30 is 31. If we use two 31’s to get the smallest composite number, we get 31*31 = 961 But 961 is not greater than 1000, so it cannot be our answer.

So, let’s find the next prime number after 31 – it is 37. Multiplying 31 and 37, we get 31*37 = 1147. This is the smallest composite number greater than 1000 with no prime factors in common with 30! – the only factor it has in common with 30! is 1. Therefore, our answer is (C).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

It’s All Greek to Me: How to Use Greek Concepts to Beat the GMAT

Aero_img084The ancient Greeks were, to put it mildly, really neat. They created or helped to create the foundations of philosophy, theater, science, democracy, and mathematics – no small accomplishment for a small war-torn civilization from over two millennia ago. Many of our contemporary ideas, beliefs, and traditions are rooted in contributions made by Greek thinkers, and the GMAT is no exception.

A few months ago, I wrote about this difficult Data Sufficiency question.

When I first encountered this problem I couldn’t help but wonder what kind of mad scientist question-writer engineered it. Where would such an idea even come from? It turns out, it wasn’t a GMAC employee at all, but Archimedes, the famous Greek geometer and coiner of the phrase “Eureka!”

The question is based on his attempt to trisect an angle with only a straight edge and a compass. (Alas, Archimedes’ work, though ingenious, was not technically a correct solution to the problem, as it provides only an approximation.) The reader is hereby invited to contemplate the kind of person who encounters a proof by Archimedes and instinctively thinks, “This would make an excellent Data Sufficiency question on the GMAT!” We’d like to believe that the good folks at GMAC are just like you and me, but perhaps not.

So this got me thinking: what other interesting Greek contributions to mathematics might be helpful in analyzing GMAT questions? In Euclid’s work Elements, he offers a simple and elegant proof for why there is no largest prime number. The proof proceeds by positing a hypothetical largest prime number “p.” We can then construct a product that consists of every prime number 2*3*5*7….*p. We’ll call this product “q.”

The next consecutive number will be q + 1. Now, we know that “q” contains 2 as a factor, as “q,” supposedly, contains every prime as a factor. Therefore q +1 will not contain 2 as a factor. (The next number to contain 2 as a factor will be q + 2.) We know that “q” contains 3 as a factor. Therefore q + 1 will not contain 3 as a factor. (The next number to contain 3 as a factor will be q + 3.)

Uh oh. If “p” really is the largest prime number, we’ve got a problem, because q + 1 will not contain any of the primes between 2 and p as factors. So either q + 1 is itself prime, or there is some prime greater than p and less than q + 1 that we’ve failed to consider. Either way, we’ve proven that p can’t be the largest prime number – I told you the Greeks were neat.

One axiom that’s worth internalizing from Euclid’s proof is the notion that two consecutive numbers cannot have any factors in common aside from 1.  When q contains every prime from 2 to p as a factor, q + 1 contains none of those primes. How would this be helpful on the GMAT? Glad you asked. Check out this question:

x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x + 1 must be:

(A) Between 1 and 10

(B) Between 11 and 15

(C) Between 15 and 20

(D) Between 20 and 25

(E) Greater than 25

We’re given information about x, and we’re asked about x + 1. If x is the product of all even numbers from 2 to 50, we can write x = 2 * 4 * 6 …* 50. This is the same as (1*2) * (2*2) * (3*2)… (25*2), which means the product consists of all the integers from 1 to 25, inclusive, and a bunch of 2’s.

So now we know that every prime number between 2 and 25 will be a factor of x. What about x + 1? (Paging Euclid!) We know that 2 is not a factor of x + 1, as 2 is a factor of x, and so the next multiple of 2 would be x + 2. We know that 3 is also not a factor of x + 1, as 3 is a factor of x, and so the next multiple of 3 would be x + 3. And once we’ve internalized that two consecutive numbers cannot have any factors in common aside from 1, we know that if all the primes between 2 and 25 are factors of x, none of those primes can be factors of x + 1, meaning that the smallest prime of x, whatever is, will be greater than 25. The answer, therefore, is E.

Takeaway: One of the beautiful things about mathematics is that fundamental truths do not change over time. What worked for the Greeks will work for us. The same axioms that allowed ancient mathematicians to grapple with problems two millennia ago will allow us to unravel the toughest GMAT questions. Learning a few of these axioms is not only interesting – though I’d caution against bringing up Archimedes’ trisection proof at a dinner party – but also helpful on the GMAT.

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTube, Google+ and Twitter!

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles by him here.

GMAT Tip of the Week

Prime Time

(This is one of a series of GMAT tips that we offer on our blog.)

A recent episode of “The Office” featured a classic, GMAT-relevant exchange, in which a cash-strapped Michael Scott asks his financial analyst to “crunch those numbers again”. The stunned analyst explains that, because the calculations were all done accurately using a computer program, there was no mechanism for “crunching” the numbers again, and even if there were, there would be no change.

Such is life in business nowadays. Sophisticated machines do a lot of the “number crunching” for us, and business managers are much more often in the business of analyzing numbers than of crunching them. The GMAT, in an attempt to determine the candidates best suited to thrive in such an environment, heavily features the analysis of numbers in similar ways, requiring you to think often about the properties of numbers.

A prime example of this is the examination of prime numbers on the GMAT. Prime numbers are those that have exactly two factors (itself and 1) – a seemingly simple definition that can often become cumbersome to employ on the GMAT. One such way in which prime numbers can lead to frustration is a question like the following:

How many prime numbers are between 110 and 120?

It’s unlikely that you’ll have memorized the list of prime numbers up in to the triple-digits, so you will probably approach this question by taking the set of numbers and eliminating any numbers that are not prime. Even numbers, by definition, are divisible by 2, so the even numbers in this set are definitely not prime, leaving us with a set of:


It’s also relatively easy to eliminate 115, because a number ending in 5 is divisible by 5, so we’re down to four numbers remaining. 111 and 117 are each divisible by 3 (there’s a trick for making that determination that we’ll probably feature in an upcoming GMAT Tip of the Week), so we’re left to test:


This is where it may get tricky, as in order to prove that a number is prime, we need to prove that it is not divisible by anything but itself and 1. With a 3-digit number, this process could be time consuming without these two principles:

1) You don’t need to test for divisibility by anything other than prime numbers.

If a number is divisible by, say, 4, it needs to also be divisible by 2, because 4 is divisible by 2. So, if you’ve already determined that 113 is not divisible by 2, you don’t need to test to see if it is divisible by 4 (or 6, or 8, etc.).

2) You don’t need to test a number by anything higher than the square root of the next-highest perfect square.

This is probably best illustrated by an example. With 113, the next highest square above it is 121, and we know that 121 is the same as 11*11. So, logically speaking, in order for a number greater than 11 to multiply with another integer to produce a number smaller than 121, that other integer must be less than 11. If it were greater than 11, the product would be higher than 121.

If we test 113 by the other primes (7 and 11), we find that it’s not divisible by 7 (7*10 = 70 and 7*6 = 42, so 7*16 is 112, meaning that 113 cannot be divisible by 7… but also that 119 is divisible by 7. So, our work with 119 is done.).

It’s also not divisible by 11 (11 * 10 = 110, so we know that 113 is not a multiple of 11).

Now, because we’ve already tested everything up to 11, we’re done…we know that 113 is prime. Again, if 113 were to be divisible by 13, it would also have to be divisible by something less than 11, because we know that 11*11 is already too high. So, there’s no need to test 113 for divisibility by anything other than what we already have, and we can prove that the answer to the overall question is 1.

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