# GMAT Tip of the Week: 3 Guiding Principles for Exponent Problems

If you’re like many GMAT examinees, you’ve found yourself in this familiar situation. You KNOW the rules for exponents. You know them cold. When you’re multiplying the same base and different exponents, you add the exponents. When you’re taking one exponent to another power, you multiply those exponents. A negative exponent? Flip that term into the denominator. A number to the zero power? You’ve got yourself a 1.

But as thoroughly and quickly as you know those rules, this exponent-based problem in front of you has you stumped. You know what you need to KNOW, but you’re not quite sure what you need to DO. And that’s an ever-important part about taking the GMAT – it’s necessary to know the core rules, facts, and formulas, but it’s also every bit as important to have action items for how you’ll apply that knowledge to tricky problems.

For exponents, there are three “guiding principles” that you should keep in mind as your action items. Any time you’re stuck on an exponent-based problem, look to do one (or more) of these things:

1) Find Common Bases
Most of the exponent rules you know only apply when you’re dealing with two exponents of the same base. When you multiply same-base exponents, you add the exponents; when you divide two same-base exponents, you subtract. And if two exponents of the same base are set equal, then you know that the exponents are equal. But keep in mind – these major rules all require you to be using exponents with the same base! If the GMAT gives you a problem with different bases, you have to find ways to make them common, usually by factoring them into their prime bases.

So for example, you might see a problem that says that:

2^x * 4^2x = 8^y. Which of the following must be true?

(A) 3x = y
(B) x = 3y
(C) y = (3/5)x
(D) x = (3/5)y
(E) 2x^2 = y

In order to apply any rules that you know, you must get the bases in a position where they’ll talk to each other. Since 2, 4, and 8 are all powers of 2, you should factor them all in to base 2, rewriting as:

2^x * (2^2)^2x = (2^3)^y

Which simplifies to:

2^x * 2^4x = 2^3y

Now you can add together the exponents on the left:

2^5x = 2^3y

And since you have the same base set equal with two different exponents, you know that the exponents are equal:

5x = 3y

This means that you can divide both sides by 5 to get x = (3/5)y, making answer choice D correct. But more importantly in a larger context, heed this lesson – when you see an exponent problem with different bases for multiple exponents, try to find ways to get the bases the same, usually by prime-factoring the bases.

2) Factor to Create Multiplication
Another important thing about exponents is that they represent recurring multiplication. x^5, for example, is x * x * x * x * x…it’s a lot of x’s multiplied together. Naturally, then, pretty much all exponent rules apply in cases of multiplication, division, or more exponents – you don’t have rules that directly apply to addition or subtraction. For that reason, when you see addition or subtraction in an exponent problem, one of your core instincts should be to factor common terms to create multiplication or division so that you’re in a better position to leverage the rules you know. So, for example, if you’re given the problem:

2^x + 2^(x + 3) = (6^2)(2^18). What is the value of x?

(A) 18
(B) 20
(C) 21
(D) 22
(E) 24

You should see that in order to do anything with the left-hand side of the equation, you’ll need to factor the common 2^x in order to create multiplication and be in a position to divide and cancel terms from the right. Doing so leaves you with:

2^x(1 + 2^3) = (6^2)(2^18)

Here, you can simplify the 1 + 2^3 parenthetical: 2^3 = 8, so that term becomes 9, leaving you with:

9(2^x) = (6^2)(2^18)

And here, you should heed the wisdom from above and find common bases. The 9 on the left is 3^2, and the 6^2 on the right can be broken into 3^2 * 2^2. This gives you:

(3^2)(2^x) = (3^2)(2^2)(2^18)

Now the 3^2 terms will cancel, and you can add the exponents of the base-2 exponents on the right. That means that 2^x = 2^20, so you know that x = 20. And a huge key to solving this one was factoring the addition into multiplication, a crucial exponent-based action item on test day.

3) Test Small Numbers and Look For Patterns
Remember: exponents are a way to denote repetitive, recurring multiplication. And when you do the same thing over and over again, you tend to get similar results. So exponents lend themselves well to finding and extrapolating patterns. When in doubt – when a problem involves too much abstraction or too large of numbers for you to get your head around – see what would happen if you replaced the large or abstract terms with smaller ones, and if you find a pattern, then look to extrapolate it. With this in mind, consider the problem:

What is the tens digit of 11^13?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Naturally, calculating 11^13 without a calculator is a fool’s errand, but you can start by taking the first few steps and seeing if you establish a pattern:

11^1 = 11 –> tens digit of 1

11^2 = 121 –> tens digit of 2

11^3 = 1331 –> tens digit of 3

And depending on how much time you have you could continue:

11^4 = 14641 –> tens digit of 4

But generally feel pretty good that you’ve established a recurring pattern: the tens digit increases by 1 each time, so by 11^13 it will be back at 3. So even though you’ll never know exactly what 11^13 is, you can be confident in your answer.

Remember: the GMAT is a test of how well you apply knowledge, not just of how well you can memorize it. So for any concept, don’t just know the rules, but also give yourself action items for what you’ll do when problems get tricky. For exponent problems, you have three guiding principles:

1) Find Common Bases
2) Factor to Create Multiplication
3) Test Small Numbers to Find a Pattern

By Brian Galvin.

# How to Solve “Hidden” Factor Problems on the GMAT

One of the interesting things to note about newer GMAC Quant questions is that, while many of these questions test our knowledge of multiples and factors, the phrasing of these questions is often more subtle than earlier versions you might have seen. For example, if I ask you to find the least common multiple of 6 and 9, I’m not being terribly artful about what topic I’m testing you on – the word “multiple” is in the question itself.

But if tell you that I have a certain number of cupcakes and, were I so inclined, I could distribute the same number of cupcakes to each of 6 students with none left over or to each of 9 students with none left over, it’s the same concept, but I’m not telegraphing the subject in the same conspicuous manner as the previous question.

This kind of recognition comes in handy for questions like this one:

All boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxes arrived and no boxes were removed, all the boxes in the warehouse were arranged in stacks of 14 boxes each, with no boxes left over. How many boxes were in the warehouse before the 60 additional boxes arrived?

(1) There were fewer than 110 boxes in the warehouse before the 60 additional arrived.
(2) There were fewer than 120 boxes in the warehouse after the 60 additional arrived.

Initially, we have stacks of 12 boxes with no boxes left over, meaning we could have 12 boxes or 24 boxes or 36 boxes, etc. This is when you want to recognize that we’re dealing with a multiple/factor question. That first sentence tells you that the number of boxes is a multiple of 12. After 60 more boxes were added, the boxes were arranged in stacks of 14 with none left over – after this change, the number of boxes is a multiple of 14.

Because 60 is, itself, a multiple of 12, the new number must remain a multiple of 12, as well. [If we called the old number of boxes 12x, the new number would be 12x + 60. We could then factor out a 12 and call this number 12(x + 5.) This number is clearly a multiple of 12.] Therefore the new number, after 60 boxes are added, is a multiple of both 12 and 14. Now we can find the least common multiple of 12 and 14 to ensure that we don’t miss any possibilities.

The prime factorization of 12: 2^2 * 3

The prime factorization of 14: 2 * 7

The least common multiple of 12 and 14: 2^2 * 3 * 7 = 84.

We now know that, after 60 boxes were added, the total number of boxes was a multiple of 84. There could have been 84 boxes or 168 boxes, etc. And before the 60 boxes were added, there could have been 84-60 = 24 boxes or 168-60 = 108 boxes, etc.

A brief summary:

After 60 boxes were added: 84, 168, 252….

Before 60 boxes were added: 24, 108, 192….

That feels like a lot of work to do before even glancing at the statements, but now look at how much easier they are to evaluate!

Statement 1 tells us that there were fewer than 110 boxes before the 60 boxes were added, meaning there could have been 24 boxes to start (and 84 once 60 were added), or there could have been 108 boxes to start (and 168 once 60 were added). Because there are multiple potential solutions here, Statement 1 alone is not sufficient to answer the question.

Statement 2 tells us that there were fewer than 120 boxes after 60 boxes were added. This means there could have been 84 boxes – that’s the only possibility, as the next number, 168, already exceeds 120. So we know for a fact that there are 84 boxes after 60 were added, and 24 boxes before they were added. Statement 2 alone is sufficient, and the answer is B.

Takeaway: questions that look strange or funky are always testing concepts that have been tested in the past – otherwise, the exam wouldn’t be standardized. By making these connections, and recognizing that a verbal clue such as “none left over” really means that we’re talking about multiples and factors, we can recognize even the most abstract patterns on the toughest of GMAT questions.

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

# Quarter Wit, Quarter Wisdom: The Power of Deduction on GMAT Data Sufficiency Questions

In a previous post, we have discussed how to find the total number of factors of a number. What does the total number of factors a number has tell us about that number? One might guess, “Not a lot,” but it actually does tell us quite a bit! If the total number of factors is odd, you know the number must be a perfect square. If the total number of factors is even, you know the number is not a perfect square.

We know that the total number of factors of a number A (prime factorised as X^p * Y^q *…) is given by (p+1)*(q+1)… etc.

So, if we know that a number has, say, 6 total factors, what can we say about the number?

6 = (p+1)*(q+1) = 2*3, so p = 1 and q = 2 or vice versa.

A = X^1 * Y^2 where X and Y are distinct prime numbers.

Today, we will look at a data sufficiency question in which we can use factors to deduce much more information than what we might first guess:

When the digits of a two-digit, positive integer M are reversed, the result is the two-digit, positive integer N. If M > N, what is the value of M?

Statement 1: The integer (M – N) has 12 unique factors.
Statement 2: The integer (M – N) is a multiple of 9.

With this question, we are told that M is a two-digit integer and N is obtained by reversing it. So if M = 21, then N = 12; if M = 83, then N = 38 (keeping in mind that M must be greater than N). In the generic form:

M = 10a + b and N =10b + a (where a and b are single-digit numbers from 1 to 9. Neither can be 0 or greater than 9 since both M and N are two-digit numbers.)

We also know that no matter what M and N are, M > N. Therefore:

10a + b > 10b + a
9a > 9b
a > b

Let’s examine both of our given statements:

Statement 1: The integer (M – N) has 12 unique factors.

First, let’s figure out what M – N is:

M – N = (10a + b) – (10b + a) = 9a – 9b

Say M – N = A. This would mean A = 9(a-b) = 3^2 * (a-b)

The total number of factors of A where A = X^p * Y^q *… can be calculated using the formula (p+1)*(q+1)* …

We know that A has 3^2 as a factor, so X = 3 and p = 2. Therefore, the total number of factors would be (2+1)*(q+1)*… = 3*(q+1)*… = 12, so (q+1)*… must be 4.

Case 1:
This means q may be 3 so that (q+1) is 4. Since a-b must be less than or equal to 9 and must also be the cube of a number, (a-b) must be 8. (Note that a-b cannot be 1 because then the total number of factors of A would only be 3.)

So, a must be 9 and b must be 1 in this case (since a > b). The integers will be 91 and 19, and since M > N, M = 91.

Case 2:
Another possibility is that (a-b) is a product of two prime factors (other than 3), both with the power of 1. In that case, the total number of factors = (2+1)*(1+1)*(1+1) = 12

Note, however, that the two prime factors (other than 3) with the smallest product is 2*5 = 10, but the difference of two single-digit positive integers cannot be 10. This means that only Case 1 can be true, therefore, Statement 1 alone is sufficient. This is certainly not what we expected to find from just the total number of factors!

Statement 2: The integer (M – N) is a multiple of 9.

M – N = (10a + b) – (10b + a) = 9a – 9b, so M – N = 9 (a-b) . This is already a multiple of 9.

We get no new information with this statement; (a-b) can be any integer, such as 2 (a = 5, b = 3 or a = 7, b = 5), etc. This statement alone is insufficient, therefore our answer is A.

Don’t take the given data of a GMAT question at face value, especially if you are expecting questions from the 700+ range. Ensure that you have deduced everything that you can from it before coming to a conclusion.

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: Divisibility by Powers of 2

We know the divisibility rules of 2, 4 and 8:

For 2 – If the last digit of the number is divisible by 2 (is even), then the number is divisible by 2.

For 4 – If the number formed by last two digits of the number is divisible by 4, then the number is divisible by 4.

For 8 – If the number formed by last three digits of the number is divisible by 8, then the number is divisible by 8.

A similar rule applies to all powers of 2:

For 16 – If the number formed by last four digits of the number is divisible by 16, then the number is divisible by 16.

For 32 – If the number formed by last five digits of the number is divisible by 32, then the number is divisible by 32.

and so on…

Let’s figure out why:

The generic rule can be written like this: A number M is divisible by 2^n if the last n digits of M are divisible by 2^n.

Take, for example, a division by 8 (= 2^3), where M = 65748048 and n = 3.

Our digits of interest are the last three digits, 048.

48 is completely divisible by 8, so we conclude that 65748048 is also divisible by 8.

A valid question here is, “What about the remaining five digits? Why do we ignore them?”

Breaking down M, we can see that 65748048 = 65748000 + 048 (we’ve separated the last three digits).

Now note that 65748000 = 65748 * 1000. Since 1000 has three 0s, it is made up of three 2s and three 5s. Because 1000 it has three 2s as factor, it also has 8 as a factor. This means 65748000 has 8 as a factor by virtue of its three 0s.

All we need to worry about now is the last three digits, 048. If this is divisible by 8, 65748048 will also be divisible by 8. If it is not, 65748048 will not be divisible by 8.

In case the last three digits are not divisible by 8, you can still find the remainder of the number. Whatever remainder you get after dividing the last three digits by 8 will be the remainder when you divide the entire number by 8. This should not be a surprise to you now – 65748000 won’t have a remainder when divided by 8 since it is divisible by 8, so whatever the remainder is when the last 3 digits are divided by 8 will be the remainder when the entire number is divided by 8.

In the generic case, the number M will be split into a number with n zeroes and another number with n digits. The number with n zeroes will be divisible by 2^n because it has n 2s as factors. We just need to see the divisibility of the number with n digits.

We hope you have understood this concept. Let’s take look at a quick GMAT question to see this in action:

What is the remainder when 1990990900034 is divided by 32 ?

(A) 16
(B) 8
(C) 4
(D) 2
(E) 0

Breaking down our given number, 1990990900034 = 1990990900000 + 00034.

1990990900000 ends in five 0’s so it is divisible by 32. 34, when divided by 32, gives us a remainder of 2. Hence, when 1990990900034 is divided by 32, the remainder will be 2. Our answer is D.

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?

Fans 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!

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!

# How to Approach Difficult GMAT Problems

My students have a hard time understanding what makes a difficult GMAT question difficult. They assume that the tougher questions are either testing something they don’t know, or that these problems involve a dizzying level of complexity that requires an algebraic proficiency that’s simply beyond them.

One of my main goals in teaching a class is to persuade everyone that this is not, in fact, how hard questions work on this test. Hard questions don’t ask you do to something you don’t know how to do. Rather, they’re cleverly designed to provoke an anxiety response that makes it difficult to realize that you do know exactly how to solve the problem.

Take this official question, for example:

Let a, b, c and d be nonzero real numbers. If the quadratic equation ax(cx + d) = -b(cx +d) is solved for x, which of the following is a possible ratio of the 2 solutions?

A) –ab/cd
B) –ac/bd
D) ab/cd

Most students see this and panic. Often, they’ll start by multiplying out the left side of the equation, see that the expression is horrible (acx^2 + adx), and take this as evidence that this question is beyond their skill level. And, of course, the question was designed to elicit precisely this response. So when I do this problem in class, I always start by telling my students, much to their surprise, that every one of them already knows how to do this. They’ve just succumbed to the question writer’s attempt to convince them otherwise.

So let’s start simple. I’ll write the following on the board: xy = 0. Then I’ll ask what we know about x or y. And my students shrug and say x or y (or both) is equal to 0. They’ll also wonder what on earth such a simple identity has to do with the algebraic mess of the question they’d been struggling with.

I’ll then write this: zx + zy = 0. Again, I’ll ask what we know about the variables. Most will quickly see that we can factor out a “z” and get z(x+y) = 0. And again, applying the same logic, we see that one of the two components of the product must equal zero – either z = 0 or x + y = 0.

Next, I’ll ask if they would approach the problem any differently if I’d given them zx = -zy – they wouldn’t.

Now it clicks. We can take our initial equation in the aforementioned problem: ax(cx +d) = -b(cx+d), and see that we have a ‘cx + d’ on both sides of the equation, just as we’d had a “z” on both sides of the previous example. If I’m able to get everything on one side of the equation, I can factor out the common term.

Now ax(cx +d) = -b(cx+d) becomes ax(cx +d) + b(cx+d) = 0.

Just as we factored out a “z” in the previous example, we can factor out “cx + d” in this one.

Now we have (cx + d)(ax + b) = 0.

Again, if we multiply two expressions to get a product of zero, we know that at least one of those expressions must equal 0. Either cx + d = 0 or ax + b = 0.

If cx + d = 0, then x = -d/c.

If ax + b = 0, then x = -b/a.

Therefore, our two possible solutions for x are –d/c and –b/a. So, the ratio of the two would simply be (-d/c)/(-b/a). Recall that dividing by a fraction is the equivalent of multiplying by the reciprocal, so we’re ultimately solving for (-d/c)(-a/b). Multiplying two negatives gives us a positive, and we end up with da/cb, which is equivalent to answer choice E.

Takeaway: Anytime you see something on the GMAT that you think you don’t know how to do, remind yourself that the question was designed to create this false impression. You know how to do it – don’t hesitate to dive in and search for how to apply this knowledge.

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

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

The 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.

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.

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

# GMAT Tip of the Week

A Trick That Might Factor In

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

The quantitative section of the GMAT has a known emphasis on factorization of numbers, asking a variety of questions about divisibility, primes, least common multiple, etc. One fairly common question type asks, “How many unique factors does (number) have?”

While there are certainly strategies to answer this question efficiently without any kind of shortcut trick, a pretty slick shortcut does exist:

The first step would likely be your first step in any factorization problem; break the number down in to its prime factors. Take a number like 24, which breaks in to the factors 2*2*2*3.

Secondly, express that factorization as the product of exponents. In this case, it’s 2^3 * 3^1.

Next, discard the bases, and add one to each of the exponents. Here we’d have 3 and 1, and add one to each to make them 4 and 2.

Then, multiply the exponents-plus-one. 4*2 is 8, and there are 8 unique factors of 24:

1, 2, 3, 4, 6, 8, 12, and 24

The proof is a bit messy, but derives from another GMAT concept — combinatorics — which relies heavily on the use of, of all things, factorials.

If you blank on a trick like this, however, note that you can systematically come up with the above list of factors on your own, by taking the prime factors (2, 2, 2, and 3) and 1 (a factor of any integer), and multiplying each possible combination of them:

1
1*2 = 2
1 * 3 = 3
2*2 = 4
2 * 2 * 2 = 8
2 * 3 = 6
2 * 2 * 3 = 12
2 * 2 * 2 * 3 = 24

It’s a bit more time consuming, but doesn’t require that you memorize a trick precisely in order to solve the problem. As with all shortcuts on the GMAT, if they “click” for you, they’ll save you time, which then you can use for those problems that simply require more thought. Please don’t spend all of your time memorizing tricks, as the GMAT is written to reward “higher order thinking”. That said, a time-saving trick or two can provide you with an additional few minutes on the exam, and the corresponding confidence that you’re primed to post a high score.

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