# 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.

# Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions Usually, cyclicity cannot help us when dealing with remainders, but in some cases it can. Today we will look at the cases in which it can, and we will see why it helps us in these cases.

First let’s look at a pattern:

20/10 gives us a remainder of 0 (as 20 is exactly divisible by 10)

21/10 gives a remainder of 1

22/10 gives a remainder of 2

23/10 gives a remainder of 3

24/10 gives a remainder of 4

25/10 gives a remainder of 5

and so on…

In the case of this pattern, 20 is the closest multiple of 10 that goes completely into all these numbers and you are left with the units digit as the remainder. Whenever you divide a number by 10, the units digit will be the remainder. Of course, if the units digit of a number is 0, the remainder will be 0 and that number will be divisible by 10 — but we already know that. So remainder when 467,639 is divided by 10 is 9. The remainder when 100,238 is divided by 10 is 8 and so on…

Along the same lines, we also know that every number that ends in 0 or 5 is a multiple of 5 and every multiple of 5 must end in either 0 or 5. So if the units digit of a number is 1, it gives a remainder of 1 when divided by 5. If the units digit of a number is 2, it gives a remainder of 2 when divided by 5. If the units digit of a number is 6, it gives a remainder of 1 when divided by 5 (as it is 1 more than the previous multiple of 5).

With this in mind:

20/5 gives a remainder of 0 (as 20 is exactly divisible by 5)

21/5 gives a remainder of 1

22/5 gives a remainder of 2

23/5 gives a remainder of 3

24/5 gives a remainder of 4

25/5 gives a remainder of 0 (as 25 is exactly divisible by 5)

26/5 gives a remainder of 1

27/5 gives a remainder of 2

28/5 gives a remainder of 3

29/5 gives a remainder of 4

30/5 gives a remainder of 0 (as 30 is exactly divisible by 5)

and so on…

So the units digit is all that matters when trying to get the remainder of a division by 5 or by 10.

Let’s take a few questions now:

What is the remainder when 86^(183) is divided by 10?

Here, we need to find the last digit of 86^(183) to get the remainder. Whenever the units digit is 6, it remains 6 no matter what the positive integer exponent is (previously discussed in this post).

So the units digit of 86^(183) will be 6. So when we divide this by 10, the remainder will also be 6.

Next question:

What is the remainder when 487^(191) is divided by 5?

Again, when considering division by 5, the units digit can help us.

The units digit of 487 is 7.

7 has a cyclicity of 7, 9, 3, 1.

Divide 191 by 4 to get a quotient of 47 and a remainder of 3. This means that we will have 47 full cycles of “7, 9, 3, 1” and then a new cycle will start and continue until the third term.

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3

So the units digit of 487^(191) is 3, and the number would look something like ……………..3

As discussed, the number ……………..0 would be divisible by 5 and ……………..3 would be 3 more, so it will also give a remainder of 3 when divided by 5.

Therefore, the remainder of 487^(191) divided by 5 is 3.

Last question:

If x is a positive integer, what is the remainder when 488^(6x) is divided by 2?

Take a minute to review the question first. If you start by analyzing the expression 488^(6x), you will waste a lot of time. This is a trick question! The divisor is 2, and we know that every even number is divisible by 2, and every odd number gives a remainder 1 when divided by 2. Therefore, we just need to determine whether 488^(6x) is odd or even.

488^(6x) will be even no matter what x is (as long as it is a positive integer), because 488 is even and we know even*even*even……(any number of terms) = even.

So 488^(6x) is even and will give remainder 0 when it is divided by 2.

That is all for today. We will look at some GMAT remainders-cyclicity questions next week!

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!

# GMAT Tip of the Week

A Pattern of Efficiency

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

A colleague recently pointed out a practice test problem he had seen that appeared to be a unique, new variety of GMAT quantitative problem. (Editor’s note: There is no need for alarm; continue reading and you’ll learn that this problem is entirely common on the exam and has been for years!) The problem asked for the test taker to sum a relatively high number of values that were displayed on a grid; the extent of the problem was similar to:

What is the sum of:

-1 + 2 – 3 + 4 – 5 + 6 – 7 + 8 – 9 + 10 – 11 + 12 – 13 + 14 – 15 + 16

Displayed in grid form, the problem in question contained a greater number of values and was asked in a slightly different way, but the takeaway is the same: The GMAT likes to ask questions that seem to require a time-consuming calculation, but can actually be solved relatively quickly through pattern recognition, leading to simpler math.

In this case, each pair of odd-then-even values sums to 1. (-1 + 2), (-3 + 4), etc. will each produce a sum of 1, meaning that the test-taker only need to recognize that pattern, determine the number of pairs* that exist in the sequence (in this case, 8), and multiply the sum of each pair by the number of pairs (1 * 8) to achieve the answer, 8.

The takeaway? When problems look to require extensive calculation, or when you detect that a pattern may be present, look to find a pattern that can make the solution quick and efficient. The GMAT rewards that style of thinking more often than not.

(*Note: be sure in this case to determine whether the sequence does, indeed, contain only distinct pairs; to make the question more difficult, the sequence could end with -17, in which case that value wouldn’t have a complementary positive number, and would have to be subtracted entirely.)

If you’re in the early stages of your GMAT preparation, Veritas Prep offers the Official Guide for GMAT Review, 12th Edition, for just \$10. It’s a great way to get started!