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

GMAT Tip of the WeekIf 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

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

By Brian Galvin.

Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions (Part 2)

Quarter Wit, Quarter WisdomLast week, we reviewed the concepts of cyclicity and remainders and looked at some basic questions. Today, let’s jump right into some GMAT-relevant questions on these topics:

 

 

If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

In this problem, we are looking for the remainder when the divisor is 5. We know from last week that if we get the last digit of the dividend, we will be able to find the remainder, so let’s focus on finding the units digit of 3^(8n + 3) + 2.

The units digit of 3 in a positive integer power has a cyclicity of: 3, 9, 7, 1

So the units digit of 3^(8n + 3) = 3^(4*2n + 3) will have 2n full cycles of 3, 9, 7, 1 and then a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7

Since the exponent a remainder of 3, the new cycle ends at 3, 9, 7. Therefore, the units digit of 3^(8n + 3) is 7. When you add another 2 to this expression, the units digit becomes 7+2 = 9.

This means the units digit of 3^(8n+3) + 2 is 9. When we divide this by 5, the remainder will be 4, therefore, our answer is E.

Not so bad; let’s try a data sufficiency problem:

If k is a positive integer, what is the remainder when 2^k is divided by 10?

Statement 1: k is divisible by 10

Statement 2: k is divisible by 4

With this problem, we know that the remainder of a division by 10 can be easily obtained by getting the units digit of the number. Let’s try to find the units digit of 2^k.

The cyclicity of 2 is: 2, 4, 8, 6. Depending on the value of k is, the units digit of 2^k will change:

If k is a multiple of 4, it will end after one cycle and hence the units digit will be 6.

If k is 1 more than a multiple of 4, it will start a new cycle and the units digit of 2^k will be 2.

If k is 2 more than a multiple of 4, it will be second digit of a new cycle, and the units digit of 2^k will be 4.

If k is 3 more than a multiple of 4, it will be the third digit of a new cycle and the units digit of 2^k will be 8.

If k is 4 more than a multiple of 4, it will again be a multiple of 4 and will end a cycle. The units digit of 2^k will be 6 in this case.

and so on…

So what we really need to find out is whether k is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4, or three more than a multiple of 4.

Statement 1: k is divisible by 10

With this statement, k could be 10 or 20 or 30 etc. In some cases, such as when k is 10 or 30, k will be two more than a multiple of 4. In other cases, such as when k is 20 or 40, k will be a multiple of 4. So for different values of k, the units digit will be different and hence the remainder on division by 10 will take multiple values. This statement alone is not sufficient.

Statement 2: k is divisible by 4

This statement tells you directly that k is divisible by 4. This means that the last digit of 2^k is 6, so when divided by 10, it will give a remainder of 6. This statement alone is sufficient. therefore our answer is B.

Now, to cap it all off, we will look at one final question. It is debatable whether it is within the scope of the GMAT but it is based on the same concepts and is a great exercise for intellectual purposes. You are free to ignore it if you are short on time or would not like to go an iota beyond the scope of the GMAT:

What is the remainder of (3^7^11) divided by 5?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

For this problem, we need the remainder of a division by 5, so our first step is to get the units digit of 3^7^{11}. Now this is the tricky part – it is 3 to the power of 7, which itself is to the power of 11. Let’s simplify this a bit; we need to find the units digit of 3^a such that a = 7^{11}.

We know that 3 has a cyclicity of 3, 9, 7, 1. Therefore (similar to our last problem) to get the units digit of 3^a, we need to find whether a is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4 or three more than a multiple of 4.

We need a to equal 7^{11}, so first we need to find the remainder when a is divided by 4; i.e. when 7^{11} is divided by 4.

For this, we need to use the binomial theorem we learned earlier in this post (or we can use the method of “pattern recognition”):

The remainder of 7^{11} divided by 4

= The remainder of (4 + 3)^{11} divided by 4

= The remainder of 3^{11} divided by 4

= The remainder of 3*3^{10} divided by 4

= The remainder of 3*9^5 divided by 4

= The remainder of 3*(8+1)^5 divided by 4

= The remainder of 3*1^5 divided by 4

= The remainder of 3 divided by 4, which itself = 3

So when 7^{11} is divided by 4, the remainder is 3. This means 7^{11} is 3 more than a multiple of 4; i.e. a is 3 more than a multiple of 4.

Now we go back to 3^a. We found that a is 3 more than a multiple of 4. So there will be full cycles (we don’t need to know the exact number of cycles) and then a new cycle with start with three digits remaining:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7

With this pattern, we see the last digit of 3^7^11 is 7. When this 7 is divided by 5, remainder will be 2 – therefore, our answer is C.

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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!