# Exponent Challenge Question Series: Day 5

This is the fifth in a series of five blog posts that we will be publishing this week highlighting challenging exponent questions!

This week on Beat The GMAT, Veritas Prep’s authors will each day contribute a difficult exponent-related challenge problem, with a solution to follow the next day. Before you begin, you may want to consider this as a way to crack the exponent code; nearly all exponent-based problems can be solved using a combination of these three guiding exponent principles:

- Exponent rules are almost all related to multiplication and division with virtually no rules that directly apply to addition and subtraction. When facing exponent problems, look for opportunities to factor and multiply to put yourself in a position to use your multiplication-heavy exponent expertise.
- Most exponent rules require you to have common bases in order to apply them, so look to break down bases into prime factors so that you have common bases with which to work.
- Exponents are very pattern-driven, so when large numbers are present you can often establish a rule using small numbers and then extrapolate it to the larger ones.

**Yesterday’s Challenge Question and Solution **

If you haven’t don’t so already, try out yesterday’s problem first before reviewing the solution.

What is the tens digit of ?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

**C**. This question may well test the upper level of GMAT difficulty, but can be solved using the large-number-with-exponents strategy of finding patterns. Look at the progression of the easier-to-calculate powers of 11:

As you’re looking at digits places, you should first note that the units digit of each is 1, which will hold true anytime you’re multiplying a units digit of 1 by a units digit of 1. Now look at the tens places. They maintain the value of the exponent and progress from 1 to 2 to 3 to 4. They simply increase by one each time. Now, you may well just assume that the pattern will continue to to hold, and after this many in a row with your knowledge that exponents are very pattern-driven, you’re likely to be correct. But for the sake of thoroughness, let’s look at why that rule holds. Each time you multiply a number by 11, you’re multiplying it by 10+1. Therefore, you can multiply it by 1, multiply it by 10, and add those products together. Multiplying by 1 is easy (it just stays the same) and multiplying by 10 is the same as multiplying by 1 and just adding a 0 at the end. Therefore, , or , is:

Watching the tens place, you should notice that the portion just takes the previous units digit (which is always 1) and shifts it to the tens place, and that the portion keeps the previous tens place. Essentially, we use the portion to add 1 to the tens digit each time, ensuring that the pattern will hold, and the **tens digit of any power of 11 will simply be the units digit of the exponent**. Therefore, the answer is C.

**Today’s Challenge Question**

If , what is the value of x?

(A) 10

(B) 11

(C) 12

(D) 13

(E) 14

## 17 comments

jackhero on September 30th, 2010 at 5:59 am

The answer is A

3^(x+2)-3^x=6^3(3^7)

= 3^x(3^2-1)=216(3^7)

=3^x(8)=216(3^7)

=3^x=27(3^7)

=3^(x-7)=27

3^3=27

x=10

Reda on September 30th, 2010 at 6:04 am

good question...

first part is 3^x*3^2-3^x=3^x(3^2-1)=3^x*8

second par is 3^3*2^3*3^7=3^10*8

so x=10

actually enjoyed solving this, hope you have more coming...

ASKI@1st on September 30th, 2010 at 6:49 am

A - 10

First of all let’s try to put it all on the same factors

3^(x+2)-3^x = 6^3*(3^7)

3^x*3^2-3^x=3^3*2^3*3^7

3^x*(3^2-1)=3^10*2^3

3^x*(8)=3^10*2^3 with 8=2^3

3^x*(2^3)=3^10*2^3 that is x=10

A = 10

Tracy Thomas on September 30th, 2010 at 11:42 am

I follow the explanation up to the second step. How does 3^x*3^2-3X become 3^x*(3^2-1)?

Ishaan Singhal on September 30th, 2010 at 3:21 pm

tracy we have 3^x common in the two terms.

we use the rule:

a*(b-c) = a*b - a*c

here

a = 3^x

b = 3^2

c = 1

i hope u understand...

Ishaan

Anahatha on September 30th, 2010 at 6:50 am

(A) 10

Divya on September 30th, 2010 at 7:51 am

Oooo, this ones easy

3^(x+2)-3^x=6^3 (3^7)

1) Breaking 3^(x+2) into 3^x.3^2

3^x.3^2-3^x=6^3(3^7)

2) Taking out 3^x which is common on left hand side

3^x(3^2-1)=6^3(3^7)

3) Solve left hand side and break 6^3 into 3^3.2^3

3^x(9-1)=(3^3)(2^3)(3^7)

4)Add exponents of 3^3 and 3^7 and Solve

3^x.8=(3^10)(8)

5) Eliminate 8 from both sides

3^x=3^10

Therefore x=10 which is Choice (A)

Nityat on September 30th, 2010 at 9:12 am

(A) 10

3^(x + 2) - 3^x = 3^x * 2^3

6^3 * 3^7 = 2^3 * 3^10

==> x=10

Kerby on September 30th, 2010 at 9:54 am

(a) 10

Siddhartha on September 30th, 2010 at 11:56 am

i dnt expect dis type of questions frm btg

anj on September 30th, 2010 at 12:26 pm

A. 10

If such are the challenge questions then everyone will get 800 in GMAT......

Jeff on September 30th, 2010 at 7:06 pm

Brian,

Thank you so much for putting this together.... This has been one of the most helpful BTG series I've seen in a while!

euro on October 1st, 2010 at 1:47 am

I liked the problem. For those who have lost touch with books, such problems help in brushing up math skills.

Yashu on October 2nd, 2010 at 5:42 pm

Good problem.

Vikas on October 3rd, 2010 at 7:10 pm

x=10.

GANDHI on October 22nd, 2010 at 5:51 am

I WANTED TO KNOW THE STRATEGIES OF SOLVING THE EXPONENTS

Pushkal on January 4th, 2011 at 5:05 am

X=10