2011 AMC 8 Problem 17

Question

Let

w,

x,

y,

and

z

be whole numbers. If

2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588,

then what does

2w + 3x + 5y + 7z

equal?

21

25

27

35

56

Match the exponents of

2

,

3

,

5

, and

7

.

Prime-factorize

588

.

To find the desired exponents, note that all the bases are prime numbers. This means that finding the prime factorization will be helpful.

We get that

588 = 2^2 \cdot 3^1 \cdot 7^2.

From this, it is clear that

w = 2,

x = 1,

y = 0,

and

z = 2

(

y = 0

since that makes the

5^y

term equal

1

).

Therefore,

\begin{gather*} 2w + 3x + 5y + 7z \\ = 2 \cdot 2 + 3 \cdot 1 + 5 \cdot 0 + 7 \cdot 2 \\ = 21. \end{gather*}

Thus, A is the correct answer.

Problem 17 in other years