2003 AMC 10B Problem 14

Attempt Problem 14 of the 2003 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10B solutions, or check the answer key.

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14.

Given that

3^8 \cdot 5^2 = a^b,

where both

a

and

b

are positive integers, find the smallest possible value for

a+b.

25

34

351

407

900

Answer: D

Concepts:prime factorization perfect power optimization

Difficulty rating: 1390

Small hint:

Since

5^2

but not

5^3

divides the number,

b

can be at most

2

Big hint:

The larger

b

is, the smaller

a

becomes; try

b=2

and write the number as a perfect square

Solution:

Because

a

must be divisible by

5,

and

3^8 \cdot 5^2

is divisible by

5^2

but not

5^3,

we need

b\le 2.

Taking

b=2

gives

a=\sqrt{3^8 \cdot 5^2}=3^4 \cdot 5=405,

so

a+b=407.

This beats

b=1,

which gives

a+b=164{,}026.

Thus, the correct answer is D.