2002 AMC 10B Problems/Problem 14

Problem

The number $25^{64}\cdot 64^{25}$ is the square of a positive integer $N$ . In decimal representation, the sum of the digits of $N$ is

$\mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 35$

Solution

Taking the root, we get $N=\sqrt{25^{64}\cdot 64^{25}}=5^{64}\cdot 8^{25}$ .

Now, we have $N=5^{64}\cdot 8^{25}=5^{64}\cdot (2^{3})^{25}=5^{64}\cdot 2^{75}$ .

Combining the $2$ 's and $5$ 's gives us $(2\cdot 5)^{64}\cdot 2^{(75-64)}=(2\cdot 5)^{64}\cdot 2^{11}=10^{64}\cdot 2^{11}$ .

This is the number $2048$ with a string of sixty-four $0$ 's at the end. Thus, the sum of the digits of $N$ is $2+4+8=14\Longrightarrow\boxed{\mathrm{ (B)}\ 14}$

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2002 AMC 10B Problems/Problem 14

Problem

Solution

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