Track / Stage 3 / 64 of 260 #64 of 2000

Problem 64

AMC 10/12, early questions

Number theory Difficulty 3.3 Find the answer

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

Your answer

Official 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 $2+4+8=14\Longrightarrow\boxed{\mathrm{ (B)}\ 14}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.