Difference between revisions of "2002 AMC 10B Problems/Problem 14"

Revision as of 14:33, 28 December 2011

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

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

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

Combing 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 $2048$ with sixty-four, $0$ 's on the end. So, the sum of the digits of $N$ is $2+4+8=14\Longrightarrow\mathrm{ (B) \ }$

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 == Solution ==
-Taking the squareroot, <math>N=5^{64}\cdot 8^{25}=5^{64}\cdot 2^{75}=10^{64}\cdot 2^{11}</math>. This is <math>2048</math> with 64 <math>0</math>'s on the end. So, the sum of the digits of <math>N</math> is <math>14\Longrightarrow\mathrm{ (B) \ }</math>
+Since, <math>N=5^{64}\cdot 8^{25}=5^{64}\cdot (2^{3})^{25}=5^{64}\cdot 2^{75}</math>.
+Combing the <math>2</math>'s and <math>5</math>'s gives us, <math>(2\cdot 5)^{64}\cdot 2^{(75-64)}=(2\cdot 5)^{64}\cdot 2^{11}=10^{64}\cdot 2^{11}</math>.
+This is <math>2048</math> with sixty-four, <math>0</math>'s on the end. So, the sum of the digits of <math>N</math> is <math>2+4+8=14\Longrightarrow\mathrm{ (B) \ }</math>