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

Revision as of 15:46, 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 $2+4+8=14\Longrightarrow\mathrm{ (B) \ }$

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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 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>
+== See also ==
+{{AMC10 box|year=2002|ab=B|num-b=13|num-a=15}}
+[[Category:Introductory Number Theory Problems]]