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

(Solution)
(Solution)
Line 12: Line 12:
  
 
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>
 
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]]

Revision as of 15:46, 28 December 2011

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

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

See also

2002 AMC 10B (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS