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

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== Problem ==
 
== Problem ==
  
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The number <math>5^{64}\cdot 64^{25}</math> is the square of a positive integer <math>N</math>. In decimal representation, the sum of the digits of <math>N</math> is
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The number <math>25^{64}\cdot 64^{25}</math> is the square of a positive integer <math>N</math>. In decimal representation, the sum of the digits of <math>N</math> is
  
 
<math> \mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 35 </math>
 
<math> \mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 35 </math>
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Now, we have <math>N=5^{64}\cdot 8^{25}=5^{64}\cdot (2^{3})^{25}=5^{64}\cdot 2^{75}</math>.
 
Now, we have <math>N=5^{64}\cdot 8^{25}=5^{64}\cdot (2^{3})^{25}=5^{64}\cdot 2^{75}</math>.
  
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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>.  
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Combining 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 the number <math>2048</math> with a string of sixty-four <math>0</math>'s at the end. Thus, the sum of the digits of <math>N</math> is <math>2+4+8=14\Longrightarrow\boxed{\mathrm{ (B)}\ 14}</math>
 
This is the number <math>2048</math> with a string of sixty-four <math>0</math>'s at the end. Thus, the sum of the digits of <math>N</math> is <math>2+4+8=14\Longrightarrow\boxed{\mathrm{ (B)}\ 14}</math>
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[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
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https://www.youtube.com/watch?v=DbL6px67Yws
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video by canada math

Latest revision as of 07:08, 27 November 2020

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

See also

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

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



https://www.youtube.com/watch?v=DbL6px67Yws video by canada math

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