Difference between revisions of "2002 AMC 10B Problems/Problem 14"
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== Solution == | == Solution == | ||
| − | + | Taking the root, we get <math>N=\sqrt{25^{64}\cdot 64^{25}}=5^{64}\cdot 8^{25}</math>. | |
| − | + | Now, we have <math>N=5^{64}\cdot 8^{25}=5^{64}\cdot (2^{3})^{25}=5^{64}\cdot 2^{75}</math>. | |
| − | This is <math>2048</math> with sixty-four | + | 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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| + | 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> | ||
== See also == | == See also == | ||
Revision as of 20:15, 25 December 2016
Problem
The number 
is the square of a positive integer 
. In decimal representation, the sum of the digits of is
Solution
Taking the root, we get 
.
Now, we have 
.
Combing the 
's and 
's gives us 
.
This is the number 
with a string of sixty-four 
's at the end. Thus, the sum of the digits of is 
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. 
