Difference between revisions of "2002 AMC 10B Problems/Problem 14"
Mariekitty (talk | contribs) (→Solution) |
|||
Line 7: | Line 7: | ||
== Solution == | == Solution == | ||
− | + | 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> |
Revision as of 14:33, 28 December 2011
Problem
The number is the square of a positive integer . In decimal representation, the sum of the digits of is
Solution
Since, .
Combing the 's and 's gives us, .
This is with sixty-four, 's on the end. So, the sum of the digits of is