Difference between revisions of "Mock AIME 5 Pre 2005 Problems/Problem 12"
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== Solution == | == Solution == | ||
| − | By the [[binomial theorem]], < | + | By the [[binomial theorem]], <cmath>\begin{aligned} 101^4 + 256 &= (100 + 1)^4 + 256 \\ &= (100^4 + 4\cdot 100^3 + 6 \cdot 100^2 + 4 \cdot 100 + 1) + 256 \\ &= 104060401 + 256 = 104060657, \end{aligned}</cmath> and so the sum of the digits is <math>1+4+6+6+5+7 = \boxed{29}.</math> |
== See also == | == See also == | ||
Revision as of 15:55, 18 August 2013
Problem
Let 
. Find the sum of the digits of 
.
Solution
By the binomial theorem,
\begin{aligned} 101^4 + 256 &= (100 + 1)^4 + 256 \\ &= (100^4 + 4\cdot 100^3 + 6 \cdot 100^2 + 4 \cdot 100 + 1) + 256 \\ &= 104060401 + 256 = 104060657, \end{aligned} (Error compiling LaTeX. Unknown error_msg) and so the sum of the digits is 
See also
| Mock AIME 5 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
