Difference between revisions of "Mock AIME 5 Pre 2005 Problems/Problem 12"

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== Solution ==
 
== Solution ==
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>
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By the [[binomial theorem]], <math>\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}</math> and so the sum of the digits is <math>1+4+6+6+5+7 = \boxed{29}.</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 19:01, 23 March 2017

Problem

Let $m = 101^4 + 256$. Find the sum of the digits of $m$.

Solution

By the binomial theorem, $\begin{aligned} 101^4 + 256 &= (100 + 1)^4 + 256 \\ &= (100^4 + 4\cdot 100^3 + 6 \cdot 100^2 + 4 \cdot 100 + 1) + 256 \\  &= 104060401 + 256 = 104060657, \end{aligned}$ and so the sum of the digits is $1+4+6+6+5+7 = \boxed{29}.$

See also

Mock AIME 5 Pre 2005 (Problems, Source)
Preceded by
Problem 11
Followed by
Problem 13
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