Mock AIME 5 Pre 2005 Problems/Problem 12

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}.$

Mock AIME 5 Pre 2005 (Problems, Source)
Preceded by Problem 11	Followed by Problem 13
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Mock AIME 5 Pre 2005 Problems/Problem 12

Problem

Solution

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