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

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 $1+4+6+6+5+7 = \boxed{29}.$

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 == Solution ==
-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><math> and so the sum of the digits is </math>1+4+6+6+5+7 = \boxed{29}.$
+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 ==

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

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

Latest revision as of 19:01, 23 March 2017

Problem

Solution

See also