Difference between revisions of "Mock AIME 5 Pre 2005 Problems/Problem 12"
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== Problem == | == Problem == | ||
| − | Let <math>m = 101^4 + 256</math>. Find the sum of digits of <math>m</math>. | + | Let <math>m = 101^4 + 256</math>. Find the sum of the digits of <math>m</math>. |
== Solution == | == Solution == | ||
| − | By the [[binomial theorem]], <math>101^4 + 256 = 104060401 + 256 = 104060657</math> | + | 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 == | ||
| − | {{Mock AIME box|year=Pre 2005|n= | + | {{Mock AIME box|year=Pre 2005|n=5|num-b=11|num-a=13|source=28368}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
. Find the sum of the digits of 
.
and so the sum of the digits is 
