Difference between revisions of "2006 AMC 10B Problems/Problem 11"
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Since <math>10!</math> is [[divisibility | divisible]] by <math>100</math>, any [[factorial]] greater than <math>10!</math> is also divisible by <math>100</math>. The last two [[digit]]s of all factorials greater than <math>10!</math> are <math>00</math>, so the last two digits of <math>10!+11!+...+2006!</math> is <math>00</math>. | Since <math>10!</math> is [[divisibility | divisible]] by <math>100</math>, any [[factorial]] greater than <math>10!</math> is also divisible by <math>100</math>. The last two [[digit]]s of all factorials greater than <math>10!</math> are <math>00</math>, so the last two digits of <math>10!+11!+...+2006!</math> is <math>00</math>. | ||
(*) | (*) | ||
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So all that is needed is the tens digit of the sum <math>7!+8!+9!</math> | So all that is needed is the tens digit of the sum <math>7!+8!+9!</math> | ||
Revision as of 17:07, 22 February 2010
Problem
What is the tens digit in the sum 
Solution
Since 
is divisible by 
, any factorial greater than is also divisible by . The last two digits of all factorials greater than are 
, so the last two digits of 
is .
(*)
So all that is needed is the tens digit of the sum 
So the tens digit is 
(*) A slightly faster method would have to take the 
residue of 
Since 
we can rewrite the sum as ![\[5040 + 8\cdot 5040 + 72\cdot 5040 \equiv 40 + 8\cdot 40 + 72\cdot 40 = 30 + 320 + 2880 \equiv 40 \pmod{100}.\]](http://latex.artofproblemsolving.com/0/f/f/0ff989ae029571bf2d68af232ff72499cbf712f2.png)
Since the last two digits of the sum is 
, the tens digit is 
