Difference between revisions of "2006 AMC 10B Problems/Problem 11"
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== Problem == | == Problem == | ||
| + | What is the tens digit in the sum <math> 7!+8!+9!+...+2006!</math> | ||
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| + | <math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 9 </math> | ||
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== Solution == | == Solution == | ||
| + | Since <math>10!</math> is divisible by <math>100</math>. Any factorial greater than <math>10!</math> is also divisible by <math>100</math>. The last two digits 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> | ||
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| + | <math>7!+8!+9!=5040+40320+362880=408240</math> | ||
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| + | So the tens digit is <math>4 \Rightarrow C</math> | ||
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== See Also == | == See Also == | ||
*[[2006 AMC 10B Problems]] | *[[2006 AMC 10B Problems]] | ||
Revision as of 21:17, 13 July 2006
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 
