Difference between revisions of "2006 AMC 10B Problems/Problem 11"

Revision as of 23:52, 13 July 2006

Problem

What is the tens digit in the sum $7!+8!+9!+...+2006!$

$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 9$

Solution

Since $10!$ is divisible by $100$ , any factorial greater than $10!$ is also divisible by $100$ . The last two digits of all factorials greater than $10!$ are $00$ , so the last two digits of $10!+11!+...+2006!$ is $00$ .

So all that is needed is the tens digit of the sum $7!+8!+9!$

$7!+8!+9!=5040+40320+362880=408240$

So the tens digit is $4 \Rightarrow C$

@@ Line 5: / Line 5: @@
 == 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>.
+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>.
 So all that is needed is the tens digit of the sum <math>7!+8!+9!</math>

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Difference between revisions of "2006 AMC 10B Problems/Problem 11"

Revision as of 23:52, 13 July 2006

Problem

Solution

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