# 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 Since the last two digits of the sum is , the tens digit is