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

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== Solution == | == Solution == | ||

− | 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> | ||

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So the tens digit is <math>4 \Rightarrow C</math> | So the tens digit is <math>4 \Rightarrow C</math> | ||

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+ | (*) A slightly faster method would have to take the <math>\pmod {100}</math> residue of <math>7! + 8! + 9!.</math> Since <math>7! = 5040,</math> we can rewrite the sum as <cmath> 5040 + 8\cdot 5040 + 72\cdot 5040 \equiv 40 + 8\cdot 40 + 72\cdot 40 = 30 + 320 + 2880 \equiv 40 \pmod{100}. </cmath> Since the last two digits of the sum is <math>40</math>, the tens digit is <math>4.</math> | ||

== See Also == | == See Also == |

## 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