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

## 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 $4 \Rightarrow C$

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

## See Also

 2006 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
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