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 $10!+11!+...+2006!$ are $00$. (*)

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 $\fbox{4}.$

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 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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