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2006 AMC 10B Problems/Problem 11

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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.

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