AoPSWiki
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
Personal tools

2006 AMC 10B Problems/Problem 11

From AoPSWiki

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

See Also

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us