Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (wikified and fixed grammer)
m (added link to previous and next problem)
Line 15: Line 15:
 
== See Also ==
 
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]
 +
 +
*[[2006 AMC 10B Problems/Problem 10|Previous Problem]]
 +
 +
*[[2006 AMC 10B Problems/Problem 12|Next Problem]]

Revision as of 13:59, 2 August 2006

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_11&oldid=9072"