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"

 
(→‎Solution)
 
(15 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
What is the tens digit in the sum <math> 7!+8!+9!+...+2006!</math>
 +
 +
<math> \textbf{(A) } 1\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 6\qquad \textbf{(E) } 9 </math>
 +
 
== Solution ==
 
== Solution ==
 +
Since <math>10!</math> is divisible by <math>100</math>, any [[factorial]] greater than <math>10!</math> is also divisible by <math>100</math>. The last two [[digit]]s of all factorials greater than <math>10!</math> are <math>00</math>, so the last two digits of <math>10!+11!+...+2006!</math> are <math>00</math>.
 +
(*)
 +
 +
So all that is needed is the tens digit of the sum <math>7!+8!+9!</math>
 +
 +
<math>7!+8!+9!=5040+40320+362880=408240</math>
 +
 +
So the tens digit is <math>\boxed{\textbf{(C) }4}</math>.
 +
 +
(*) A slightly faster method would be to take the <math>\pmod {100}</math> residue of <math>7! + 8! + 9!.</math> Since <math>7! = 5040,</math> we can rewrite the sum as <cmath> 5040 + 8\cdot 5040 + 72\cdot 5040 \equiv 40 + 8\cdot 40 + 72\cdot 40 = 40 + 320 + 2880 \equiv 40 \pmod{100}. </cmath> Since the last two digits of the sum is <math>40</math>, the tens digit is <math>\boxed{\textbf{(C) }4}</math>.
 +
 
== See Also ==
 
== See Also ==
−
*[[2006 AMC 10B Problems]]
+
{{AMC10 box|year=2006|ab=B|num-b=10|num-a=12}}
 +
 
 +
[[Category:Introductory Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 13:00, 26 January 2022

Problem

What is the tens digit in the sum $7!+8!+9!+...+2006!$

$\textbf{(A) } 1\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 6\qquad \textbf{(E) } 9$

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 $\boxed{\textbf{(C) }4}$.

(*) A slightly faster method would be 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 $\boxed{\textbf{(C) }4}$.

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_11&oldid=170279"