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2006 AMC 10B Problems/Problem 11: Difference between revisions

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What is the tens digit in the sum <math> 7!+8!+9!+...+2006!</math>
What is the tens digit in the sum <math> 7!+8!+9!+...+2006!</math>


<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 9 </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 [[divisibility | 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> is <math>00</math>.
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>
So all that is needed is the tens digit of the sum <math>7!+8!+9!</math>
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<math>7!+8!+9!=5040+40320+362880=408240</math>
<math>7!+8!+9!=5040+40320+362880=408240</math>


So the tens digit is <math>4 \Rightarrow C</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

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