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

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== Solution ==
 
== Solution ==
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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>.
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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> are <math>00</math>.  
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(*)
  
 
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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So the tens digit is <math>4 \Rightarrow C</math>
 
So the tens digit is <math>4 \Rightarrow C</math>
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(*) A slightly faster method would have 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>\fbox{4}.</math>
  
 
== See Also ==
 
== See Also ==
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*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=10|num-a=12}}
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Revision as of 07:12, 27 September 2020

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 $10!+11!+...+2006!$ 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 $4 \Rightarrow C$

(*) 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 $\fbox{4}.$

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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