Difference between revisions of "2006 AMC 10B Problems/Problem 11"
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== 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> | + | 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> | ||
− | (*) 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 = | + | (*) 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 == | ||
− | + | {{AMC10 box|year=2006|ab=B|num-b=10|num-a=12}} | |
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[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 06:12, 27 September 2020
Problem
What is the tens digit in the sum
Solution
Since is divisible by , any factorial greater than is also divisible by . The last two digits of all factorials greater than are , so the last two digits of are . (*)
So all that is needed is the tens digit of the sum
So the tens digit is
(*) A slightly faster method would have to take the residue of Since we can rewrite the sum as Since the last two digits of the sum is , the tens digit is
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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