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| − | == Problem ==
| + | #REDIRECT [[2010_AMC_12A_Problems/Problem_23]] |
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| − | The number obtained from the last two nonzero digits of <math>90!</math> is equal to <math>n</math>. What is <math>n</math>?
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| − | <math>\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68</math>
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| − | == Solution ==
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| − | We will use the fact that for any integer <math>n</math>,
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| − | <cmath>\begin{align*}(5n+1)(5n+2)(5n+3)(5n+4)&=[(5n+4)(5n+1)][(5n+2)(5n+3)]\\
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| − | &=(25n^2+25n+4)(25n^2+25n+6)\equiv 4\cdot 6\\
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| − | &=24\pmod{25}\equiv -1\pmod{25}.\end{align*}</cmath>
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| − | First, we find that the number of factors of <math>10</math> in <math>90!</math> is equal to <math>\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21</math>. Let <math>N=\frac{90!}{10^{21}}</math>. The <math>n</math> we want is therefore the last two digits of <math>N</math>, or <math>N\pmod{100}</math>. Since there is clearly an excess of factors of 2, we know that <math>N\equiv 0\pmod 4</math>, so it remains to find <math>N\pmod{25}</math>.
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| − | If we divide <math>N</math> by <math>5^{21}</math> by taking out all the factors of <math>5</math> in <math>N</math>, we can write <math>N</math> as <math>\frac M{2^{21}}</math> where
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| − | <cmath>M=1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 6\cdot 7\cdot 8\cdot 9\cdot 2\cdots 89\cdot 18,</cmath>
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| − | where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form <math>5n</math> is replaced by <math>n</math>, and every number in the form <math>25n</math> is replaced by <math>n</math>.
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| − | The number <math>M</math> can be grouped as follows:
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| − | <cmath>\begin{align*}M= &(1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots(86\cdot 87\cdot 88\cdot 89)\\
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| − | &\cdot (1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots (16\cdot 17\cdot 18) \\
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| − | &\cdot (1\cdot 2\cdot 3).\end{align*}</cmath>
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| − | Using the identity at the beginning of the solution, we can reduce <math>M</math> to
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| − | <cmath>\begin{align*}M&\equiv(-1)^{18} \cdot (-1)^3(16\cdot 17\cdot 18) \cdot (1\cdot 2\cdot 3) \\
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| − | &= 1\cdot -21\cdot 6\\
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| − | &= -1\pmod{25} =24\pmod{25}.\end{align*}</cmath>
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| − | Using the fact that <math>2^{10}=1024\equiv -1\pmod{25}</math> (or simply the fact that <math>2^{21}=2097152</math> if you have your powers of 2 memorized), we can deduce that <math>2^{21}\equiv 2\pmod{25}</math>. Therefore <math>N=\frac M{2^{21}}\equiv \frac {24}2\pmod{25}=12\pmod{25}</math>.
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| − | Finally, combining with the fact that <math>N\equiv 0\pmod 4</math> yields <math>n=\boxed{\textbf{(A)}\ 12}</math>.
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| − | == See also ==
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| − | {{AMC10 box|year=2010|num-b=23|num-a=25|ab=A}}
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| − | [[Category:Intermediate Number Theory Problems]]
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