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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 1 ==
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| − | We will use the fact that for any integer <math>n</math>, <cmath>\begin{align*}(5n+1)(5n+2)(5n+3)(5n+4)&=[(5n+4)(5n+1)][(5n+2)(5n+3)]\\ &=(25n^2+25n+4)(25n^2+25n+6)\equiv 4\cdot 6\\ &=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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| − | We can write <math>N</math> as <math>\frac M{2^{21}}</math> where <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 = \frac{90!}{5^{21}},</cmath> where every number in the form <math>5n</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)\\ &\cdot (1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots (16\cdot 17\cdot 18) \\ &\cdot (1\cdot 2\cdot 3).\end{align*}</cmath>
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| − | Hence, 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) \\ &= 1\cdot -21\cdot 6\\ &= -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>,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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| − | == Solution 2(bash) ==
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| − | First, we list out all the numbers<br \>
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| − | <math>90!=2^{65}3^{44}5^{21}7^{13}11^813^617^519^423^329^331^237^241^243^2\cdot47\cdot53\cdot\cdot\cdot89</math><br \>
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| − | Since we must get rid of ending <math>0</math>s, we get rid of <math>5^{21}</math> and the corresponding <math>2^{21}</math><br \>
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| − | Next, we note that <math>2^22\equiv2^2</math>,<math>3^20\equiv3</math>, and <math>7^4\equiv7</math>, so it can be simplified to<br \>
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| − | <math>2^43^47\cdot11^813^617^5\cdot\cdot\cdot89</math><br \>
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| − | Then, we bash using these following <math>4</math> difference of squares tricks and multiplication<br \>
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| − | 1. <math>(10k-1)(10k+1)\equiv-1</math><br \>
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| − | 2. <math>(10k+3)(10k+7)\equiv21=3\cdot7</math><br \>
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| − | 3. <math>(10k+1)(10k+9)\equiv9=3^2</math><br \>
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| − | 4. <math>(10k-3)(10k+3)\equiv-9\equiv7\cdot13</math><br \>
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| − | Handling of powers of <math>3</math> and <math>7</math> are especially critical<br \>
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| − | In the end, we get <math>\boxed{(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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| − | {{MAA Notice}}
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