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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 3]] |
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| − | Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 of the original integer.
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| − | == Solution ==
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| − | Note that the product of the first <math>\displaystyle 100</math> positive odd integers can be written as <math>\displaystyle 1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}</math>
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| − | Hence, we seek the number of threes in <math>\displaystyle 200!</math> decreased by the number of threes in <math>\displaystyle 100!.</math>
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| − | There are
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| − | <math>\displaystyle \left\lfloor \frac{200}{3}\right\rfloor+\left\lfloor\frac{200}{9}\right\rfloor+\left\lfloor \frac{200}{27}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor =66+22+7+2=97</math>
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| − | threes in <math>\displaystyle 200!</math> and
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| − | <math>\displaystyle \left\lfloor \frac{100}{3}\right\rfloor+\left\lfloor\frac{100}{9}\right\rfloor+\left\lfloor \frac{100}{27}\right\rfloor+\left\lfloor\frac{100}{81}\right\rfloor=33+11+3+1=48 </math>
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| − | threes in <math>\displaystyle 100!</math>
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| − | Therefore, we have a total of <math>\displaystyle 97-48=049</math> threes.
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| − | For more information, see also [[Factorial#Prime factorization| prime factorizations of a factorial]].
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| − | == See also ==
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| − | *[[2006 AIME II Problems]]
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| − | [[Category:Intermediate Number Theory Problems]]
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