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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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| − | The number can be represented as <math>10^na+b</math>, where <math> a </math> is the leftmost digit, and <math> b </math> is the rest of the number. We know that <math>b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na</math>. Thus <math> a </math> has to be 7 since <math> 10^n </math> can not have 7 as a factor, and the smallest <math> 10^n </math> can be and have a factor of <math> 2^2 </math> is <math> 10^2=100. </math> We find that <math> b </math> is 25, so the number is 725.
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| − | == See also ==
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| − | {{AIME box|year=2006|n=II|num-b=2|num-a=4}}
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| − | [[Category:Intermediate Number Theory Problems]] | |
