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Difference between revisions of "2006 AIME I Problems/Problem 3"
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== See also ==
 
== See also ==
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* [[2006 AIME I Problems/Problem 2 | Previous problem]]
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* [[2006 AIME I Problems/Problem 4 | Next problem]]
 
* [[2006 AIME I Problems]]
 
* [[2006 AIME I Problems]]
 
* [[Number Theory]]
 
* [[Number Theory]]
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 19:44, 11 March 2007

Problem

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 of the original integer.

Solution

The number can be represented as $10^na+b$, where $a$ is the leftmost digit, and $b$ is the rest of the number. We know that $b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na$. Thus $a$ has to be 7 since $10^n$ can not have 7 as a factor, and the smallest $10^n$ can be and have a factor of $2^2$ is $10^2=100.$ We find that $b$ is 25, so the number is 725.

See also

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