Difference between revisions of "2006 AIME I Problems/Problem 3"

Revision as of 19:05, 4 July 2013

Problem

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ 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 $b=25$ , so the number is $\boxed{725}$ .

It is quite obvious that $n=2$ , since the desired number can't be single or double digit, and cannot exceed $999$ . From $100a+b=29b$ , proceed as above.

Revision as of 16:52, 8 March 2012 (view source) Hli (talk \| contribs) (→‎Solution: Added "boxed" so that users can more easily see the answer without reading through the whole solution) ← Older edit		Revision as of 19:05, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	[[Category:Intermediate Number Theory Problems]]		[[Category:Intermediate Number Theory Problems]]
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Difference between revisions of "2006 AIME I Problems/Problem 3"

Revision as of 19:05, 4 July 2013

Problem

Solution

See also