2006 AIME I Problems/Problem 3

Revision as of 05:43, 15 August 2011 by Neutrinonerd3333 (talk | contribs) (Solution)


Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.


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 $725$.

  • 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.

See also

2006 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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