2015 AIME II Problems/Problem 1


Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$.


If $N$ is $22$ percent less than one integer $k$, then $N=\frac{78}{100}k=\frac{39}{50}k$. In addition, $N$ is $16$ percent greater than another integer $m$, so $N=\frac{116}{100}m=\frac{29}{25}m$. Therefore, $k$ is divisible by 50 and $m$ is divisible by 25. Setting these two equal, we have $\frac{39}{50}k=\frac{29}{25}m$. Multiplying by $50$ on both sides, we get $39k=58m$.

The smallest integers $k$ and $m$ that satisfy this are $k=1450$ and $m=975$, so $N=1131$. The answer is $\boxed{131}$.

See also

2015 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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