# 2015 AIME II Problems/Problem 1

## Problem

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 $1000$.

## Solution 1

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 $\boxed{131}$.

## Solution 2

Continuing from Solution 1, we have $N=\frac{39}{50}k$ and $N=\frac{29}{25}m$. It follows that $k=\frac{50}{39}N$ and $m=\frac{25}{29}N$. Both $m$ and $k$ have to be integers, so, in order for that to be true, $N$ has to cancel the denominators of both $\frac{50}{39}$ and $\frac{25}{29}$. In other words, $N$ is a multiple of both $29$ and $39$. That makes $N=\operatorname{lcm}(29,39)=29\cdot39=1131$. The answer is $\boxed{131}$.

## Solution 3

We can express $N$ as $0.78a$ and $1.22b$, where $a$ and $b$ are some positive integers. $N=0.78a=1.22b\implies100N=78a=122b.$ Let us try to find the smallest possible value of $100N$, first ignoring the integral constraint.

Obviously, we are just trying to find the LCM of $78$ and $116.$ They have no common factors but $2$, so we multiply $78$ and $116$ and divide by $2$ to get $4524.$ This is obviously not divisible by $100$, so this is not possible, as it would imply that $N=\dfrac{4524}{100},$ which is not an integer. This can be simplified to $\dfrac{1131}{25}$.

We know that any possible value of $N$ will be an integral multiple of this value; the smallest such $N$ is achieved by multiplying this value by $25.$ We arrive at $1131$, which is $\boxed{131}\mod1000.$

~Technodoggo

## Video Solution

~MathProblemSolvingSkills.com

## See also

 2015 AIME II (Problems • Answer Key • Resources) Preceded byFirst Problem Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.