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 $\boxed{131}$ .

Video Solution

https://www.youtube.com/watch?v=9re2qLzOKWk&t=7s

~MathProblemSolvingSkills.com

2015 AIME II (Problems • Answer Key • Resources)
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2015 AIME II Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

Video Solution

See also