AIME 2015 Problem 1

Number Theory Level 5

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

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The answer is 131.

1 solution

Brandon Monsen
Dec 3, 2015

we know that $\frac{78}{100}k=\frac{116}{100}j$ for some positive integers $j$ and $k$ , or $\frac{j}{k}=\frac{39}{58}$ .

We can then deduce that $\frac{(39)(116)}{100}a$ is an integer for some integer $a$ .

The least integer $a$ that satisfies those conditions is $a=25$ , and so our smallest integer is $\frac{39 \times 116}{4}=1131$ , so the remainder when divided by 1000 is $\boxed{131}$

Did the same way. Why is this question 170 points??

A Former Brilliant Member - 5 years, 6 months ago

AIME 2015 Problem 1

Try more problems of AIME from this set AIME 2015

The answer is 131.

1 solution

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