# Find the smallest number who gives remainders $75,60,105$ when divided by $90,75,120$

Which is the least number when divided by $90,75,120$ leaves remainder $75,60,105$ respectively? How should approach this question

• Use the Chinese Remainder Theorem and all possible modular information you can get by breaking these numbers into prime factors. E.g your number should be odd, $3$ mod $9$, $10$ mod $25$, etc. Feb 6 '18 at 12:00
• Welcome to Mathematics Stack Exchange! Could you show us what you have tried so far?
– Jan
Feb 6 '18 at 12:02

$x \equiv \ 75 \bmod 90 \\ x \equiv \ 60 \bmod 75 \\ x \equiv 105 \bmod 120$
$x+15 \equiv 0 \bmod 90 \\ x+15 \equiv 0 \bmod 75 \\ x+15 \equiv 0 \bmod 120$
Therefore, $x+15$ is a common multiple of $90,75,120$ and so is a multiple of $lcm(90,75,120)=1800$.