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

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  • $\begingroup$ 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. $\endgroup$ Feb 6 '18 at 12:00
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    $\begingroup$ Welcome to Mathematics Stack Exchange! Could you show us what you have tried so far? $\endgroup$
    – Jan
    Feb 6 '18 at 12:02
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The Chinese Remainder Theorem gives a systematic way to solve such problems.

For this particular problem, there is a shortcut. We want

$ x \equiv \ 75 \bmod 90 \\ x \equiv \ 60 \bmod 75 \\ x \equiv 105 \bmod 120 $

or, equivalently,

$ 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 $lcm(90,75,120)=1800$.

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