Find the least number which is a multiple of | vedclass

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(B) Let the required number be $N$.Given that $N$ is a multiple of $31$,so $N = 31k$.Also,$N$ leaves remainders $2, 11, 19$ when divided by $15, 24, 3

Vedclass · Accepted Answer

$2387$

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(B) Let the required number be $N$.Given that $N$ is a multiple of $31$,so $N = 31k$.Also,$N$ leaves remainders $2, 11, 19$ when divided by $15, 24, 32$ respectively.Note that $15 - 2 = 13$,$24 - 11 = 13$,and $32 - 19 = 13$.Thus,$N + 13$ is exactly divisible by $15, 24,$ and $32$.$L.C.M.$ of $15, 24, 32$ is $480$.So,$N + 13 = 480m$,which implies $N = 480m - 13$.Since $N$ is a multiple of $31$,$480m - 13 \equiv 0 \pmod{31}$.$480 = 31 	imes 15 + 15$,so $15m - 13 \equiv 0 \pmod{31}$.For $m = 5$,$15(5) - 13 = 75 - 13 = 62$,which is divisible by $31$.Therefore,$N = 480(5) - 13 = 2400 - 13 = 2387$.

Find the least number which is a multiple of $31$ and when divided by $15, 24$ and $32$ leaves the remainders $2, 11$ and $19$,respectively.

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