Find the least number which when divided by 2 | vedclass

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Question

(B) Let the number be $x$. The conditions are $x \equiv 1 \pmod 2$,$x \equiv 2 \pmod 3$,$x \equiv 3 \pmod 4$,$x \equiv 4 \pmod 5$,and $x \equiv 5 \pmo

Vedclass · Accepted Answer

$119$

Vedclass · Answer

(B) Let the number be $x$. The conditions are $x \equiv 1 \pmod 2$,$x \equiv 2 \pmod 3$,$x \equiv 3 \pmod 4$,$x \equiv 4 \pmod 5$,and $x \equiv 5 \pmod 6$.This implies $x+1$ is divisible by $2, 3, 4, 5$ and $6$.The $L.C.M.$ of $2, 3, 4, 5, 6$ is $60$.So,$x+1 = 60k$,which means $x = 60k - 1$.For $k=1$,$x = 59$. For $k=2$,$x = 119$. For $k=3$,$x = 179$.We are given that $x$ is divisible by $7$.Checking the values: $59/7$ (remainder $3$),$119/7 = 17$ (remainder $0$).Thus,the least number is $119$.

Find the least number which when divided by $2, 3, 4, 5$ and $6$ leaves $1, 2, 3, 4$ and $5$ as remainders,respectively,but when divided by $7$ leaves no remainder.

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