Question

Find the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case.

Answer 1

If the number is divisible by $35,56$ and $91$ then it is the LCM of these numbers.

$\therefore$ Find the by prime factorisation

$35=5\times 7$ $56=2^3\times 7$ $91=7\times 13$

$\therefore$ $LCMof$ $35,56,91=2^3\times 5\times 7\times 13=3,640$

$\therefore$ The least number divisible by $35,56$ and $91$ is $3,640.$

Since it leaves a remainder $7$, the required number is $3,640+7=3,647$