Question:
What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?
Solution:
TO FIND: Smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case
L.C.M OF 35, 56 and 91
$35=5 \times 7$
$56=2^{3} \times 7$
$91=13 \times 7$
L.C.M of 35,56 and $91=2^{3} \times 5 \times 7 \times 13$
$=3640$
Hence 84 is the least number which exactly divides 28, 42 and 84 i.e. we will get a remainder of 0 in this case. But we need the smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case
Therefore
$=3640+7$
$=3647$
Hence $=3647$ is smallest number that, when divided by 35,56 and 91 leaves remainder of 7 in each case.