What is the smallest number that, when divided by 35,
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.

 

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