What is the smallest number which is divisible by 35 56 and 91 and leaves remainder 7 in each case?

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= 5xx7`

`56=2^2xx7`

`91=13xx7`

L.C.M of 35,56 and 91 = `2^2xx5xx7xx13`

=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 

= 3640 

Hence 3640  is smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case.