Question

Find the smallest number which when divided by 39,52 and 65, leaves a
remainder of 5 in each case.

Answer 1

Answer:

785

Step-by-step explanation:

Solution :-

To find the smallest number which when divided by 39, 52 and 65

leaves a remainder of 5 in each case.

First let us find the L.C.M. of 39, 52 and 65.

Prime factorization of 39 = 3 x 13

Prime factorization of 52 = 2 x 2 x 13

Prime factorization of 65 = 5 x 13

L.C.M. = 2 x 2 x 3 x 5 x 13

L.C.M. of 39, 52 and 65 = 780

The required smallest number = 780 + 5 = 785

So, the smallest number is 785, which when divided by 39, 52 and 65 leaves a remainder of 5.

Answer 2

Step-by-step explanation:

To find the smallest no. which when divided by 39,52 and 65

leaves a remainder of 5 in each case .

First let us find the L.C.M. of 39 , 52 and 65 .

Prime factorization of 39 = 3 x 13

Prime factorization of 52 = 2 x 2 x 13

Prime factorization of 65 = 5 x 13

L.C.M. = 2 x 2 x 3 x 5 x 13

L.C.M. of 39 , 52 and 65 = 780

The required smallest number = 780 + 5 = 785