Question

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

Answer 1

Hi! Here is your answer.

We have to find the smallest number which when divided by 39, 52 and 65

leaves a remainder of 5 in each case. For this first, we have to compute the

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

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


Prime factorization of 39 = 3*13

Prime factorization of 52 = 2*2*13

Prime factorization of 65 = 5*13

L.C.M. = 2*2*3*5*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 in each case.

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Let us check our answer -

1) 785 ÷ 39

Quotient = 20

Remainder = 5

2) 785 ÷ 52

Quotient = 15

Remainder = 5

3) 785 ÷ 65

Quotient = 12

Remainder = 5

So, when 785 is divided by 39, 52 and 65, the remainder is 5 in each case.

Answer 2

Answer:  785

Step-by-step explanation:

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

leaves a remainder of 5 in each case we will first 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.