Question

what is the smallest number which when divided by 39,52 and 91 leaving a remainder of 13 in each case

Answer 1

find the lcm of39,52,91
then add 13 to the lcm obtained. then the answer will be 1105

Answer 2

Answer:

1105 is the smallest no when divided by 39, 52, 91 and leaves 13 as remainder in each case.

Step-by-step explanation:

Given: Given nos. 39 , 52 , 91

To find: Smallest No. which divided by given no. leaves remainder 13

             in each case.

To find the Required No. we follow following Steps:

Step 1: First Find LCM of Given Numbers by any method

          I'm using Prime factorization method.

39 = 3 × 13

52 = 2 × 2 × 13

91 = 7 × 13

LCM ( 39, 52, 91 ) = 13 × 3 × 2 × 2 × 7

                             = 1092

So, 1092 is the smallest no which is divisible by 39, 52 , 91

Step 2: Now we add 13 in 1092 as it leaves 13 as remainder in each case.

          ⇒ Required No. = 1092 + 13

                                      =  1105

Therefore, 1105 is the smallest no when divided by 39, 52, 91 and leaves 13 as remainder in each case.