Question

find the smallest number which when divides by 25 ,40,60 leave remainder 7in each case

Answer 1

Hi!
Here is the answer to your question.

The given numbers are 25, 40 and 60.
The smallest number which when divided by 25, 40 and 60 leaves remainder 7 is obtained by adding 7 to the LCM of 25, 40 and 60.
Prime factorization of 25 = 5 × 5
Prime factorization of 40 = 2 × 2 × 2 × 5
Prime factorization of 60 = 2 × 2 × 3 × 5
LCM of 25, 40 and 60 = 2 × 2 × 2 × 3 × 5 × 5 = 600
∴Smallest number which when divided by 25, 40 and 60 leaves remainder 7 = 600 + 7 = 607

HOPE IT HELPS..! ^o^