Question

Find the smallest number which when divided by 32,48 and 60 leaves 5 as the remainder in each case

Answer 1

Answer:

To find the smallest number, we first find LCM ( least common multiple) of 24,30,48 & 60 and then add remainder 2 in the LCM. Hence, the smallest number which when divided by 24 ,30 ,48 and 60 leaves a remainder 2 in each case is 242.

PLS MARK ME AS BRAINLIEST