Problems on H.C.F and L.C.M
The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 0, 62,98 and 130 as remainders respectively is .
Let the numbers be 37a and 37b.Then, 37a x 37b = 4107 ab = 3.Now, co-primes with product 3 are (1, 3).So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.
EXAMIANS