Problems on H.C.F and L.C.M
Find the least which when divided y 16,18,20, and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
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