Problems on H.C.F and L.C.M
Let the least number of six digits, which when divided by 4, 6, 10 and 15, leaves in each case the same remainder of 2, be N. The sum of the digits in N is:
To express any fraction in simple (small form), we have to find the HCF of numerator and denominator. 1095)1168(1 1095 73)1095(15 1095 ——— 0 HCF of 1095 and 1168 is 73Therefore (1168/73)/(1095/73) = 16/15
EXAMIANS