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 :
The greatest number that will divide 172,205 and 304 leaving the same remainder in each case is HCF ((205-172), (304-205), (304-172)) = HCF of 33, 99, and 132 33=31x11 99=3 x 3 x 11 132=3 x 2 x 2 x11 Required number =3 x 11=33