Problems on H.C.F and L.C.M
Let the leas 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
Smallest number of five digits is 10000. Required number must be divisible by L.C.M. of 16,24,36,54 i.e 432, On dividing 10000 by 432,we get 64 as remainder. Required number = 10000 +( 432 – 64 ) = 10368.
EXAMIANS