Problems on H.C.F and L.C.M
The maximum number of student amoung them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is :
L.C.M. of 6, 9, 15 and 18 is 90.Let required number be 90k + 4, which is multiple of 7.Least value of k for which (90k + 4) is divisible by 7 is k = 4.Required number = (90 x 4) + 4 = 364.
EXAMIANS