Problems on H.C.F and L.C.M
The maximum number of students among 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:
The least number which when divided by 4,5,6,8 and 10 is the LCM of these numbers. But each time to get 3 as remainder, we have to add the remainder 3 to the obtained LCM.LCM of (4, 5, 6, 8, 10)+ 3=120+3=123
LCM of 2, 3, 4 and 5 is 60.The smallest number of two digits is 10.But we also see that the smallest number divisible by 2, 3, 4 and 5 is 60. Therefore, 10 has no significance here.The difference between divisor and remainder is 2 in each case. So, we subtract 2 from 60.60 - 2 = 58
EXAMIANS