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 :
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
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
Required length = H.C.F. of 495 cm, 900 cm and 1665 cm. 495 = 3² x 5 x 11, 900 = 2² x 3² x 5², 1665 = 3² x 5 x 37. H.C.F. = 32 x 5 = 45. Hence, required length = 45 cm.
EXAMIANS