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 :
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.
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
EXAMIANS