Problems on H.C.F and L.C.M
Find the least number which when divided by 16, 18, 20 and 2 leaves 4 as remainder in each case , but when divided by 7 leaves no. remainder
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.
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.
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