Problems on H.C.F and L.C.M
Find the least number which when divided by 16, 18, 20 and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
Let the numbers be 37a and 37b.Then, 37a x 37b = 4107 ab = 3.Now, co-primes with product 3 are (1, 3).So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.
The largest 4 digit number is 9999.We know that , if LCM of given numbers divide a number N , then N is exactly divisible by all the given numbers .LCM of 12,15,18,27 is 540.On dividing 9999 by 540 we get 279 as remainder.Therefore number =9999 - 279 =9720.
EXAMIANS