Problems on H.C.F and L.C.M The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: 1045 777 1745 364 1045 777 1745 364 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 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.
Problems on H.C.F and L.C.M LCM of 87 and 145 is: 48 435 875 1305 48 435 875 1305 ANSWER DOWNLOAD EXAMIANS APP
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 : 6 5 3 4 6 5 3 4 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the largest number of four digits which is exactly divisible by 27,18,12,15 11720 10720 9720 8720 11720 10720 9720 8720 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP LCM of 27-18-12-15 is 540.After dividing 9999 by 540 we get 279 remainder.So answer will be 9999-279 = 9720
Problems on H.C.F and L.C.M Find the largest number of four digits exactly divisible by 12, 15, 18 and 27. 8760 9620 8720 9720 8760 9620 8720 9720 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Four bells begin to toll together respectively at the intervals of 8, 10, 12 and 16 seconds. After how many seconds will they toll together again? 242 seconds 246 seconds 243 seconds 240 seconds 242 seconds 246 seconds 243 seconds 240 seconds ANSWER DOWNLOAD EXAMIANS APP
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