Problems on H.C.F and L.C.M 252 can be expressed as a product of primes as: 2 x 2 x 2 x 3 x 7 2 x 3 x 3 x 3 x 7 2 x 2 x 3 x 3 x 7 3 x 3 x 3 x 3 x 7 2 x 2 x 2 x 3 x 7 2 x 3 x 3 x 3 x 7 2 x 2 x 3 x 3 x 7 3 x 3 x 3 x 3 x 7 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Clearly, 252 = 2 x 2 x 3 x 3 x 7.
Problems on H.C.F and L.C.M In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are? 820,369 800,500 700,400 820,360 820,369 800,500 700,400 820,360 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: 3 5 4 6 3 5 4 6 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the least number which when divided by 6,7,8,9, and 12 leave the same remainder 1 each case 505 606 707 404 505 606 707 404 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 3 | 6 - 7 - 8 - 9 - 12----------------------------- 4 | 2 - 7 - 8 - 3 - 4 --------------------------- 2 | 2 - 7 - 2 - 3 - 1 -------------------------- | 1 - 7 - 1 - 3 - 1 L.C.M = 3 X 2 X 2 X 7 X 2 X 3 = 504.Hence required number = (504 +1) = 505
Problems on H.C.F and L.C.M What will be the LCM of 8, 24, 36 and 54 345 112 216 444 345 112 216 444 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP LCM of 8-24-36-54 will be2*2*2*3*3*3 = 216
Problems on H.C.F and L.C.M Let the leas 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 3 4 5 6 3 4 5 ANSWER DOWNLOAD EXAMIANS APP
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