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.
L.C.M. of 5,6,7,8 = 840. Required number is of the form 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2. Required number = (840 X 2 + 3)=1683
To solve this question quickly, first remove decimal by multiplying each term with 100,Then terms become 210, 105, 63Then HCF of above terms is 21, So Answer is 0.21
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