Question 26
Find the least number required to be added to 3105 so that it is exactly divisible by 3, 4, 5 and 6.
Solution
$$LCM of 3, 4, 5 and 6 = 2\times 3\times 2\times 5\times 1\times 1 = 60$$
On dividing 60 with 3105, we get remainder 45
∴ Number to be added = 60 - 45 = 15
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