Question
If 'x' is the smallest positive integer divisible by 9,
15, and 28, then determine the second smallest positive integer that is divisible by all three of these numbers.Solution
Lowest positive integer divisible by 9, 15 and 28 = LCM (9, 15, 28) So, LCM of (9, 15, and 28) = 22Â X 32Â X 51Â X 71Â = 1260 Therefore, the next larger number which is divisible by all the three given numbers = 2 X 1260 = 2520
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