Question
If 'x' is the lowest positive integer divisible by 10,
18 and 21, then find the second smallest positive integer which is divisible by all the three given numbers?Solution
Lowest positive integer divisible by 10, 18 and 21 = LCM (10, 18, 21) So, LCM of (10, 18, and 21) = 2¹ × 3² × 5¹ × 7¹ = 630 Therefore, the next larger number which is divisible by all the three given numbers = 2 × 630 = 1260
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