Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

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Solution

First, we have to find the LCM of 8, 15, and 21.

Prime factorisation of 8 = 2 × 2 × 2
Prime factorisation of 15 = 3 × 5
Prime factorisation of 21 = 3 × 7
Therefore, required LCM = 2 × 2 × 2 × 3 × 5 × 7 = 840.

The number nearest to 1,00,000 and exactly divisible by each of 8, 15, and 21 should also be exactly divisible by their LCM (i.e. 840).

We have to divide 1,00,000 by 840.

Remainder = 40
∴ Number just greater than 1,00,000 and exactly divisible by 840 = 1,00,000 + (840 − 40)
= 1,00,000 + 800 = 1,00,800
∴ Required number = 1,00,800

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