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Mathematics

Find the greatest 4-digit number which is exactly divisible by each of 8, 12 and 20.

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Answer

First, we find the LCM of 8, 12 and 20.

28, 12, 20
24, 6, 10
22, 3, 5
31, 3, 5
51, 1, 5
1, 1, 1

LCM of 8, 12 and 20 = 2 × 2 × 2 × 3 × 5 = 120.

The greatest 4-digit number is 9999.

We divide 9999 by 120 and find the remainder.

x283120)9999‾x()−960‾x2+(399x+−360‾x2+2(39\begin{array}{l} \phantom{x^2 }{\quad 83} \ 120\overline{\smash{\big)}9999} \ \phantom{x^(}\phantom{)}\underline{-960} \ \phantom{{x^2 } + (} 399 \ \phantom{{x} +}\underline{-360} \ \phantom{{x^2 } + 2(} 39 \ \end{array}

The remainder is 39.

The required greatest 4-digit number = 9999 − 39 = 9960.

Hence, the greatest 4-digit number divisible by 8, 12 and 20 is 9960.

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