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Mathematics

Find the least number of five digits which is exactly divisible by 32, 36 and 45.

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Answer

First, we find the LCM of 32, 36 and 45.

232, 36, 45
216, 18, 45
28, 9, 45
24, 9, 45
22, 9, 45
31, 9, 45
31, 3, 15
51, 1, 5
1, 1, 1

LCM of 32, 36 and 45 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440.

The smallest 5-digit number is 10000.

We divide 10000 by 1440 and find the remainder.

x21+61440)10000‾x+)))−8640‾x2+2a1360\begin{array}{l} \phantom{x^2 1+}{\quad 6} \ 1440\overline{\smash{\big)}10000} \ \phantom{x^ + )}\phantom{))}\underline{-8640} \ \phantom{{x^2 } + 2a} 1360 \ \end{array}

The remainder is 1360.

The required least 5-digit number = 10000 + (1440 − 1360) = 10000 + 80 = 10080.

Hence, the least 5-digit number divisible by 32, 36 and 45 is 10080.

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