Find the smallest number which when reduced by 3 is divisible by 24, 36 and 40.

Solution:

The prime factorization of 14,28,36,45 is:

Factors of $14 = 2 \times 7$

Factors of $28 = 2 \times 2 \times 7$

Factors of $36 = 2 \times 2 \times 3 \times 3$

Factors of $45 = 3 \times 3 \times 5$.

LCM(14,28,36,45) = $2 \times 2 \times 7 \times 3 \times 3 \times 5$

= > 1260.

Therefore, the smallest number = 1260 + 3 =1263.

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Updated on: 10-Oct-2022

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