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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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