Find the smallest number which when divided by 11, 12,
15, 20 gives a remainder of 3 every time.
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Step-by-step explanation:
Prime factorizations first.12 = 2 x 2 x 3
15 = 3 x 5
18 = 2 x 3 x 3
27 = 3 x 3 x 3
LCM = 2 x 2 x 3 x 3 x 3 x 5 = 540
Since all of the remainders are 4 less than the divisors, the number will be 540 - 4 = 536
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