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Question Find the smallest number that is divisible by 3, 4, 5, and 7, but leaves a remainder of 10 when divided by 11.
LCM(3, 4, 5,7) = 3 × 4 × 5 × 7 = 420 The number must be a multiple of 420, so the number can be written as 420k, where k is a whole number. N = 420 × 1 = 420 When divided by 11 leaves a remainder of 2 11 × 38 + 2 = 420 But we require 10 as a remainder ∴ 2k = 10 ⇒ k = 5 Hence number = 420 × 5 = 2100