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What is the smallest number which when divided by 6, 7 and 8 leaves 2 as the remainder in each case?

A.	169	B.	165
C.	166	D.	170

Answer and Explanation

Answer:166

Explanation:
Step 1: A. Let the required number and the remainder in each case be
N. Using Rule 3, N + 2 is divisible by 6, 7 and 8.
B. N + 2 must be a multiple of 6, 7 and 8
C. The smallest possible value of N + 2 must be the LCM of (6, 7, 8). Any such value of N + 2 can be written as follows.
N + 2 = K × LCM(6, 7, 8) where K = 1, 2, 3
Step 2: × LCM(6, 7, 8) = K × 23 × 3 × 7 = 168K
Step 3: To find out the least possible value of N + 2, Substituting K = 1in (1), we get N + 2 = 168. N = 166

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