A) 106
B) 301
C) 309
D) 400
Correct Answer: B
Solution :
Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4, 5 or 6 and no remainder when divided by 7. Thus, the number must be of the form (LCM of 3, 4, 5, 6) x+1 i.e., \[(60x+1)\] and a multiple of 7. Clearly, for x = 5, the number is a multiple of 7. So, the number is 301.You need to login to perform this action.
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