Find the least number which when divide by 2, 3, 4, 5 and 6 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 7 leaves no remainder?
Correct answer: B. 119
- A. 210
- B. 119
- C. 126
- D. 154
Explanation
Again, every remainder is one less than its divisor, so n + 1 is divisible by 2, 3, 4, 5 and 6. Their least common multiple is 60, so n = 60k - 1; requiring divisibility by 7 gives k = 2 and n = 120 - 1 = 119. The likely mistake is option a, which is divisible by 7 but leaves zero rather than the stated remainders when divided by 2 through 6.
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