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Find the least number which when divided by 6, 7, 8, 9 and 10 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 19 leaves no remainder?

Correct answer: D. 5035

  • A. 5073
  • B. 5016
  • C. 5054
  • D. 5035

Explanation

The remainders are one less than the divisors, so n + 5 must be divisible by 6, 7, 8, 9 and 10. Their least common multiple is 2520, hence n = 2520k - 5; requiring divisibility by 19 gives k = 2 as the least positive solution, so n = 5040 - 5 = 5035. The likely mistake is option a, which has the required pattern for the first divisors but is not divisible by 19.

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