Find the greatest number of four digits which when divided by 10, 15, 21 and 28 leaves 4, 9, 15 and 22 as remainders respectively?
Correct answer: A. 9654
- A. 9654
- B. 9666
- C. 9664
- D. 9864
Explanation
Every stated remainder is 6 less than its divisor, so n + 6 must be divisible by 10, 15, 21 and 28. Their least common multiple is 420, hence n = 420k - 6; the greatest four-digit value occurs at k = 23, giving n = 9660 - 6 = 9654. The likely mistake is option b, which is close but does not leave all four required remainders.
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