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30 June, 18:07

Answer only. What is the greatest possible sum of the digits of a six-digit number that is a multiple of 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11?

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  1. 30 June, 19:43
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    Answer: 36

    The LCM of 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11 is 2 * 3 * 2 * 5 * 7 * 2 * 3 * 11 = 27720.

    A multiple of 27720 will end in 0, and the sum of its digits will be a multiple of 9.

    The largest 6-digit number which is multiple of 9 with last digit of 0 is 999990, and the sum of its digits is 45. 45 is not a multiple of 27720, and there is no other way to get a digit sum of 45.

    We now look at the next biggest multiple of 9, 9 * 4

    = 36.

    The largest 6-digit multiple of 27720 is 27720 * 36 = 997920, and the sum of its digits is 36.
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