Problem:
Prove that, however one chooses 18 consecutive positive integers less than or equal to 2005, there is at least one of them divisible by the sum of its digits.
Solution
Solution:
Among the 18 numbers under consideration there are two consecutive multiples of 9. By the divisibility criterion for 9, the sum of the digits of these numbers is a multiple of 9. Considering that the maximum digit sum of a number less than 2005 is 28 (in the case of 1999), let us analyze two cases.
If the digit sum of one of these numbers is 27, then this number must necessarily be 999, 1899, 1989 or 1998. The numbers 999 and 1998 are divisible by 27, while the multiples of 9 immediately preceding and immediately following 1899 and 1989, namely 1890, 1908, 1980 and 1998, are all divisible by the sum of their digits. If instead the digit sums of the two consecutive multiples of 9 are equal to 9 or to 18, then it suffices to note that one of them is even, and therefore divisible not only by 9 but also by 2, and hence by 18.
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