Problem 1251

A sequence of eighteen consecutive numbers always has two terms that are multiples of 9. The sum of the digits of a multiple of 9 is always a multiple of 9. Therefore, every sequence of eighteen consecutive numbers always has two terms that are divisible by 9 and whose sum of digits is also divisible by 9. Now, each of these two numbers has three digits, so the only multiples of 9 that can be the sum of the digits are 9, 18, and 27. However, 999 is the only three-digit number whose sum of digits is 27, and the only sequence of eighteen consecutive three-digit numbers that includes it is the sequence from 982 to 999, which does not include a three-digit number with a digit sum of 9 and only one with a digit sum of 27. Thus, the only possibilities for the sums of the digits of the two multiples of 9 in the sequence are:
(i) 9 and 9;
(ii) 9 and 18;
(iii) 18 and 18;
(iv) 18 and 27.

Let's look at some examples of each of these four cases.

(i) 9 and 9: one of the numbers is 144 and the other is $135 = 144 - 9$ or $153 = 144 + 9$. Two possible sequences are

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(ii) 9 and 18: one of the numbers is 900 and the other is 891 or 909. Two possible sequences are

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$\underbrace{}_{10 \varrho}, \underbrace{}_{11 \varrho}, \underbrace{909}_{12 \varrho}, \underbrace{}_{13 \varrho}, \underbrace{}_{14 \varrho}, \underbrace{}_{15 \varrho}, \underbrace{}_{160}, \underbrace{}_{17 \varrho}, \underbrace{915}_{189}$.

(iii) 18 and 18: one of the numbers is 828 and the other is 819 or 837. Two possible sequences are

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!

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(iv) 18 and 27: one of the numbers is 999 and we have a single option for the sequence, namely,

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Now, let's analyze each case. In cases (i) and (ii), one of the numbers is divisible by 9, which is the sum of its digits. In case (iv), one of the numbers is 999, which is divisible by 27. Finally, in case (iii), one of the two multiples of 9 is necessarily even, as they are two consecutive multiples of 9. Therefore, this number is a multiple of 2 and 9, so it is a multiple of the sum of its digits, which is 18.