Maths Olympiad Prep

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Number theory Difficulty 5.3 AIME, harder Prove it Romania

Find the natural numbers m,nm, n so that
n(n+1)=3m+s(n)+1182, n \cdot (n + 1) = 3^m + s(n) + 1182,
where s(n)s(n) represents the sum of the digits of the positive integer nn.

Solution

n2=3m+1182(ns(n)).(1) n^2 = 3^m + 1182 - (n - s(n)). \qquad (1)
If m2m \ge 2, because ns(n)n - s(n) is divisible by 9, but 311823 \mid 1182 and 911829 \nmid 1182, from (1) follows that 3n23 \mid n^2 and 9n29 \nmid n^2, false. Similarly, if m=1m = 1, we obtain that n2n^2 is divisible by 3 and not by 9, false. Therefore m=0m = 0, and n2+(ns(n))=1183n^2 + (n - s(n)) = 1183. (2)
If n9n \le 9, then s(n)=ns(n) = n, whence n2=1183n^2 = 1183, false. Consequently, we have n10n \ge 10. From (2) we deduce that n34n \le 34, therefore n=abn = \overline{ab}, with a{1,2,3}a \in \{1, 2, 3\}. Studying all the situations, we find that the only solution is m=0,n=34m = 0, n = 34.

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