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Romanian Mathematical Olympiad

Algebra Difficulty 5.4 AIME, harder Prove it Romania

Find all the positive integers $p$ with the property that the sum of the first $p$ positive integers is a four-digit positive integer whose decomposition into prime factors is of the form $2^m3^n(m+n)$ , where $m, n \in \mathbb{N}^*$ .

Solution

The number $m+n$ is prime and, obviously, $m+n \ge 5$ . If $m+n=5$ , the largest value for $N = 2^m3^n(m+n)$ is $2^1 \cdot 3^4 \cdot 5 = 810$ , which has only three digits.
Suppose now that $m+n \ge 11$ . Then $N \ge 2^{10} \cdot 3 \cdot 11 > 10000$ , hence $N$ cannot have four digits.
So, $m+n=7$ . In this case, the four-digit numbers are: $2^6 \cdot 3^1 \cdot 7 = 1344$ , $2^5 \cdot 3^2 \cdot 7 = 2016$ , $2^4 \cdot 3^3 \cdot 7 = 3024$ , $2^3 \cdot 3^4 \cdot 7 = 4536$ and $2^2 \cdot 3^5 \cdot 7 = 6804$ .
The equality $N = \frac{p(p+1)}{2}$ can be fulfilled only for $N = 2016$ , when $p = 63$ .

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