Competition problem — Number theory

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Number theory Difficulty 4.7 AIME Prove it Romania

Call a positive integer balanced if the number of its distinct prime factors is equal to the number of its digits in the decimal representation; for example, the number $385 = 5 \cdot 7 \cdot 11$ is balanced, while $275 = 5^2 \cdot 11$ is not. Prove that there exist only a finite number of balanced numbers.

Solution

Let $p_1 = 2$ , $p_2 = 3$ , $p_3 = 5$ , $\ldots$ be the sequence of primes. Any balanced number $a$ with $n$ digits satisfies $a \ge p_1 p_2 \cdots p_n$ . Since $p_1 p_2 \cdots p_{11} = 2 \cdot 3 \cdot 5 \cdots 29 \cdot 31 > 10^{11}$ and $p_k > 10$ , for any $k > 11$ , it follows that there are no balanced numbers having more than 10 digits.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.