Which are those natural numbers that, when divided by any smaller positive integer, result in a finite decimal (or whole number)?
Solution. It is known that the decimal representation of a fraction is finite if and only if, in its simplest form, the prime factorization of the denominator contains only the primes 2 and 5. If \( n \) is a suitable positive integer, then since \( n \) and \( n-1 \) are coprime, it follows that \( n-1 = 2^u \cdot 5^v \). Since the greatest common divisor of \( n \) and \( n-2 \) is 1 or 2, \( n-2 = 2^p \cdot 5^q \). However, \( n-1 \) and \( n-2 \) are also coprime, so (depending on the parity of \( n \)) either \( n-1 = 2^u \) and \( n-2 = 5^q \), or \( n-1 = 5^v \) and \( n-2 = 2^p \).

If \( n \leq 5 \), then from the above, \( n = 3 \) or \( n = 2 \) follows, and both are solutions to the problem. If \( n > 5 \), then \( n-1 \) and \( n-2 \) are proper powers of 5; thus, \( n-3 \) is certainly not divisible by 5. On the other hand, both \( n \) and \( n-3 \) are divisible by 3, so only \( n-3 = 3 \cdot 2^a \) is possible.

In the second case (if \( n \) is even), \( n-3 \) is odd, so \( 2^a = 1 \), i.e., \( n-3 = 3 \), thus \( n = 6 \), which is indeed a solution.

If \( n \) is odd, then from \( n-1 = 2^u \) and \( n-3 = 3 \cdot 2^a \),

\[
2(2^{u-1} - 1) = 2^u - 2 = 3 \cdot 2^a
\]

so \( a = 1 \), i.e., \( n = 9 \). However, in this case, the decimal representation of \( 9 / 7 \) is not finite. (This is actually the only exception.)

With this, we have solved the problem, and the sought numbers are: 2, 3, and 6.

Csikai Szabolcs (Kecskemét, Katona J. Gymnasium, 1st year)

Note. In the investigation of the numbers \( n \) (among the numbers smaller than \( n \)), it was sufficient to go as far as examining \( n-3 \). Only the \( n \) values thus identified could be solutions. For these, the numbers \( n-4, n-5, \ldots \) then only needed to be checked. This is how \( n = 9 \) was eliminated due to \( n-2 = 7 \).