Which is the smallest positive integer whose half is a square number, and whose third is a cube number?
The sought number's half and third are both integers, so the former is a square number divisible by 3, and the latter is a cube number divisible by 2. The half of the number is thus divisible by $3^2 = 9$. Therefore, the number itself is divisible by $2 \cdot 9 = 18$ and $3 \cdot 8 = 24$, or the least common multiple of 18 and 24, which is 72.

The half of 72 is 36, which is a square number, so the sought number must be a multiple of 72 by a square number. We need to find the smallest square number that, when multiplied by $72 \div 3 = 24$, results in a cube number. Since $24 = 2^3 \cdot 3$, any suitable multiplier must be divisible by $3^2 = 9$, and the smallest such multiplier is 9.

Thus, the sought number is $9 \cdot 72 = 648$. Its half is 324, which is the square of 18, and its third is 216, which is the cube of 6.

Based on the work of Klug Róbert (Budapest XI., Leiningen u. Ált. Isk., 8th grade)

Note. During the solution, we used several times the not entirely obvious fact that the prime factors in the square and cube of a number are the same as in the number itself, and in the former, each prime factor's exponent is even, while in the latter, each exponent is divisible by 3.

The above are consequences of the "fundamental theorem of arithmetic," which states that positive integers can be uniquely factored into prime numbers.