Thank you to Julia from Langley Park School for Girls, Bromley for this solution. Well done Julia.

For $N / 2$ to be a perfect cube, $N$ must have $2^{3 x + 1}$ as a factor so that the 2 in the denominator will cancel, leaving only a perfect cube in the numerator.

For $N / 3$ to be a perfect 5th power, $N$ must have $3^{5 y + 1}$ as a factor for a similar reason to the one given above for $N / 2$ .

Similarly, $N / 5$ must have $5^{7 z + 1}$ as a factor for it to be a perfect 7th power.

So we have now made $N / 2$ into a perfect cube, but it also needs to be a perfect 5th and 7th power to satisfy the other conditions. Thus $(3 x + 1)$ must be a multiple of 5 and 7 and therefore a multiple of 35 (since 5 and 7 are relatively prime). The smallest case of this is when $x = 23$ and $(3 x + 1) = 70$ . So $2^{70}$ is a factor of $N$ .

Continuing the same method and reasoning, $(5 y + 1)$ must be a multiple of 3 and 7, and thus a multiple of 21. The smallest case of this is when $y = 4$ and $(5 y + 1) = 21$ . So $3^{21}$ is another factor of $N$ .

Similarly $(7 z + 1)$ must be a multiple of 3 and 5, and thus a multiple of 15. The smallest case of this is when $z = 2$ and $(7 z + 1) = 15$ . So $5^{15}$ is another factor of $N$ .

Combining all the factors we conclude that: $N = 2^{70} \times 3^{21} \times 5^{15}$ .

$N / 2$ is a perfect cube because $2^{69} \times 3^{21} \times 5^{15}$ has a multiple of 3 in every power. $N / 3$ is a perfect 5th power because $2^{70} \times 3^{20} \times 5^{15}$ has a multiple of 5 in every power. $N / 5$ is a perfect 7th power because $2^{70} \times 3^{21} \times 5^{14}$ has a multiple of 7 in every power.

Pierce Geoghegan, Tarbert Comprehensive , Ireland and Ang Zhi Ping, River Valley High, Singapore also sent excellent solutions for this problem.