A Biggy

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

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

So we have now made $N/2$ into a perfect cube, but it also needs to be a perfect $5$th and $7$th power to satisfy the other conditions. Thus $(3x +1)$ 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 $(3x+1)=70$ . So $2^{70}$ is a factor of $N$ .

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

Similarly $(7z+1)$ must be a multiple of 3 and 5, and thus a multiple of 15. The smallest case of this is when $z=2$ and $(7z+1)=15$ . 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 $5$th power because $2^{70} \times 3^{20} \times 5^{15}$ has a multiple of $5$ in every power. $N/5$ is a perfect $7$th 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.