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 5th 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 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 (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 2⁷⁰ 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 3²¹ 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 5¹⁵ is another factor of N .

Combining all the factors we conclude that: N=2⁷⁰ × 3²¹×5¹⁵ .

N/2 is a perfect cube because 2⁶⁹ ×3²¹ × 5¹⁵ has a multiple of 3 in every power. N/3 is a perfect 5th power because 2⁷⁰ ×3²⁰ ×5¹⁵ has a multiple of 5 in every power. N/5 is a perfect 7th power because 2⁷⁰ ×3²¹ ×5¹⁴ 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.