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# 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.

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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.

a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.