Let S be a set of 9 distinct integers all of whose prime factors are at most 3.
Prove that S contains 3 distinct integers such that their product is a perfect cube.
Using pigeonhole might help( ).
APMO-2007(1)
- Sazid Akhter Turzo
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Re: APMO-2007(1)
There are quite similar problems in BdMO 2010 H.Sec 10, IMO (maybe) 1984 - 4.
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Nur Muhammad Shafiullah | Mahi
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Nur Muhammad Shafiullah | Mahi
- nafistiham
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Re: APMO-2007(1)
there are two prime factors here $2,3$
so the numbers are such $2^{a_k} \cdot 3^{b_k}$
where $a,b\geq 0$ and $1 \leq k \geq 9$
these powers can be in $3$ forms like $3p,3p+1,3p-1$
as there is $a,b$ there can be $9$ cases
from arguments of PHP we can say there will be at least $3$ numbers so that there product comes in the form
$2^{3r}\cdot 3^{3q}$ or $(2^r\cdot 3^q)^3$ which is a cube
so the numbers are such $2^{a_k} \cdot 3^{b_k}$
where $a,b\geq 0$ and $1 \leq k \geq 9$
these powers can be in $3$ forms like $3p,3p+1,3p-1$
as there is $a,b$ there can be $9$ cases
from arguments of PHP we can say there will be at least $3$ numbers so that there product comes in the form
$2^{3r}\cdot 3^{3q}$ or $(2^r\cdot 3^q)^3$ which is a cube
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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