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amilapsn
 one year ago
Hey guy this is a fun question:
Given any nine integers whose prime factors lie in the set {3,7,11}, prove that there must be two whose product is a square....
amilapsn
 one year ago
Hey guy this is a fun question: Given any nine integers whose prime factors lie in the set {3,7,11}, prove that there must be two whose product is a square....

This Question is Closed

rational
 one year ago
Best ResponseYou've already chosen the best response.0pigeonhole principle ?

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0you're in the right path....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh so it's 3,7 and 11 without their respective powers ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Damn that makes things a lot easier.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The box of Dirichlet.

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0hey @rational not only 1 and 0 there can be many

rational
 one year ago
Best ResponseYou've already chosen the best response.0i read the question wrong haha, let me erase

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's a good answer though. This could pose for a nice 5'th grade question or something.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We study the powers and the cases for odd and even. Any such number will have the form of 3^a * 7*b * 11^c. The cases for (a,b,c) are : o o o o o e o e o o e e e e e e e o e o e e o o Where o=odd and e=even. Any combination of two numbers that set will of yield an odd number at least at one of the powers of either 3,7 or 11. However, when we add in a ninth number, regardless of the (odd,even) combination at the powers of 3,7 and 11  this combination will be found in that list of 8 numbers. Which means that when we multiply those two numbers whose (odd,even) combination at the powers is the same the resulting combination at the powers will be (e e e) and thus a perfect square.
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