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ritvik1512
 2 years ago
Peter has recently been fascinated by numbers that can be expressed in the form f(a,b,c)=a(b−c)^3+b(c−a)^3+c(a−b)^3 for some integers a,b and c. He thinks that we should use this system to do arithmetic in future. How many positive integers N≤1000 can be expressed in the form of f(a,b,c)?
ritvik1512
 2 years ago
Peter has recently been fascinated by numbers that can be expressed in the form f(a,b,c)=a(b−c)^3+b(c−a)^3+c(a−b)^3 for some integers a,b and c. He thinks that we should use this system to do arithmetic in future. How many positive integers N≤1000 can be expressed in the form of f(a,b,c)?

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goformit100
 2 years ago
Best ResponseYou've already chosen the best response.0f(a,b,c)=a(b−c)^3+b(c−a)^3+c(a−b)^3

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.1Oh! Peter should try it first.

PeterPan
 2 years ago
Best ResponseYou've already chosen the best response.1Unless it's Spiderman or Peter Griffin :D

goformit100
 2 years ago
Best ResponseYou've already chosen the best response.0Oh! Peter should definitely try it first.

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.1@ritvik1512 I meant to say that, can you show us what you tried so that we can get to your mistake and problem and help accordingly.

electrokid
 2 years ago
Best ResponseYou've already chosen the best response.0does anybody think this might give only the even numbers?

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1If you manipulate the equation a bit (by expanding the cubes, and simplifying/factoring), you can show that:\[f(a,b,c)=ab(b+a)(ba)+bc(c+b)(cb)+ac(ac)(a+c)\]From here is is easier to show (using a modular arithmetic argument) that no matter what integers you pick, this will always be divisible by both 2 and 3 (so divisible by 6). My guess is that you can only get the multiples of 6, and nothing else.

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1In fact, note that for any integer a:\[f(a,a+1,a+2)=a(1)^3 +(a+1)(2)^3 +(a+2)(1)^3 \]\[a+8a+8a2=6a+6=6(a+1)\]So you can get any multiple of 6 you want.
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