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ritvik1512
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1Oh! Peter should try it first.

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

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

mathslover
 one year 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.

goformit100
 one year ago
Best ResponseYou've already chosen the best response.0This is Truly LIE.

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

joemath314159
 one year 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
 one year 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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