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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)?
 one year ago
 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)?
 one year ago
 one year ago

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goformit100Best ResponseYou've already chosen the best response.0
f(a,b,c)=a(b−c)^3+b(c−a)^3+c(a−b)^3
 one year ago

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

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

goformit100Best ResponseYou've already chosen the best response.0
Oh! Peter should definitely try it first.
 one year ago

mathsloverBest 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.
 one year ago

goformit100Best ResponseYou've already chosen the best response.0
This is Truly LIE.
 one year ago

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

joemath314159Best ResponseYou've already chosen the best response.1
If 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.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.1
In 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.
 one year ago
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