Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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)?

Precalculus
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

f(a,b,c)=a(b−c)^3+b(c−a)^3+c(a−b)^3
Oh! Peter should try it first.
This is a lie.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Unless it's Spider-man or Peter Griffin :D
Oh! Peter should definitely try it first.
@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.
This is Truly LIE.
does anybody think this might give only the even numbers?
If you manipulate the equation a bit (by expanding the cubes, and simplifying/factoring), you can show that:\[f(a,b,c)=ab(b+a)(b-a)+bc(c+b)(c-b)+ac(a-c)(a+c)\]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.
In fact, note that for any integer a:\[f(a,a+1,a+2)=a(-1)^3 +(a+1)(2)^3 +(a+2)(-1)^3 \]\[-a+8a+8-a-2=6a+6=6(a+1)\]So you can get any multiple of 6 you want.

Not the answer you are looking for?

Search for more explanations.

Ask your own question