A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

Question P2 Theorem: Not sure how to prove the theorem that if x,...x+5 sets are possible, then given 6,9,20 packs it is possible to buy any number >=x?

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Suppose that you find a way to get x, x+1, ..., x+5, but there is some number larger number, call it k, that you can't get. Now, let r be the remainder of k divided by 6, aka k mod 6, and notice: (k - x) = n * 6 + r or equivalently x + r + n*6 = k but r is the remainder when dividing by 6, so r must be 0, 1, 2, 3, 4, or 5. Thus you can take the way you already found to get (x + r), and toss in n additional sixes, and you're done.

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oops. I meant 'let r be the remainder of (k-x) divided by 6, aka (k-x) mod 6.'

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So we got six b/c it is the smallest of the pack sizes? If we were to generalize, you could change 6 to a variable which represents the smallest of the pack sizes? So if Mcdonalds changed McNugget sizes to an 8 pack, 9 pack, 20 pack, would the equation change to (k-x) = n*8 +r. meaning that R can only be from 0 to 7, so now we have to have x,...x+6 in order to ensure that every integer above it can be solved? Maybe i'm not 100% understanding, but how would the generalized formula look so we can any value of pack sizes, and still find the maximum unsolvable number? Thanks!

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You got it! if you have several different pack sizes and the smallest is S then any set of S continuous sizes implies that all larger sizes can be found. So in general you want to find solutions for x, x+1, x+2, ..., x+S where S is the smallest of the pack sizes, and you're guaranteed to be able to find solutions to all larger pack sizes. Of course, you're not guaranteed to be able to find S contiguous solutions. For example, if all of your pack sizes are even, you won't be able to get any odd numbers.

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Love it, thanks for the help.

  6. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.