anonymous
  • anonymous
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?
MIT 6.00 Intro Computer Science (OCW)
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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.
anonymous
  • anonymous
Oops. I meant 'let r be the remainder of (k-x) divided by 6, aka (k-x) mod 6.'
anonymous
  • anonymous
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!

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anonymous
  • anonymous
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.
anonymous
  • anonymous
Love it, thanks for the help.

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