2
$\begingroup$

Just confused as to how I'm suppose to set up the sets. Since there's an upper bound on each of the integers, could we use the Inclusion Exclusion Principle? I'm not too sure how to set that up because each integer depends on the one before it.

Anyway, any help would be appreciated. Thanks.

  • 2
    Hint: if you have 15 distinct positive integers, what is the smallest number they can add to? – 2017-02-05
  • 0
    Hint: suppose that none of the integers are the same. What's the smallest the sum can be? – 2017-02-05
  • 1
    The smallest sum would be $1+2+3...+15=120$, so that would be the size of the set of pigeons? And the 100 is the number of pigeonholes? – 2017-02-05
  • 0
    @user2965071 I wouldn't bother describing this using pigeon-hole principle, this is more fundamental and immediate than that. This is simply a result of properties of positive integers and their sum. – 2017-02-05
  • 0
    @user2965071 if you have to use the pigeon metaphor, yes: 120 is the number of pigeons and 100 is the holes, the problem with using it here is that you've added the requirement that groups of pigeons must have different sizes, whereas the classical case doesn't allot for grouping together, so ultimately you're stretching it in what one might reasonably argue is not necessary. – 2017-02-05

1 Answers 1

6

The smallest the sum of $15$ distinct, positive integers can be is

$$1+2+\ldots + 15 = \displaystyle{16\choose 2}=120>100.$$

So it must be that there are repeats in the list.