1
$\begingroup$

Say you have 5 buckets. Each bucket contains a distinct number $[1, 10]$.

So why can it be guaranteed that the total sum of 1 or more numbers from the 5 buckets will equal the total sum of 1 or more OTHER numbers from the 5 buckets?

Example: 6, 7, 8, 9, 10. We know that $6+9=15$ and $7+8= 15$. Regardless of the distinct numbers we fill the buckets with, we will always be able to find cases where this condition holds. Another Example: 1,2,3,5,10 We know $1+2= 3$ and $3 = 3$

I find this extremely interesting but I'm not quite sure how to explain this observation.

In my reasoning, I know the lowest value that can be obtained is 1, and the highest, $10+9+8+7=34$ (assuming we must leave one number for the other group).

What principles explain simply why this works?

  • 0
    Is it that both the groups should be disjoint? – 2017-02-18
  • 0
    Yes, for example, all integers in the group are distinct and when creating groups, in the first example, you could not use one number in both sums. Such as in the first example, 6 could not be used in both sums, only one, or not at all. – 2017-02-18
  • 0
    @MayankDeora: the requirement that the groups be disjoint is not important. If you find two groups that have equal sum, you can delete the common elements and get two disjoint groups with equal sum. – 2017-02-18
  • 0
    @RossMillikan actually the question is bit confusing. – 2017-02-18
  • 0
    @MayankDeora The question basically states, given any 5 distinct numbers from 1 to 10 inclusive, why is it guaranteed that 2 separate groups, not sharing any elements, of 1 or more numbers will have the same total? – 2017-02-18

2 Answers 2

1

This is the pigeonhole principle at work, although it takes a little digging to get it to work.

Firstly we can always assume that the numbers are different, or we could just match two that are identical. So although you gave me that the numbers were distinct, that's actually not needed to make this work.

So given the smallest value chosen, $s$, the largest sum possible is $s+7+8+9+10 = s+34$, giving a range of $35$ possible sums, although the all-five-numbers can't possibly match anything.

Now the analysis has to devolve into cases, to get the range small enough that the pigeonhole principle applies, of more possible subset sums than the range that contains them. Discarding the empty set and the five-number set, there are 30 subset sums of interest.

Looking at the top four numbers, we review cases to drive them below $30$ by attempting to avoid a common sum [noting that $s$ cannot equal a difference in the proposed top-$4$ set]:

Sum: $34$, combination: $7,8,9,10$ - $7+10=8+9$
Sum: $33$, combination: $6,8,9,10$ - $s$ cannot be any of $1,2,3,4$ and $s=5$ gives $5+10=9+6$ Sum: $32$, combination (1): $5,8,9,10$ - $s$ cannot be any of $1,2,3,4$
Sum: $32$, combination (2): $6,7,9,10$ - $6+10=7+9$
Sum: $31$, combination (1): $4,8,9,10$ - $s$ cannot be any of $1,2$ and $s=3$ gives $3+10=4+9$
Sum: $31$, combination (2): $5,7,9,10$ - $s$ cannot be any of $1,2,3,4$
Sum: $31$, combination (3): $6,7,8,10$ - $s$ cannot be any of $1,2,3,4$ and $s=5$ gives $5+8=6+7$ Sum: $30$, combination (1): $3,8,9,10$ - $s$ cannot be any of $1,2$
Sum: $30$, combination (2): $4,7,9,10$ - $s$ cannot be any of $1,2,3$
Sum: $30$, combination (3): $5,6,9,10$ - $5+10=6+9$
Sum: $30$, combination (4): $5,7,8,10$ - $5+10=7+8$
Sum: $30$, combination (5): $6,7,8,9$ - $6+9=7+8$

And otherwise the top numbers sum below $30$ and the pigeonhole principle guarantees matching sums.

  • 0
    This definitely helped clear up my confusion. My main question is in your third paragraph. Why does discarding the empty set and five number set leave us with only 30 subset sums of interest? – 2017-02-18
  • 0
    @user7090180 Each number can either belong to a subset or not, so there are $2^5=32$ subsets including the empty set and the full set. See [power set](https://en.wikipedia.org/wiki/Power_set) – 2017-02-18
1

It is the pigeonhole principle with a bit of work to utilize it. There are $31$ nonempty sets of buckets. The sums of subsets can range from $1$ to $40$. If the range were only $1$ to $30$ we would be done because we would know that two sets had the same sum. If $1$ is a possible sum, the maximum sum is $35$. In any case there at most $35$ possible sums. Then note that if $1$ is included, you can't have neighboring numbers at the top end, so the maximum sum is $10+8+6+4+1=29$ and there must be two sums the same, here $10+4=8+6$. Now we know that $1$ is not in a bucket. If $2$ is in a bucket, the maximum sum is $10+9+6+5+2=32$ and there must be two that match, here $10+5=9+6$ or $10+8+7+6+2=33$ and we have $8=6+2$ and otherwise the range is too small. Now we know neither $1$ nor $2$ is in a bucket, so see what happens if you try $3$. Soon you will always have two pairs of neighboring numbers and you can use that for your sum.

  • 0
    This makes a great deal of sense for the most part. The one thing I may need some clarification on is "If 1 is a possible sum, the maximum sum is 35. In any case there at most 35 possible sums. " – 2017-02-18
  • 0
    If $1$ is a possible sum, $1$ must be in one bucket, so the maximum sum is at most $1+10+9+8+7=35$ As we increase the maximum, we must increase $1$ so the range top to bottom is at most $34$ – 2017-02-18
  • 0
    And say 1 is included, why can we not have a situation 10 + 9 + 8 + 7 +1? Wouldn't 1 + 9 give us the value 10? Sorry for the continued comments, just trying to clarify. Thank you! – 2017-02-18
  • 0
    There are stages of analysis. The first is that the maximum sum is $10+9+8+7+6=40$. We find we have too many pigeonholes so we need to work harder. The next is that if $1$ is one of the numbers the maximum sum is $35$, but that is still too large. Then we notice that getting $35$ requires $10,9,8,7$ and we have $10=9+1$, so we can't have any two numbers in a row. Now the max sum is $10+8+6+4+1=29$ and there are few enough holes to apply the pigeonhole principle. This lets us conclude that no bucket has $1$ in it. Now we turn our gaze to what happens is a bucket has $2$ in it. – 2017-02-18