Pigeonhole question about distinct sums How do I show with the pigeonhole principle that no seven positive integers not exceeding $24$ can have sums of all subsets different. 
As observed by Ross Millikan, the simplest possible approach does not work. The maximum possible subset sum is $24+23+\cdots+18=147$, and the minimum possible subset sum is $0$. So as there are only $2^7=128$ subsets the pigeonhole principle does not force a collision. At least not without some sharpening.
 A: Suppose you had seven integers in $\{1,2,\dots,24\}$ so that all subsets have different sums.
That means that there are $2^7=128$ different sums.
The set of seven integers cannot contain a quadruplet of the form $\{a,a+b,a+c,a+b+c\}$ (with all four numbers distinct) because otherwise it would contain two pairs with the same sum.
Therefore the sum of the whole set cannot exceed
$$
24+23+22+20+18+13+7=127.
$$
If the set does not contain the number 1, then the sums of different subsets are in the set $\{0,2,3,\dots,127\}$ which has strictly less than 128 elements, which is impossible.
If the set does contain 1, the sum of the whole set has to be strictly less 127.
This is impossible, too, by the pigeonhole principle.
A: Obviously it is necessary that the 7 integers are distinct. Let $x$ be the minimum integer. Then the minimum sum of any (non-empty) subset is $x$ and the maximum value is $$x+19+20+21+22+23+24 = 129 + x.$$ 
Hence there are at most 130 different possible values for the sum of the 7 integers.
Suppose the number 24 was not part of our 7 integers, then the maximum sum would be 
$$x+18+19+20+21+22+23 = 123+x.$$
Then there would only be 124 different possible sums within our 2^7 -1 = 127 different subsets, contradicting uniqueness of each subset sum. So 24 must be one of our 7 numbers. An identical argument shows 23 and 22 must also members of our set of 7. 
If we suppose 21 is not a member, we get a maximum sum of
$$x + 18+19+20+22+23+24 = 126+x$$
which gives exactly 127 possible different sums for our 127 subsets. Thus for this to be possible, every number between $x$ and $126+x$ must be achieved as the sum of a subset of $\{x,18,19,20,22,23,24\}$. Since $\{x+1\}$ is such a number, we deduce $x = 17$. But then the sum $21$ can't be achieved (among many others).
Finally, we conclude that it must be the case that all of $\{21,22,23,24\}$ are among our 7 integers. But since $21+24 = 22+23$ we have another contradiction. Thus 127 distinct subsets are not possible.
