User:Matt Kwan/MATH5505

render math: http://checkmyworking.com/misc/mathjax-bookmarklet/

PS1
pigeonhole principle

1
We can assume all ages are distinct as otherwise two singleton sets with the same age can be found (although this isn't necessary for thee first part of the Q). We also ignore the condition that the two sets are disjoint, since if $A$ and $B$ have the same age sum and are distinct, then $A' = A - B$, $B' = B-A$ still have the same age sum and are distinct. Note that both $A'$ and $B'$ are non-empty as otherwise one must be a nonempty set with zero sum - impossible, as all people are at least 1 year old. Now the 'pigeons' are \[ \text{number of subsets of the 10 people} = 2^{10} - 1 = 1023, \] and the possible set of age sums for nonempty subsets (ie. the pigeonholes) is $\{ 1,2,\ldots, 600 \}$. By the pigeonhole principle, since $1023 > 1\times 600$, there is at least one age sum attained by more than 1 subset. That is, there are two distinct subsets with the same age sum, as required. $\Box$\\

Does it work with only 9 people? Yes, if you use the different ages assumption. In this case there are $2^{9}-1 = 511$ possible subsets and $\{1,2,\ldots, 504 \}$ possible age sums, so the pigeonhole principle still works.

Exercise: Construct a set of 6 with no two age sums the same.