So I know this is a pigeon hole problem, however, I am struggling with what the holes and pigeons should be. I originally was thinking that there should be 13 holes because there's 27 numbers and we needed to fit at least two into each hole (26) plus another number to get the third subset. But this doesn't make sense to me any more because that would mean we are acting as if there's only 27 possible subsets.
Now I believe there are 55 possible sums for each group of 27 six-digit numbers. I got this by doing 9+9+9+9+9+9 + 1 for 0 this would make sense if we now use the numbers as holes and the possible sums as pigeons because there are 27 numbers and 27*2=54 and then the +1 means that there is a third number in the subset. However this doesn't make sense to me as much now because then we are acting as if the last 5 digits can't be zero. So, does anyone have some ideas of what I should use for pigeons/holes?
Note: I suspect you're not stating the problem correctly / as intended. If you've rephrased the problem, let me know.
(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)
There are $ 2^{27}$ subsets of these 27 six-digit positive integers. This is our pigeons.
For our holes, they are the possible values of these subsets, which range from $0$ to $27 \times 999999$.
Hence, by PP, one of these holes must contain at least $X$ pigeons. Hence we are done.