Every member of S is of the form
, and S can be sorted in increasing order as follows: sort by greater z, then by greater y, and then by greater x. Let us consider the number of elements sorting by greater z:
1 element for z = 2
3 elements for z = 3...
elements for z = k.
Thus, up to z = k, there are a total of
The first k such that
is 9, so for the 100th element, z = 9.
Now the 120th element is
, so we simply work backwards from there:
The 113th element is
The 112th element is
The 106th element is
The 105th element is
The 100th element is