Pigeonhole principle and finding the total ways for this The question is:

Given any $7$ positive integers whose sum is $100$, show there are $3$ of the numbers whose sum is at least $43$.

So I'm having trouble identifying the pigeonholes. However, I think I know the pigeons: the total number of solutions $\{x_1,x_2,\ldots,x_7\}$ of the equation 
$$x_1 + x_2 + \ldots + x_7 = 100$$
where $x_k\geq 1$ for all $k=1,2,\ldots,7$.

This looks like a partitioning problem (since it says "any $7$ positive integers", so $1 + 1 + 1 + 1 + 1 + 1 + 94 = 100$ is the same as
$94 + 1 + 1 + 1 + 1 + 1 + 1 = 100$),
so I'm not sure if I interpreted it correctly.
Divide the given values into 3 sets, two of three elements and one with a single element. Note that at least one of the given elements is less than or equal to 14. Suppose the set with only one member contains this value. Thus, the sum of the two sets of 3 members is greater than or equal to 86. In fact, at least one of the two groups had a sum of their members greater or equal (in the case of both have the same sum) than 43.
Reworded to work with pigeons and pigeonholes, this question is asking: "Say you have 100 pigeons and you wished to stuff them in $7$ [pigeon]holes. Show that there must be some set of $3$ holes such that there are at least $43$ pigeons in them."

We can start by attempting to evenly distribute the pigeons into the holes, with $14$ per hole and $2$ left over. Since all possible sets of $3$ pigeonholes have only $42$ pigeons in them, when we add the $2$ leftover pigeons, there must be at least one set of $3$ pigeonholes that have at least $43$ pigeons (at least one pigeonhole will have $15$, plus $2$ others of $14$).