I have a math puzzle book, and in the chapter I'm working in right now, there's is this type of question that gets asked in one form or another. An example of such a puzzle:
How many darts are needed to score exactly 100?
How could one go about without checking every combination? Yes, a simple observation shows that one could not throw 1 dart to score 100, neither 2 darts, since the sum of the two highest numbers does not exceed 100. But with 3, 4, or 5 the number of combinations to check gets... well... out of hand.
Another puzzle of this form is:
Can you find 6 odd numbers that add up to 100?
But the question to finding answers to these questions remains the same: is there a good way of efficiently finding these numbers?
A) Check for $3$-darts and $4$-darts solutions.
This method is simple, but works mainly for $3$ or $4$ darts.
Make the set of possible sums of $2$ darts:
$\color{red}{A = \{a_j\}} = \{16+16, 16+17, 16+23, \ldots, 39+40 \}$;
then sort it and remove repeating values (for more comfortable usage):
$\color{red}{A} = \{32,33,34,\;39,40,41,\;46,47,48,\;55,56,57,\;62,63,64,\;78,79,80 \}$;
Then look for remainders to $100$:
$\color{green}{B = \{100 - a_j\} } = \{68,67,66,\; 61,60,59,\;52,53,54,\; 45,44,43,\;38,37,36,\; 22,21,20\}$.
If there is solution of $3$ darts, then set $\color{green}{B}$ would contain one of given numbers $16,17,23,24,39,40$.
If there is solution of $4$ darts, then set $\color{green}{B}$ would have common element with the set $\color{red}{A}$.
There are no $3$-darts solutions, no $4$-darts solutions.
B) Consider $5$- and $6$-darts combinations.
If solution has large number like $39$ or $40$, then the sum of all other ($4$ or $5$) numbers will be $61$ or $60$. Impossible, since even $16+16+16+16$ is greater than $61$.
So, solution consists of small numbers like $16,17,23,24$.
If it has number $23$ or $24$, then the sum of all other numbers is $77$ or $76$. Since $16+16+16+16+16>77$, then this sum of other numbers consists of $4$ numbers.
(Show that it is impossible to compose number $76$ or $77$ of four numbers of the set $\{16,17,23,24\}$).
So, solution consists of numbers $16$ and $17$ solely.
Now: