# How to get all natural solutions of the equation?

I have this statement:

In how many ways, you can pay $12,000$ dollars, with bills of $10,000$ , $5,000$ and $1,000$ dollars?

I can do combinatorics like:

$1)$ 1 bill of $10,000$ and two of $1,000$

$2)$ 2 bills of $5,000$ and two of $1,000$

Well, but i read a other solution that says:

Get all natural solutions of:

$12 = 10a + 5b + c$

then, $a$, $b$ and $c$ can be $\{0, 1, 2, 3..\}$ $\rightarrow \mathbb{N}$, example:

$12 = 10*1 + 5*0 + c*2$, with $a=1$, $b = 0$ and $c = 2$

But how can I find all naturals solutions? Like determinant in quadratic equations.

• So rather than figuring out the specific answer to this specific problem (which is trivial), you are asking for a general formula/method/equation to figure out the answer for these general types of question, right? – Bram28 Mar 13 '18 at 14:28
• Look at your second example again. – user535339 Mar 13 '18 at 14:28

We must have $a \in \{0,1\}$,

If $a=1,$ the problem becomes $$2 = 5b+c$$

Hence $b=0, c=2$.

If $a=0$, the problem become $$12=5b+c$$

Only $3$ cases $b \in \{0,1,2\}$ and for each case we can get our $c$.

$10a+5b+c=12$, because $a,b,c$ are natural numbers, $a$ can only be $0$ or $1$ (if $a \ge 2$ than the total would be more than or equal to $20$, contradiction).

With $a=0$, we have $5b+c=12$ and with $a=1$, we have $5b+c+10=12$, for both cases we can get the solution easily, using the method on this question.

This cannot be obtained by a formula, but using a recursive algorithm:

• order the bills by decreasing value;

• try all possible numbers of the largest bill that fit, and solve the same problem on the remainder and without using the largest bill.

In this case, $12$ using $10,5,1$:

• $1\times10$ and solve for $2$ using $5,2$;

• $0\times5$ and solve for $2$ using $1$;
• $0\times10$ and solve for $12$ using $5,2$;

• $2\times 5$ and solve for $2$ using $1$;

• $1\times5$ and solve for $7$ using $1$;

• $0\times5$ and solve for $12$ using $1$.

(All the "using $1$" subproblems are trivial.)