find solutions to following equations?

Question

This is an interview question. There are a total of 100 coins of values 5, 1, 0.2. I need to find a solution that satisfies the following constraints.

1) The total number of coins must be 100.

2) The total value of coins must be 100.

3) There must be at least 1 coin of each type.

So basically i need to solve these equations. x+y+z = 100 and 5x+1y+0.2z=100.

I've written a java program to brute force the values and found that 16 solutions exist.

1) 1, 94, 5

2) 2, 88, 10

3) 3, 82, 15

4) 4, 76, 20

5) 5, 70, 25

6) 6, 64, 30

7) 7, 58, 35

8) 8, 52, 40

9) 9, 46, 45

10) 10, 40, 50

11) 11, 34, 55

12) 12, 28, 60

13) 13, 22, 65

14) 14, 16, 70

15) 15, 10, 75

16) 16, 4, 80

How do i go about solving this manually?

Answer 1

Let $A$ be the number of $5's$, $B$ the number of $1's$, and $C$ the number of $.2's$. Clearly $C$ is divisible by $5$. Also we have $1≤C≤95$, which implies $5≤C≤95$.

Choose $C=5D$, for $1≤D≤19$. For a given $D$ we then seek to solve the pair of equations $$A+B=100-5D \quad \&\quad 5A+B=100-D$$

Subtracting the former from the latter we see that $$4A=4D\implies A=D$$.

Thus your triples are of the form $(D,100-6D,5D)$. Since we require $100-6D≥1$ we must have $99≥6D$ or $D≤16$ and we are done.

Answer 2

Start from $100$ coins of $1$.

If you turn $5$ of them to a coin of $5$, the number of coins decreases by $4$. You can compensate by turning a coin of $1$ to $5$ coins of $0.2$. In total, you consume $6$ coins of $1$.

After $k$ such operations, the configuration is $(k,100-6k,5k)$. There are $16$ solutions, from $(1,94,5)$ to $(16,4,80)$. (Note that $4=100\bmod6$.)