A jar holding 1 cent, 5 cent and 10 cent coins contains twelve coins with a total value of 86 cent. How many coins of each type are in the box? (Preferably using an augmented matrix)
My attempt so far is (let x= no. of 1 cent coins, y = etc...): $$ x+y+z=12\\ x+5y+10z=86\\ x\geq0, y\geq0, z\geq0 $$ but I'm not sure how to formulate the last condition correctly in an augmented matrix.
On
Here is a (somewhat) structured way to reach an answer:
Start with the greedy method to reach the sum $86$, regardless of number of coins. That's eight $10$'s, one $5$ and one $1$. Ten coins in total.
Now, repeatedly exchange one $10$ into two $5$'s until you have twelve coins. You end up with six $10$'s, five $5$'s and one $1$.
There are no other solutions. To see this, first see that if you have exactly one $1$, then the remaining eleven coins must be divided as above to make a sum of $85$. If you don't have one $1$, the next possibility is to have six of them (anywhere between one and six and you can't make the $6$ in $86$). And with six $1$'s, the largest sum you can possibly make is $66$. Finally, for the same reason, you can't have eleven $1$'s.
Hint: Subtract the first from the second to get: $$4y+9z=74 \Rightarrow z=\frac{74-4y}{9}=\frac{2(37-2y)}{9}.$$