Make $\$100$ by taking $\$1$, $\$5$, and $\$10$... but we can take only $21$ notes

Question

I am weak in mathematics, but I need to know if this is possible. I will take $\$100$ from my friend, but he will give me only $21$ notes of $\$1$, $\$5$, $\$10$. I need to tell him the numbers of each notes that will make $100$ dollars.

Answer 1

Here's a solution..

You need $10$ $\cdot$ $1$ notes...$7$ $\cdot$ $10$ notes and $4$ $\cdot $ $5$ notes

Let

No. of $1$ dollar notes $=x$

No. of $5$ dollar notes $=y$

No. of $10$ dollar notes $=z$

Note that $$0\leq x\leq 100$$ $$0\leq y\leq20$$ $$0\leq z\leq10$$ and $$x,y,z \in \mathbb{Z^+}$$

So$$x+y+z=21$$ and $$x+5y+10z=100$$

Subtracting the first equation from second gives $$4y+9z=79$$ or $$z=\frac{79-4y}{9}$$

which has integer solution at $y=4$ within given bounds EDIT: One other solution occurs at $y =13 , z=3 , x=5$

Answer 2

Expanding the answer by @user3558 , we can show, that these two solutions are the only solutions.

We have: $$\begin{cases}x,y,z \geq 0 \\ x,y,z \in \mathbb{Z} \\ z=t\\ y=19+\frac{3-9t}{4}\\ x=2-\frac{3-5t}{4} \end{cases}$$

We want $y$ to be integer, so if $z$ is an integer, then we have

$3-9t \equiv 0 \mod 4$

$t \equiv -1 \mod 4$

$t=4k-1, \, t \in \mathbb{Z}$

Because $z\geq 0$:

$z=t=4k-1\geq0$

$k \geq \frac{1}{4}$

Because $y \geq 0$:

$y=19+\frac{3-9t}{4}=19+\frac{3-9(4k-1)}{4} \geq 0$

$-9k +22 \geq 0$

$ k\leq \frac{22}{9} (=3+\frac{4}{9})$

Because $x\geq 0$:

$x=2-\frac{3-5t}{4} = 2-\frac{3-5(4k-1)}{4}$

$ 2- 5k-2\geq 0$

$k \geq 0$

So if we now get these three conditions, we have

$$k\in \{1,2\}$$

• k=1: (x,y,z)=(5,13,3)
• k=2: (x,y,z)=(10,4,7)