Make $\$100$ by taking $\$1$, $\$5$, and $\$10$... but we can take only $21$ notes 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.
 A: 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 $y =13 , z=3 , x=5$
A: 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)

