Solving a linear Diophantine equation .

Question

In one month a store sold pads of paper, some for $\$\,11$, some for $\$\,3$, and some for $50$ cents. A total of $98$ pads were sold for a total of $\$\,98$ evaluate the total number of $50$ cent pads sold minus the number of $3$ pads sold.

Answer 1

Let the number of $\$ 11$ pads be $a$ , $\$ 3$ pads be $b$ and $50$ cents pads be $c$ . Then according to the question , we have the following two equations :

$$a+b+c = 98 \quad \text{ (1.)}$$

$$11a+3b+\dfrac c2 = 98 \quad \text{ (2.)}$$

Multiplying $(2.)$ by $2$ and subtracting $(1.)$ from it :

$$21a+5b=98$$

By hit and trial , we get a trivial solution $(3,7)$ . So , all the pairs of solutions are given by :

$$\color{#d05}{a = 3+5t\quad , \quad b = 7 - 21t}$$

Since the number of pads is always positive , we must have $b \gt 0$ , which happens only for $t= 0$.

So the value of $a = 3$ , $b = 7$ . Substituting these in the $(1.)$ , we get $c = 88$.

Thus , total number of $50$ cent pads sold minus the number of $\$3$ pads sold : $$\color{#2dd}{c-b = 88-7=81 }$$