coiny two

x/10+y/4=13, 2x+5y=260 ,x=260-5y/2 no of ordered pairs, y=0,2,4.......52............27 terms ,hence the solution.........................

Using $D$ dimes and $Q$ quarters, the problem gives us the equation

\begin{align} \frac{1}{10} D + \frac{1}{4} Q = 13 \end{align} because dimes are tenths of a dollar and quarters are fourths of a dollar.

Adjusting our equation, we find $2D + 5Q = 260$ . Because we have $2$ variables and only $1$ equation, it is not a definite system. Letting one of the variables become an integer parameter, let's suppose, $D = n, \; n \in \mathbb{Z}$ , we can find "general solution" for these cases.

$D = n$ yields $Q = 52 - \frac{2}{5} n$ , which is an ugly form for finding $Q$ (remember $Q$ can only be a natural value, and for some integers $n$ , this $Q$ equality does not satisfy it). Modifying $D = n = 5k, \: n \; \text{and} \; k \in \mathbb{Z}$ , we get $Q = 52 - 2k$ , which now always gives integer values for $Q$ . Restricting $Q$ to be a non-negative number, we must have $52 - 2k \geq 0$ . This means the $k$ parameter can only go from $k = 0$ to $26$ . This yields $26 - 0 + 1$ possible values for it, and therefore, 27 ways to make $13$ dollars.

Phew!

Creating a multiple of 10 requires 2 quarters. Thus, with 0, 2, 4, 6, ..., 50, 52 quarters, you could create multiples of 10, so the remainder could be created using dimes. This is a total of 27 possibilities.

$13 = 1300 cents

Let $x$ be the number of dimes; let $y$ be the number of quarters. Then, from the given information:

$10x + 25y = 1300$

$2x+5y = 260$

This Diophantine equation can be solved by considering the possible values of $y$ . Obviously, if $y$ is odd, $x$ cannot be an integer. So, $y$ has to be even. But the maximum value of $y$ occurs when $x = 0$ , where $y = 52$ .

Hence, $y$ can take the all even values from 0 to 52, giving us a total of 27 possible values for $y$ , which is also the number of solutions.

2 quarters are needed to become a multiple of 10, so 1300 / (25 x 2) = 26. 26 + 1 (for the possibility that not a single quarter is used) = 27.

1300/10 = 130x and 1300/25 = 52y

5x = 2y; 130/5 = 52/2 = 26

(0+26), (1+25), (2+24)......., (26+0) = 27 total

Let's take the number of dimes x and number of quarters y. We get this equation 4x + 10y = 520, Solving with respect to x we get x = 5(26-y/2). As x is an integer y must be even. Let y = 2k On substitution of y we get x = 5(26-k). This equation has 27 solutions in integers by putting values of k from 0 to 26.