Question

2x + 5y = 103. Find the number of pairs of positive integers x and y that satisfy this equation.
9
10
12
20

Answer 1

Here 2x always ends in any of the even integers 0, 2, 4, 6 and 8, and 5y always ends in either 5 or 0.

In the equation 2x + 5y = 103, if 5y ends in 0, then 2x shall always end in 3. But there are no multiples of 2 which end in 3. Thus 5y can't end in 0.

But if 5y ends in 5, then 2x shall always end in 8. This can be done!!!

So, when we find the no. of possible integers for either 2x or 5y, we get the answer.

Okay, we're going to find the no. of possible values of 2x.

As 2x + 5y = 103,  2x < 103

As 2x must end in 8, the possible values of 2x are,

8, 18, 28, 38, 48, 58, 68, 78, 88, 98

but can't 108 or higher because 108 > 103.

Here are 10 possible values for 2x, hence the answer is 10.

Now let's find the no. of possible values of 5y.

As 5y must end in 5, the possible values are,

5, 15, 25, 35, 45, 55, 65, 75, 85, 95

but not 105 or higher because 105 > 103.

Here are also 10 possible values, hence the answer is 10.

$\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6}& x & y & 2x & 5y & 2x+5y \\ \cline{1-6}1.&4&19&8&95&103\\ \cline{1-6}2.&9&17&18&85&103\\ \cline{1-6}3.&14&15&28&75&103\\ \cline{1-6}4.&19&13&38&65&103\\ \cline{1-6}5.&24&11&48&55&103\\ \cline{1-6}6.&29&9&58&45&103\\ \cline{1-6}7.&34&7&68&35&103\\ \cline{1-6}8.&39&5&78&25&103\\ \cline{1-6}9.&44&3&88&15&103\\ \cline{1-6}10.&49&1&98&5&103\\ \cline{1-6} n. & 5n-1 & 21-2n & 10n-2& 105-10n &103\\ \cline{1-6}\end{tabular}$