If 2x + 5y = 103, then how many pairs of (x, y), where x and y are pos

Solution

• x and y are positive integers.
• \(2x + 5y = 103\)

o Or, \(x = \frac{(103 – 5y)}{2}⟹ x = \frac{ (102 – 4y)}{2} +\frac{(1-y)}{2} ⟹ x = 51 - 2y + \frac{(1 – y)}{2}\)
o Now y can be 1, in that case \( x = 51 - 2y + \frac{(1 – y)}{2} = 51-2*1 + \frac{1-1}{2} = 49\)
o Next value of y can be 3, in the case \(x = 51 -2*3 + \frac{(1-3)}{2} = 44\)
o Next value of y can be 5, in that case \(x = 51 -2*5 +\frac{(1- 5)}{2} = 39\)….. And so on.
• We can notice that value of x is decreasing by 5 and maximum possible value of x = 49.
• Since, x is a positive integer, minimum value of x can be 4

o In that case \(y = \frac{103 – 4*2}{5} = 19\)
• Thus, the values of x are in A.P. such that first term is 4, last term is 49 and the common difference is 5.

o Now, let us assume that the total possible positive integral values of x is n.

 Then, \(49 = 4 + (n – 1)*5 ⟹ n = \frac{45}{5} + 1 = 10\)
• Therefore, total possible values of x will be 10 and correspondingly y will also have 10 values.
• So, the required number of (x,y ) pairs = 10

5 can be subtracted 9 times from 49 to keep it positive and least

so total solutions = 1 (for x=49) + 9 (9 more solutions) = 10 SOlutions

Solution

• \(2x + 5y = 103\), x and y are positive integers.

• Pairs of (x,y) that satisfy the equation.

• The expression can be rearranged as \(2x = 103 –5y.\)
• Thus \(x = (103-5y)/2\). Now since x and y are positive integers, the term \((103-5y)\) should be a multiple of 2. In other words, \(103-5y\) should be even.

o Since odd – 5y should be even, therefore 5y should be odd as well and thus y should be odd (since odd *odd = odd).
• Therefore, the problem has been reduced to finding the possible values of positive integer ‘x’ when ‘y’ is an odd positive integer and \(x =(103-5y)/2\), where x is a positive integer.

• Similarly, when \(y= 3, x = (103-15)/2 = 88/2 = 44.\)
• Since \(103 –5y >0\), therefore \(y < 103/5\). Therefore \(y <=20 \) (since ‘y’ is a positive integer)

o But since y is odd, therefore maximum possible value of \(y = 19\).

• Least value of \(y = 1\), greatest \(y = 19\). ‘y’ has to be odd, therefore number of possible values of \( y = 1, 3, 5, 7, 9,… 19. = 10\) total possible values.

o We can find ‘x’ for each such possible value of ‘y’.