Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that are are only 16 integer solutions to the following equation: $$11x + 8y + 17 = xy$$ What I tried: I took a modulo 2, and I got that $y$ must be even and $x$ must be odd. But beyond that, I don't really know how to start.

share|improve this question
2  
$17=xy-11x-8y$, then $105=xy-11x-8y+88$, then $105=(x-8)(y-11)$. I will leave it for you to finish. –  chubakueno Aug 17 '13 at 2:50

3 Answers 3

up vote 13 down vote accepted

Here is an algebraic method: $$17=xy-11x-8y$$ $$\iff 105=xy-11x-8y+88$$ $$\iff 105=(x-8)(y-11)$$

share|improve this answer
2  
Just a note: this is not an ad-hoc trick specific to these numbers, it really does work for any $ax + by + c = xy$: we have $c = xy - ax - by$ iff $c + ab = (x-b)(y-a)$. –  ShreevatsaR Aug 17 '13 at 3:14
    
@ShreevatsaR Exactly, I learnt about this method from calculating the number of integer solutions to $x^{-1}+y^{-1}=2010^{-1}$. It works pretty well in a variety of situations! –  chubakueno Aug 17 '13 at 3:19

Hint:

  1. Solve for $x$ to see, that there is an integer solution for a given $y\in\mathbb Z$, iff: $$y-11\mid 8y+17$$
  2. Conclude $y-11\mid 105$.
  3. Check the remaining (finitely many) possibilites.
share|improve this answer
11x + 17 = y(x - 8)
y = (11x + 17)/(x - 8)
y = 11(x - 8) + {105/(x - 8)}

Now, 11(x - 8) can have infinite x. But, 105/(x - 8) should be an integer too for y to be an integer. So, we see that 105 is divisible by these numbers = 1 x 105, 3 x 35, 5 x 21, 7 x 15. So, 105 is divisible by 1,3,5,7,15,21,35,105. Also, negative of these numbers. So we have 8 + 8 = 16 numbers.

So, possible x will be solved for like this:

x - 8 = 1
x - 8 = 3
...
x - 8 = 105
x - 8 = -1
x - 8 = -3
...
x - 8 = -105

These are the values of x for which y will be an integer! (16)

share|improve this answer
    
Factors of 105 from here: mathsisfun.com/numbers/factors-all-tool.html –  Aayush Aug 17 '13 at 3:09
    
There is a formula for the number of divisors. You factor your number and then $$\text{number of divisors}=(e_1+1)(e_2+2)...(e_i+1)$$ where e_k is the exponent of the kth prime. So you don't even need to know who they are! $105=2*3*5$ implies $$\text{number of divisors}=(2)(2)(2)=8$$. However, the data you presented deserves an upvote :) –  chubakueno Aug 17 '13 at 3:16
    
Yes, I know that formula but I thought that the actual data might help. Thank you! –  Aayush Aug 17 '13 at 3:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.