1. ## Pairs of (x,y)

How many pairs (x,y) of positive integers satisfiy 2x+7y=1000

2. Originally Posted by ginafara
The answer is all of the points x,y in Quadrant I, that lie on the line...

y = (-2/7)x +1000

when y = 0, x = 500 so all of the integers from 0 -> 500, since it is defined over its domain, it would be 500 pair (?)

Hello,

I don't want to pick at you but from

$\displaystyle 2x + 7y = 1000 \ \Longleftrightarrow \ y=-\frac{2}{7}x + \frac{1000}{7}$

Since you were looking for integers for x and y you have to calculate how often 1000 is divided by 14. Thus I got 71 pairs.

3. Originally Posted by Dragon
How many pairs (x,y) of positive integers satisfiy 2x+7y=1000
$\displaystyle y=\frac{1000-2x}{7}$

so for $\displaystyle y$ to be an integer $\displaystyle 1000-2x$ must be a multiple of $\displaystyle 7$,
but as this is even $\displaystyle 500-x$ must be a multiple of $\displaystyle 7$.

But as $\displaystyle x$ ranges over the integers $\displaystyle 1, .. , 500$ (which is the range
that $\displaystyle x$ is constrained to) $\displaystyle 500-x$ ranges over $\displaystyle 500, .. , 1$.

Of the numbers $\displaystyle 500, .. , 1$ exactly $\displaystyle \lfloor 500/7 \rfloor=71$ are divisible by 7,
so there are $\displaystyle 71$ positive integer pairs $\displaystyle (x,y)$ that satisfy the
given equation.

RonL

4. Hello, Dragon!

My approach is a variation of CaptainBlack's . . .

How many pairs (x,y) of positive integers satisfy $\displaystyle 2x+7y\:=\:1000$

We have: .$\displaystyle y \:=\:\frac{1000-2x}{7}$
Since $\displaystyle y$ is a positive integer, we have:. $\displaystyle y \;=\;\frac{\text{even number }\leq 1000}{\text{divisible by 7}}$

There are: $\displaystyle \left[\frac{1000}{7}\right] = 142$ multiples-of-seven which are less than 1000

. . and half of them are even.

Therefore, there are 71 solutions.

5. Originally Posted by Dragon
How many pairs (x,y) of positive integers satisfiy 2x+7y=1000
$\displaystyle 2x+7y=1000$

We can solve this using Bezout's Identity.

Trivially we see that,
$\displaystyle 2(-3)+7(1)=1$
Thus,
$\displaystyle 2(-3000)+7(1000)=1000$.

Hence, all integer solutions are given by,
$\displaystyle x=-3000+7t$
$\displaystyle y=1000-2t$.

We want positive solutions, meaning,
$\displaystyle x>0 \to -3000+7t >0 \to 7t > 3000 \to t> 428.5...$
$\displaystyle y>0 \to 1000-2t >0 \to 2t <1000 \to t<500$
Since, $\displaystyle t$ is an integer, we need to find all integers so that,
$\displaystyle 428.5...<t<500$
We see that,
$\displaystyle 429,430,431,....,499$
Are all these integers.
In total we have 71 positive solutions.