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

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

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

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

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

Of the numbers $500, .. , 1$ exactly $\lfloor 500/7 \rfloor=71$ are divisible by 7,
so there are $71$ positive integer pairs $(x,y)$ 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 $2x+7y\:=\:1000$

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

There are: $\left[\frac{1000}{7}\right] = 142$ 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
$2x+7y=1000$

We can solve this using Bezout's Identity.

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

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

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