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
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.
$\displaystyle 2x+7y=1000$
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.