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

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- May 27th 2007, 08:46 PMDragonPairs of (x,y)
How many pairs (x,y) of positive integers satisfiy 2x+7y=1000

- May 27th 2007, 10:07 PMearboth
- May 27th 2007, 11:35 PMCaptainBlack
$\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 - May 28th 2007, 05:31 AMSoroban
Hello, Dragon!

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

Quote:

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.

- May 28th 2007, 07:11 AMThePerfectHacker
$\displaystyle 2x+7y=1000$

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.