Diophantine Equation

[bghalmeida]

The following Diophantine equation has exactly one solution where [MATH]x [/MATH] and [MATH]y[/MATH] are integers, and [MATH]4 \leq y≤6[/MATH]. What is the value of [MATH]x[/MATH] in that solution?
[MATH]6x+9y=27[/MATH]
Hey guys, I tried to solve this equation but I didn't undertand why [MATH]x=-6[/MATH] or [MATH]x=-9[/MATH] are wrong, could someone help me?

 

[lookagain]

6x is even. 27 is odd. What must 9y be, even or odd, for the sum to be odd? Depending
on what 9y is, what must y be, even or odd? You have three choices among the y-values
given in the problem, and notice that exactly one of them is odd.

 

[bghalmeida]

6x is even. 27 is odd. What must 9y be, even or odd, for the sum to be odd? Depending
on what 9y is, what must y be, even or odd? You have three choices among the y-values
given in the problem, and notice that exactly one of them is odd.

I realized which was the correct answer during my time thinking about it, but I still have some doubts.

Using the Extended Euclidean Algorith I got:
[MATH]6(-9)+9(9)=27[/MATH]And continuing the math:
[MATH]6(6k)+9(-4k)=0[/MATH]So:
[MATH]6(6k-9)+9(-4k+9)=27[/MATH]
With that in mind and using what the exercise gave me, I know that [MATH]4\leq-4k+9\leq 6[/MATH], because [MATH]y=-4k+9[/MATH].

So I calculated the inequation and it gave me [MATH]x\in{\left[\frac{3}{4},\frac{5}{4}\right]}[/MATH]
But it doesn't give me my answer. I understand what is the correct answer, but I don't want to do the "math", I want to really understand what I'm doing wrong here, could you show me?

 

[JeffM]

[MATH]6x + 9y = 27 \implies \dfrac{2}{3} * x + y = 3.[/MATH]
But both y and 3 are integers so (2/3) * x must be an integer such that

[MATH]\dfrac{2}{3} * x = 3 - y \implies 2x = 9 - 3y \implies x = 4.5 - 1.5y.[/MATH]
If y is even, then X is not an integer. Thus, y must be odd. But y is limited to three succesive integers, namely 4, 5, or 6. Only one is odd, namely 5.

[MATH]\therefore x = 4.5 - 1.5(5) = 4.5 - 7.5 = -3.[/MATH]
Lookagain explained this in post #2.

 

[Deleted member 4993]

The following Diophantine equation has exactly one solution where [MATH]x [/MATH] and [MATH]y[/MATH] are integers, and [MATH]4 \leq y≤6[/MATH]. What is the value of [MATH]x[/MATH] in that solution?
[MATH]6x+9y=27[/MATH]
Hey guys, I tried to solve this equation but I didn't undertand why [MATH]x=-6[/MATH] or [MATH]x=-9[/MATH] are wrong, could someone help me?

If x = -6 then we get:

6*(-6) + 9y = 27 \(\displaystyle \ \ \to \ \ \) y = 7 \(\displaystyle \ \ \to \ \ \) beyond the allowed domain of y \(\displaystyle \ \ \to \ \ \) [MATH]4 \leq y≤6[/MATH]
If x = -9 then we get:

6*(-9) + 9y = 27 \(\displaystyle \ \ \to \ \ \) y = 9 \(\displaystyle \ \ \to \ \ \) beyond the allowed domain of y \(\displaystyle \ \ \to \ \ \) [MATH]4 \leq y≤6[/MATH]

 

[Steven G]

I am confused about the difficulty here. If y is an integer and 4≤y≤6, then can only be 4, 5 or 6
Lookagain clearly explained why y must be odd. That is y = 5.
So you solve 6x+9y=27, where y = 6. This is trivial.

6x + 9*5 = 27 so 6x = -18. Then x = -3

If you did not realize what Lookagain pointed out then you try to find a solution for x for y=4, 5 and 6.

 

[bghalmeida]

I am confused about the difficulty here. If y is an integer and 4≤y≤6, then can only be 4, 5 or 6
Lookagain clearly explained why y must be odd. That is y = 5.
So you solve 6x+9y=27, where y = 6. This is trivial.

6x + 9*5 = 27 so 6x = -18. Then x = -3

If you did not realize what Lookagain pointed out then you try to find a solution for x for y=4, 5 and 6.

I wasn't looking for an answer, I was looking for the real math of the problem. I know the correct answer, but what if I had to find a number that satisfies the equation and the set of numbers was bigger? But I already solved it, thanks.

 

[JeffM]

@Jomo

I suspect that the OP has learned the technique for dealing with a class of Diophantine equations using the extended Euclidean algorithm. Thus, he was thinking about division being involved and so did not see the relevance of Lookagain's post. (This is not a criticism of Lookagain's post, which was right on point.) So I started my post with division to accord with the OP's initial mindset. I acknowledge that is a roundabout way of making Lookagain's argument.

 

[JeffM]

What the OP was thinking involves this:

[MATH]6x + 9y = 27[/MATH]
has a solution in integers if and only if the greatest common divisor of 6 and 9 is a divisor of 27.

3 is the magic number in this case. There is a non-empty solution set.

I am not sure how the OP concluded

[MATH]\dfrac{3}{4} \le x \le \dfrac{5}{4} \implies x = 1[/MATH]
because that is an error. Possibly what was meant was that (3, 1) is a valid solution. Therefore,

[MATH]k \in \mathbb Z \text { and } x = 3 + 3k \text { and } y = 1 - 2k.[/MATH]
Given the extra condition that

[MATH]4 \le y \le 6 \implies 4 \le 1 - 2k \le 6 \implies - \dfrac{6 - 1}{2} \le k \le - \dfrac{4 - 1}{2} \implies \\ k = - 2 \implies x = 3 + 3(-2) = -3 \text { and } y = 1 - 2(-2) = 5.[/MATH]And obviously that works [MATH]6(-3) + 9(5) = 45 - 18 = 27.[/MATH]
To answer the OP's most recent question, if there is a magic number and one variable has the further restriction that it must be one of a magic number of successive integers, then there is a unique solution. My guess is that a binary search will find that solution relatively quickly, but I have not demonstrated that.

 

[bghalmeida]

What the OP was thinking involves this:

[MATH]6x + 9y = 27[/MATH]
has a solution in integers if and only if the greatest common divisor of 6 and 9 is a divisor of 27.

3 is the magic number in this case. There is a non-empty solution set.

I am not sure how the OP concluded

[MATH]\dfrac{3}{4} \le x \le \dfrac{5}{4} \implies x = 1[/MATH]
because that is an error. Possibly what was meant was that (3, 1) is a valid solution. Therefore,

[MATH]k \in \mathbb Z \text { and } x = 3 + 3k \text { and } y = 1 - 2k.[/MATH]
Given the extra condition that

[MATH]4 \le y \le 6 \implies 4 \le 1 - 2k \le 6 \implies - \dfrac{6 - 1}{2} \le k \le - \dfrac{4 - 1}{2} \implies \\ k = - 2 \implies x = 3 + 3(-2) = -3 \text { and } y = 1 - 2(-2) = 5.[/MATH]And obviously that works [MATH]6(-3) + 9(5) = 45 - 18 = 27.[/MATH]
To answer the OP's most recent question, if there is a magic number and one variable has the further restriction that it must be one of a magic number of successive integers, then there is a unique solution. My guess is that a binary search will find that solution relatively quickly, but I have not demonstrated that.

You explained it very well, I understood everything. Thank you.