Number of Solutions for ##\frac{xy}{x+y}=2^3\times 3^4 \times 5^6##

[Saitama]

Homework Statement

Consider the equation $\frac{xy}{x+y}=2^3\times 3^4 \times 5^6$ then the number of positive integral solutions of equation are
A)140
B)819
C)72
D)None

Homework Equations

The Attempt at a Solution

Honestly, I am out of ideas on this one. I don't even have a clue about how should I start. -_-

Any help is appreciated. Thanks!

 

[mfb]

Hint: if (a,b) with x!=y is a solution for (x,y), what about (b,a)?
What about x=y?

Based on that, is the number of solutions even or odd?

 

[Saitama]

Hint: if (a,b) with x!=y is a solution for (x,y), what about (b,a)?
What about x=y?

Based on that, is the number of solutions even or odd?

Are you sure about the factorial sign or is it a typo? -.-'

 

[NascentOxygen]

I believe mfb is using a common programming symbol ❲viz., !=❳ that equates to
the mathematician's ≠[/size] sign.

 

[Saitama]

I believe mfb is using a common programming symbol that equates to the mathematician's ≠[/size] sign.

Ah yes, now I recognise it, thanks!

I am confused about mfb's first question. If (a,b) is a solution when x!=y, then (b,a) is surely a solution or am I missing something?

 

[NascentOxygen]

I am confused about mfb's first question. If (a,b) is a solution when x!=y, then (b,a) is surely a solution or am I missing something?

No, no, it sounds as though you do understand what he is saying there.

 

[haruspex]

I am confused about mfb's first question. If (a,b) is a solution when x!=y, then (b,a) is surely a solution

Yes, that's exactly mfb's point. So are there an odd or even number of such solutions? What about solutions with x=y?

But to me this is a somewhat unsatisfactory way of answering the question. It would be more educational to come up with the number as if it were not multiple choice. The trick is not to be put off by the huge number on the right. If you were given xy/(x+y) = a, how might you simplify that?

 

[Saitama]

Yes, that's exactly mfb's point. So are there an odd or even number of such solutions? What about solutions with x=y?

Equal number of odd and even solutions?

With x=y, there is only one solution.

But to me this is a somewhat unsatisfactory way of answering the question. It would be more educational to come up with the number as if it were not multiple choice. The trick is not to be put off by the huge number on the right. If you were given xy/(x+y) = a, how might you simplify that?

$xy=xa+ya \Rightarrow xy-xa-ya=0 \Rightarrow xy-xa+a^2-ya=a^2$
$\Rightarrow x(y-a)-a(y-a)=a^2 \Rightarrow (x-a)(y-a)=a^2$

Does this look good?

 

[Dick]

Equal number of odd and even solutions?

With x=y, there is only one solution.$xy=xa+ya \Rightarrow xy-xa-ya=0 \Rightarrow xy-xa+a^2-ya=a^2$
$\Rightarrow x(y-a)-a(y-a)=a^2 \Rightarrow (x-a)(y-a)=a^2$

Does this look good?

Yes, that's great! The best thing you can do with an integer problem is factor it. Now the next step is to figure out the number of ways to factor (2^3*3^4*5^6)^2, right?

 

[Saitama]

Yes, that's great! The best thing you can do with an integer problem is factor it. Now the next step is to figure out the number of ways to factor (2^3*3^4*5^6)^2, right?

Do I need to use the Divisor function?

 

[Dick]

Do I need to use the Divisor function?

Great idea!

 

[Saitama]

Great idea!

Okay so that gives me a total of (6+1)(8+1)(12+1)=7*9*13=819 solutions.

Thank you everyone!

 

[haruspex]

Do I need to use the Divisor function?

No, it's much simpler than that. How many possible values are there for u = x-a given that uv = 2⁶3⁸5¹²?

 

[Dick]

No, it's much simpler than that. How many possible values are there for u = x-a given that uv = 2⁶3⁸5¹²?

That IS the divisor function, isn't it?

 

[haruspex]

That IS the divisor function, isn't it?

Sorry, I was thinking of something else.

 

[NascentOxygen]

I gave the original equation to wolfram alpha, but it spins its wheels without making progress. I think the task must be exceeding the slice of time it allocates to a freeloader.

 

[mfb]

But to me this is a somewhat unsatisfactory way of answering the question. It would be more educational to come up with the number as if it were not multiple choice.

Well, it will not work for arbitrary given options, but it is the first solution I found and the "proper" solution takes more time.