# Find the number

• Jun 30th 2009, 01:49 PM
perash
Find the number
Find the total number of positive integer solutions of:

$\frac{2}{n} + \frac{3}{m} = \frac{1}{2006}$
• Jun 30th 2009, 08:35 PM
Prove It
Quote:

Originally Posted by perash
Find the total number of positive integer solutions of:

$\frac{2}{n} + \frac{3}{m} = \frac{1}{2006}$

$\frac{2}{n} + \frac{3}{m} = \frac{2m}{mn} + \frac{3n}{mn} = \frac{2m + 3n}{mn}$

So you have $\frac{2m + 3n}{mn} = \frac{1}{2006}$.

Therefore $2m + 3n = 1$ and $mn = 2006$.

Try to solve these equations simultaneously.
• Jun 30th 2009, 09:54 PM
dhiab
Quote:

Originally Posted by Prove It
$\frac{2}{n} + \frac{3}{m} = \frac{2m}{mn} + \frac{3n}{mn} = \frac{2m + 3n}{mn}$

So you have $\frac{2m + 3n}{mn} = \frac{1}{2006}$.

Therefore $2m + 3n = 1$ and $mn = 2006$.

Try to solve these equations simultaneously.

HEllo : remark
http://www.mathramz.com/xyz/latexren...320f38cb95.png
is not :
http://www.mathramz.com/xyz/latexren...45b369c0d5.png
conclusion :
http://www.mathramz.com/xyz/latexren...bec49c6522.png
THANKS (Evilgrin)
• Jul 1st 2009, 12:03 AM
Moo
$\frac ab=\frac cd \Leftrightarrow \exists k \in \mathbb{Q} ~:~ \begin{cases} a=ck \\ b=dk\end{cases}.$

So here, we have $\exists k \in\mathbb{Q} ~:~ \begin{cases} 2m+3n=k \\ mn=2006k\end{cases}.$

And since m and n are positive integers, we can choose $k\in\mathbb{N}$

We get $2m+3n-\frac{mn}{2006}=0$

$\Rightarrow m=\frac{3n}{2-\frac{n}{2006}}=\frac{3\cdot 2006 \cdot n}{4012-n}$

For m to be an integer, we need 4012-n to divide the numerator.

I conjecture that the number of possibilities is the number of divisors of $3\cdot 4012$ (more or less, because we have to remember that n<4012)
But I really don't know yet how to prove it :(

Quote:

$2m + 3n = 1$
I bet you can't find 2 positive integers such that it works (Wink)
• Jul 1st 2009, 04:22 AM
running-gag
A little mistake (Happy)

Quote:

Originally Posted by Moo
We get $2m+3n-\frac{mn}{2006}=0$

$\Rightarrow m=\frac{-3n}{2-\frac{n}{2006}}=\frac{3\cdot 2006 \cdot n}{n-4012}$

• Jul 1st 2009, 07:37 AM
BobP
$\frac{2}{n} + \frac{3}{m} = \frac{1}{2006}$
To begin with, since $m$ and $n$ are positive integers, neither fraction on the LHS can be greater than $\frac{1}{2006}$, (because then, the other would have to be negative).
So, if $\frac{2}{n} < \frac{1}{2006}$ then $n>4012$ and similarly $m>6018$.
Multiplying throughout by $mn$, $2m+3n=\frac{mn}{2006}$, so $mn$ must be an integer multiple of 2006, (since the LHS is an integer).
Let $mn = 2006k$ where $k$ is an integer, then $n=\frac{2006k}{m}$.
Substituting, $2m + \frac{3.2006k}{m} = k$, so $2m^2 - km + 6018k = 0$.
Solving this for $m,\\\\\ m = \frac{k \pm \sqrt{k^2 - 48144k}}{4}$. What we need now is that $\sqrt{k^2 - 48144k}$ should be an integer and in such a way that the resulting values of $m$ and $n$ are both integers.
One obvious possibility is $k = 0$, leading to $m = 12036$ and $n = 8024$, and this produces the solution $\frac{2}{8024}+\frac{3}{12036}=\frac{1}{2006}$.
Sadly, at this point I grind to a halt (Worried), I don't know of a systematic way of finding further values of $k$ such that $k^2-48144$ is a perfect square. (Help anyone ?)
However, notice that the calculated solution cancels down to $\frac{1}{4012}+\frac{1}{4012}=\frac{1}{2006}$, which is kind of obvious and does suggest a second solution.
$\frac{1}{2006} = \frac{5}{10030} = \frac{2}{10030} + \frac{3}{10030}$.
• Jul 1st 2009, 07:48 AM
BobP
Ooops !! (Giggle).
I should have said that one obvious value of $k$ was $k = 48144$ leading to $m = 12036$.
(Itwasntme)
• Jul 2nd 2009, 12:34 AM
malaygoel
Quote:

Originally Posted by BobP
Sadly, at this point I grind to a halt (Worried), I don't know of a systematic way of finding further values of $k$ such that $k^2-48144k$ is a perfect square. (Help anyone ?)

Let $k^2-48144k$ be $r^2$.

Then $k^2-48144k+24072^2=r^2+24072^2$

now, if $24072=2uv$
or $12036=uv$

then, $(k-24072)^2=(u^2+v^2)^2$

hence, it comes down down to factorise 12036 into two factors(u,v), and for each pair we will get two values of k(one positive and one negative).
Since we are interested only in positive values,
$k=(u-v)^2+48144$

Prime factorisation of 12036 will be helpful for finding factors:

$12036=2^2.3.17.59$

Example:
u=102
v=118
gives k=48400,r=3520
• Jul 4th 2009, 12:58 AM
BobP
Thanks for that, a useful idea that I don't recall seeing before.
It does produce further solutions, m=12980 n=7480, m=2018036 n=4024, m=8054090 n =4015. (I don't know whether I got all of them !)
I've still not got my head round it totally though, since this method doesn't seem to pick up my original solution of m=12036 n=8024, and I'm not sure whether there might not be others.
• Jul 4th 2009, 01:16 AM
malaygoel
Quote:

Originally Posted by BobP
I've still not got my head round it totally though, since this method doesn't seem to pick up my original solution of m=12036 n=8024, and I'm not sure whether there might not be others.

Your solutions were trivial...i.e. the discriminant is zero for your solutions. I am sure that solutions from my method(it gives 12 solutions) and your solutions make the complete set...or at least I can hope so.(Wondering)
• Jul 4th 2009, 08:53 AM
Soroban
Hello, perash!

I had a good start on it, but I haven't finished it yet . . .

Quote:

Find the number of positive integer solutions of: . $\frac{2}{n} + \frac{3}{m} = \frac{1}{2006}$

Multiply by $2006mn\!:\quad 4012m + 6018 n \:=\:mn$

Solve for $n\!:\quad n \;=\;\frac{4012m}{m - 6018}$

Since $n$ is a positive integer, then there are two cases:

. . [1] $m-6018$ is a factor of 4012

. . [2] $m-6018$ is a factor of $m$

Case [1]: $4012 \:=\:2^2\!\cdot\!17\!\cdot\!59$ . . . which has 12 divisors.
. . So we have 12 solutions:

. . $\begin{array}{|c||c|c|} \hline
\text{divisor} & m & n \\ \hline \hline
1 & 6019 & 24,\!148,\!228 \\
2 & 6020 & 12,\!088,\!156 \\
4 & 6022 & 6,\!040,\!066 \\
17 & 6035 & 1,\!424,\!260 \\
34 & 6052 & 714,\!136 \\
59 & 6077 & 413,\!236 \\ \hline\end{array}$
. . . $\begin{array}{|c||c|c|} \hline
\text{divisor} & m & n \\ \hline \hline
68 & 6086 & 359,\!074 \\
118 & 6136 & 208,\!624 \\
236 & 6254 & 106,\!318 \\
1003 & 7021 & 28,\!084 \\
2006 & 8024 & 16,\!048 \\
4012 & 10,\!030 & 10,\!030 \\ \hline \end{array}$

Case [2]: I found three more solutions . . . so far.

. . $\begin{array}{|c|c|} \hline m & n \\ \hline \hline
6021 & 8,\!052,\!080 \\ 6024 & 4,\!028,\!048 \\ 6069 & 477,\!428 \\ \vdots & \vdots \\\hline \end{array}$

At this point, I desperately needed a nap . . .