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 idontknow December 30th, 2018 02:26 PM

Number of solutions

Find the number of pairs $\displaystyle (n_1,n_2)$ that satisfies the equation below:
$\displaystyle n_1 + n_2 = x\; \;$ where $\displaystyle n_1 ,n_2 ,x \in \mathbb{N}$

 topsquark December 30th, 2018 03:16 PM

Quote:
 Originally Posted by idontknow (Post 603844) Find the number of pairs $\displaystyle (n_1,n_2)$ that satisfies the equation below: $\displaystyle n_1 + n_2 = x\; \;$ where $\displaystyle n_1 ,n_2 ,x \in \mathbb{N}$
Let's rephrase the problem statement a bit.
$\displaystyle n_1 + n_2 = x$

$\displaystyle n_2 = x - n_1$

So
$\displaystyle n_1 = 1$, $\displaystyle n_2 = x - 1$.

$\displaystyle n_1 = 2$, $\displaystyle n_2 = x - 2$.

And now it's a counting problem. How many values can $\displaystyle n_1$ take on?

-Dan

 idontknow December 31st, 2018 02:37 AM

It is seen that $\displaystyle x\geq 2$
Number of pairs is dependent on $\displaystyle x$
Example , $\displaystyle x=10^{10}$

 topsquark December 31st, 2018 12:17 PM

Quote:
 Originally Posted by idontknow (Post 603861) It is seen that $\displaystyle x\geq 2$ Number of pairs is dependent on $\displaystyle x$ Example , $\displaystyle x=10^{10}$
Yes. Though I'm not going to try to list all that with your x value!

-Dan

 idontknow December 31st, 2018 04:57 PM

What i got is $\displaystyle f(x)=x-1$
Where f(x) is the number of pairs

 topsquark December 31st, 2018 06:00 PM

Quote:
 Originally Posted by idontknow (Post 603886) What i got is $\displaystyle f(x)=x-1$ Where f(x) is the number of pairs
Yes.

-Dan

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