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 December 30th, 2018, 02:26 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91 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 $\displaystyle n_1 ,n_2 ,x \in \mathbb{N}$ Last edited by idontknow; December 30th, 2018 at 02:32 PM.
December 30th, 2018, 03:16 PM   #2
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Quote:
 Originally Posted by idontknow 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

 December 31st, 2018, 02:37 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91 It is seen that $\displaystyle x\geq 2$ Number of pairs is dependent on $\displaystyle x$ Example , $\displaystyle x=10^{10}$
December 31st, 2018, 12:17 PM   #4
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Quote:
 Originally Posted by idontknow 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

 December 31st, 2018, 04:57 PM #5 Senior Member   Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91 What i got is $\displaystyle f(x)=x-1$ Where f(x) is the number of pairs
December 31st, 2018, 06:00 PM   #6
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Quote:
 Originally Posted by idontknow What i got is $\displaystyle f(x)=x-1$ Where f(x) is the number of pairs
Yes.

-Dan

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