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  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,270 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle n_1 + n_2 = x$ $\displaystyle n_2 = x  n_1$ 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 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,270 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  
December 31st, 2018, 04:57 PM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91 
What i got is $\displaystyle f(x)=x1$ Where f(x) is the number of pairs 
December 31st, 2018, 06:00 PM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,270 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  

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