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June 16th, 2018, 07:26 AM  #1 
Newbie Joined: Jun 2018 From: Nepal Posts: 3 Thanks: 0  Find the number of possible choices for x and y when A=100?
I have some questions during the writing of this post, please read it via https://www.physicslog.com/numberofchoices/ 
July 4th, 2018, 11:56 PM  #2 
Member Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 57 Thanks: 3 Math Focus: Number theory, Applied maths 
I agree with your answer = 49 and think your algorithm is OK for any value of A. Hope you get more replies to this and your other questions with A being the sum of more than two different numbers/ 
July 7th, 2018, 04:25 PM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Yes, there are 49. The positive integer pairs, (x, y), such that x+ y= 100 are (1, 19), (2, 98 ), (3, 97), ... (97, 3), (98, 2), (99, 1). My first thought was that there were 99 such pairs but then I noticed the conditions that $x\ne y$ and that order is not to be considered. Dropping the pair (50, 50) leave 98 pairs and half of that is 49. 
July 16th, 2018, 01:49 AM  #4 
Member Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 4 
If you include the pair 0 + 100 = 100, then you have a half pairs for even, and half  1 pairs for odd.
Last edited by skipjack; July 16th, 2018 at 03:45 AM. 
July 22nd, 2018, 04:58 AM  #5  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
Quote:
 
July 22nd, 2018, 05:00 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  
July 22nd, 2018, 11:31 AM  #7 
Member Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 4 
Ok, sorry. Sent from my HUAWEI SCLL04 using Tapatalk 
July 22nd, 2018, 11:59 AM  #8 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
The point I think is to prove a theorem about the number of pairs of distinct positive integers that sum, without regard to order, to a given positive integer. This is a slightly informal proof. Let a be the given positive integer. Let f(x) be the number of pairs of distinct positive integers that sum to a. First, consider any a that is odd and not less than 3 $\implies \exists \ b \text { such that } b \in \mathbb Z^+,\ a = 2b  1, \text { and } b \ge 2.$ $ \text {Consider } p \in \mathbb Z^+ \text { and } 1 \le p \le b  1.$ $ \therefore a  p \in Z \text { and } p + ( a  p) = a $ $ \text {Furthermore, } \ (b  1) \le \ p \implies a  (b  1) \le a  p \implies$ $2b  1  (b  1) \le a  p \implies b \le a  p \implies$ $ 1 \le p < a  p \in \mathbb Z \implies a  p \in \mathbb Z^+.$ How many such pairs (p, a  p) are there? Clearly b  1. $\therefore f(a) = b  1.$ $\text {But } 2b  1 = a \implies 2b  2 = a  1 \implies b  1 = \dfrac{a  1}{2} \implies b  1 = \left \lfloor \dfrac{a  1}{2} \right \rfloor \implies$ $f(a) = \left \lfloor \dfrac{a  1}{2} \right \rfloor.$ And there is no pair of positive integers that add to 1. $f(1) = 0 = \left \lfloor \dfrac{1  1}{2} \right \rfloor \implies$ $f(a) = \left \lfloor \dfrac{a  1}{2} \right \rfloor \text { if } a \text { is an odd positive integer.}$ Next, consider any a that is even and not less than 4 $\implies \exists \ b \text { such that } b \in \mathbb Z^+,\ a = 2b, \text { and } b \ge 2.$ $ \text {Consider } p \in \mathbb Z^+ \text { and } 1 \le p \le b \le b  1.$ $ \therefore a  p \in Z \text { and } p + ( a  p) = a$ $ \text {Furthermore, } \ (b  1) \le \ p \implies a  (b  1) \le a  p \implies$ $2b  (b  1) \le a  p \implies b  1 < b + 1 \le a  p \implies $ $ 1 \le p < a  p \in \mathbb Z \implies a  p \in \mathbb Z^+.$ How many such pairs (p, a  p) are there? Clearly b  1. $\therefore f(a) = b  1.$ $\text {But } 2b = a \implies 2b  2 = a  1  1 \implies b  1 = \dfrac{a 1}{2}  \dfrac{1}{2} \implies b  1 = \left \lfloor \dfrac{a  1}{2} \right \rfloor \implies $ $f(a) = \left \lfloor \dfrac{a  1}{2} \right \rfloor.$ And there is no pair of distinct positive integers that add to 2 $f(2) = 0 = \left \lfloor \dfrac{2  1}{2} \right \rfloor \implies$ $f(a) = \left \lfloor \dfrac{a  1}{2} \right \rfloor \text { if } a \text { is an even positive integer.}$ $\therefore f(a) = \left \lfloor \dfrac{a  1}{2} \right \rfloor \text { for any positive integer } a.$ Last edited by JeffM1; July 22nd, 2018 at 12:06 PM. Reason: Fixed some typos 

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