My Math Forum Find the number of possible choices for x and y when A=100?

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 June 16th, 2018, 08:26 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/number-of-choices/
 July 5th, 2018, 12:56 AM #2 Member   Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 54 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/ Thanks from topsquark
 July 7th, 2018, 05:25 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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. Thanks from topsquark
 July 16th, 2018, 02:49 AM #4 Member   Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 3 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 04:45 AM.
July 22nd, 2018, 05:58 AM   #5
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Quote:
 Originally Posted by Country Boy Yes, there are 49. The positive integer pairs, (x, y), such that x+ y= 100 are (1, 19)
Typo. This was, of course, supposed to be (1, 99).

Quote:
 , (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 22nd, 2018, 06:00 AM   #6
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Quote:
 Originally Posted by 1ucid If you include the pair 0 + 100 = 100, then you have a half pairs for even, and half - 1 pairs for odd.
The initial post specifically said pairs of positive integers.

 July 22nd, 2018, 12:31 PM #7 Member   Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 3 Ok, sorry. Sent from my HUAWEI SCL-L04 using Tapatalk
 July 22nd, 2018, 12:59 PM #8 Senior Member   Joined: May 2016 From: USA Posts: 1,189 Thanks: 489 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 01:06 PM. Reason: Fixed some typos

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