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June 16th, 2018, 07:26 AM   #1
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Wink 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/
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July 4th, 2018, 11:56 PM   #2
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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/
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July 7th, 2018, 04:25 PM   #3
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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.
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July 16th, 2018, 01:49 AM   #4
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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.
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July 22nd, 2018, 04:58 AM   #5
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Quote:
Originally Posted by Country Boy View Post
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.
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July 22nd, 2018, 05:00 AM   #6
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Quote:
Originally Posted by 1ucid View Post
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.
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July 22nd, 2018, 11:31 AM   #7
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Ok, sorry.

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July 22nd, 2018, 11:59 AM   #8
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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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