Problem: Take any number $A \in \Bbb{N} = \{1, 2, 3, \dots\}$, and then take $x, y \in \Bbb{N}$, where $x \ne y$ and $x + y = A$. Find the number of possible choices for $x$ and $y$ when $A=100$. Order doesn't matter, e.g., $x = 80$ and $y = 20$ is the same as $x = 20$ and $y = 80$.

(Proposed by me and I solved it.) I want to know, Is there is another method that can solve this problem? Here is my method:

From the pattern I have seen, we can say;

  • When $A$ is odd, number of possibilities is $\frac{A-1}{2}$
  • When $A$ is even, number of possibilities is $\frac{A-2}{2}$

And, the answer is 49 because $A = 100$ is even.

  • $\begingroup$ I think $x+y=y+x,$ isn't it? So if $x+y=A,$ then $y+x=A$ as well? $\endgroup$ – awllower Jul 18 '16 at 7:12
  • $\begingroup$ I edited your formatting so one can read your question, but I didn't understand a single bit... Could you please edit to make it more clear ? $\endgroup$ – BusyAnt Jul 18 '16 at 7:13
  • $\begingroup$ If you are not comfortable using English, you might as well write the question in your own language and someone here can translate that for you. As of now, this question makes no sense. $\endgroup$ – Roby5 Jul 18 '16 at 7:17
  • 1
    $\begingroup$ How can we know if there is another method, when we don't know your method? $\endgroup$ – Gerry Myerson Jul 18 '16 at 7:29
  • $\begingroup$ I have edited the question! Please mention if there is any difficulty while reading the question. $\endgroup$ – user354230 Jul 18 '16 at 7:29

I think you are asking: Given a natural number $A$, how many pairs of distinct natural numbers $0<x<y<A$ satisfy $x+y = A$?

I believe the answer must be $\lfloor (A-1) / 2\rfloor$.

  • 3 can be written as 1+2 only.
  • 4 can be written as 1+3 only.
  • 5 can be written as 1+4 or 2+3, etc.

In general, the rule is that $x$ can vary between $1\ldots (A-1)/2$, in which case $y=A-x$ is an integer distinct from $x$ which makes $A=x+y$. In the case that $A$ is even, we must specifically exclude the case $x=y=A/2$, so we round down.

Therefore the answer to your particular question is: when $A=100$, the number of ways is $$\lfloor(100-1)/2\rfloor=49.$$

Is that what you were looking for?

  • $\begingroup$ $x,y $ can be any natural number without the restriction as $0<x<y<A$. $\endgroup$ – user354230 Jul 18 '16 at 7:43
  • $\begingroup$ You have did mistake, it's not $\frac{100-1}{2}$. It's $\frac{100-2}{2}$ ! $\endgroup$ – user354230 Jul 18 '16 at 8:04
  • $\begingroup$ @DamodarRajbhandari. I understand. Because you wanted to count each pair $\{x,y\}$ exactly once (without counting $x+y$ and $y+x$ as separate sums), I supposed without loss of generality that I will use $x$ to refer to whichever number in the pair is smaller. $\endgroup$ – user326210 Jul 18 '16 at 8:05
  • $\begingroup$ @DamodarRajbhandari. I am not sure there's a mistake. The brackets $\lfloor \cdot \rfloor$ indicate rounding down, so when I take (100-1)/2, I get 49.5. Rounding downward gives 49 as my overall answer, which seems correct. (1+99, 2+98, 3+97, ..., 47+53, 48+52, 49+51). $\endgroup$ – user326210 Jul 18 '16 at 8:07
  • $\begingroup$ Okey! Thanks for your solution sir! $\endgroup$ – user354230 Jul 18 '16 at 8:07

You've got a good intuition, but how to prove it ? Here's what I'd go with, it's voluntarily very detailed :

For all $A \in \mathbb{N}$, we'll be looking for $x, y \in \left[1; A-1\right]$ so that

$$ \cases{0<x<y \\ x+y = A} $$

And we'll try to determine how many values can $x$ take. If we know this, then we know how many combinations there are.

Let's start with finding the maximum value for $x$.

If $x=\frac{A+1}2$, then :

$$ x+y>x+x=A+1>A $$

So this $x$ does not fit our case. Therefore we must have $x<\frac{A+1}2$. Since $x$ is a natural integer, this is equivalent to $x\le\frac{A+1}2-1=\frac{A-1}2$.

We have a maximum bound $M=\frac{A-1}2$.

Now, does it work for every integer $x$ between $1$ and $\left\lfloor M\right\rfloor$ ?

It's pretty obvious that for any integer $x \in \left[1; \left\lfloor M\right\rfloor\right]$, there exists another integer $y \in \left[\left\lfloor M\right\rfloor+1; A-1\right]$ so that $x+y=A$, and $x \ne y$

Let's see for the limit cases :

  • $x=1$ : choose $y=A-1$, and $x+y=1+(A-1)=A$
  • $x=2$ : choose $y=A-2$, and $x+y=2+(A-2)=A$
  • ...
  • $x=\left\lfloor M\right\rfloor-1$ : choose $y=\left\lceil M\right\rceil+2$, and $x+y=2M+1=(A-1)+1=A$
  • $x=\left\lfloor M\right\rfloor$ : choose $y=\left\lceil M\right\rceil+1$, and $x+y=2M+1=(A-1)+1=A$

We have found that all $x$ between $1$ and $M$ works, and that $x$ cannot be lower than $1$ or greater than $M$. Thus we have found that there are exactly $\left\lfloor M\right\rfloor=\left\lfloor\frac{A-1}2\right\rfloor$ valid combinations.

For $A=100$, this gives : $$ \left\lfloor M\right\rfloor=\left\lfloor\frac{100-1}2\right\rfloor=\left\lfloor49.5\right\rfloor=49 $$

  • $\begingroup$ I will suggest you, Try to see the pattern in that problem by addressing $x+y=A$. $\endgroup$ – user354230 Jul 18 '16 at 15:40
  • $\begingroup$ Well you already did find the pattern, no need to repeat it (other answers explain it well, too). I wanted to give a mathematical proof, without searching for patterns but by counting how many combinations are possible for sure. You asked for another way, didn't you ? $\endgroup$ – BusyAnt Jul 19 '16 at 7:38
  • $\begingroup$ I thought you asked how I did! Sorry for that! $\endgroup$ – user354230 Jul 19 '16 at 11:36
  • $\begingroup$ Well, thanks for your concept to deal with this problem! Its great by using celing and floor concept! $\endgroup$ – user354230 Jul 19 '16 at 11:38
  • $\begingroup$ If any of the provided answers were useful to you, you can upvote them to let the author know about it. In the end, please think about accepting one of them so that we know your problem is resolved. :) $\endgroup$ – BusyAnt Jul 19 '16 at 12:02

$x+(100-x)=100$ so $x$ can go from $1$ to $49$ since $x+(100-x)$ and $(100-x)+x$ are considered as just one case and $x\ne 100-x$

  • $\begingroup$ Oh yes! It can be answered from this too. Can you explain it for $ A=2003 $? $\endgroup$ – user354230 Jul 18 '16 at 15:38
  • $\begingroup$ In $x+(2003-x)$ you can go till $x=1001$ because from $x=1002$ all the precedent sums are repeated by commutativity. $\endgroup$ – Piquito Jul 18 '16 at 15:53
  • $\begingroup$ How can you say "you can go till $x=1001$"? $\endgroup$ – user354230 Jul 18 '16 at 16:17
  • $\begingroup$ I mean predict? $\endgroup$ – user354230 Jul 18 '16 at 16:18
  • $\begingroup$ My English is weak, sorry. I want to help you. I mean $x=1,2,3,,,,,,1001$ so your solution is $1001$ $\endgroup$ – Piquito Jul 18 '16 at 16:42

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