Find the number of sums that you will get A, when A=100?

Question

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.

Answer 1

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?

Answer 2

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.

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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 $M=\frac{A-1}2$.

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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 $x+y=2M+1=(A-1)+1=A$

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

Answer 3

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