How many pairs of $(x,y)$ satisfy $\lvert x\rvert + \lvert y\rvert < 100$

Question

How many integer pairs of $(x,y)$ satisfy $ \lvert x\rvert + \lvert y\rvert < 100$?

$(1,2)$ and $(2,1)$ are considered different pairs. I don't even know how to start.

Answer 1

The highest possible integer for any $x$ or $y$ is $99$.

If $x=99$ then $y=0$.
If $x=98$ then $y=-1, 0, 1$.
If $x=97$ then $y=-2, -1, 0, 1, 2$.
...
If $x=1$ then $y=-98, -97,... 97, 98$.
If $x=0$ then $y=-99, -98,... 98, 99$.
If $x=-1$ then $y=-98, -97,... 97, 98$.
...
If $x=-97$ then $y=-2, -1, 0, 1, 2$.
If $x=-98$ then $y=-1, 0, 1$.
If $x=-99$ then $y=0$.

So, for $-99≤x≤-1$ there are $99$ integer values of $x$, same goes for $1≤x≤99$. The answer can be stated like this.
$$\bigg(2\sum_{n=1}^{99}(2n-1)\bigg)+199$$

Answer 2

HINT:

$x=0, n(y) = 2 \times 99+1\\ x=\pm1, n(y)=2 \times 98+1\\ .\\.\\ x=\pm98, n(y)=2\times1+1\\ x=\pm99, n(y)=2\times 0+1$

Answer 3

I'll guess both $x$ and $y$ must be in $\mathbb{Z}$? Assuming that, here's an elementary way to do it:

An idea might be to count how many solutions you have with $x=1$ and $y$ positive. Then with $x=2$, and so on until $x=99$.

For $x=1$ you have $y=1,...,98$. That is 98 solutions. For $x=2$ you have $y=1,...,97$. That is 97 solutions.

Iterating for $x$ from $1$ to $98$, you get that you have to sum $98+97+...+2+1$. You can do via the trick that Gauss invented. Do you know it?

Once you do it, you'll have to multiply the result by $2$ because $y$ could also be negative for each $x$. And you must also consider that $y$ might be zero.

After everything, you do the same for $x$ negative (or just multiply what you got before by $2$) and add the case $x=0$ to compute the final result.

Answer 4

Guide:

For $x=-99$, $|y|<1$, $y=0$.

For $x=-98$, $|y|<2$, $y \in \{ -1,0,1\}$

For $x=-97$, $|y|<3$, $y \in \{ -2,-1,0,1,2\}$.

You should be able to see an arithmetic progression up to $x=0$.

From $x=1$ to $x=99$, you shoud see another arithmetic progression.

Answer 5

We first consider the case when both $x$ and $y$ are positive. Obviously, $1\le x\le 99$ and $1\le y\le 99$. There are totally $99^2=9801$ pairs of possible $(x,y)$.

Among them, $99$ pairs ($(x,y)=(1,99), (2,98),\dots,(99,1)$) will give $x+y=100$.

Half of the remaining pairs give $x+y<100$ and the other half give $x+y>100$.

So, the number of pairs of positive $(x,y)$ such that $x+y<100$ is $(9801-99)\div 2=4851$.

By symmetry, the number of pairs of $(x,y)$ such that both $x$ and $y$ are non-zero and $|x|+|y|<100$ is $4851\times 4=19404$.

Take $x=0$, $y$ can be any among $-99,-98,\dots,99$. So we have $199$ more solutions.

There are also $199$ solutions with $y=0$.

Obviously, $(0,0)$ is double counted. So the number of solutions is $19404+2\times 199-1=19801$.

Answer 6

It is equivalent to $$ \left\{ \matrix{ x = \pm u\quad y = \pm v \hfill \cr 0 \le u \le 99 \hfill \cr 0 \le v \le 99 \hfill \cr 0 \le u + v \le 99 \hfill \cr} \right. $$

Now, $$ \left\{ \matrix{ 0 \le u \le 99 \hfill \cr 0 \le v \le 99 \hfill \cr 0 \le u + v \le 99 \hfill \cr} \right. $$ represents the points, with integer coordinates that lies in a cube $100 \times 100$, below and on the diagonal plane.

So $$ cube = \left\{ {(u,v)} \right\}:\left\{ \matrix{ 0 \le u \le 99 \hfill \cr 0 \le v \le 99 \hfill \cr 0 \le u + v \le 98 \hfill \cr} \right. \cup \left\{ {(u,v)} \right\}:\left\{ \matrix{ 0 \le u \le 99 \hfill \cr 0 \le v \le 99 \hfill \cr u + v = 99 \hfill \cr} \right. \cup \left\{ {(u,v)} \right\}:\left\{ \matrix{ 0 \le u \le 99 \hfill \cr 0 \le v \le 99 \hfill \cr 100 \le u + v \le 198 \hfill \cr} \right. $$ that is $$ 10^{\,4} = 2X + 100\quad \Rightarrow \quad C = X + 100 = 5050\;couples\,(u,v) $$

Of these, the couple $(0,0)$ will correspond to $(x,y)=0$, the couples $(0,\{1,\cdots,99\})$and $(\{1,\cdots,99\}, 0)$ to two $(x,y)$ couples and the remaining to four. Therefore the answer is $$ N = 1 + 2\cdot 198 +4 \cdot (5050-199) = 19801 $$