5. The absolute value of a number $x$ is equal to the distance from 0 to $x$ along a number line and is written as $|x|$. For example, $|8|=8,|-3|=3$, and $|0|=0$. For how many pairs $(a, b)$ of integers is $|a|+|b| \leq 10$ ?
5. Since $|a|$ is at least 0 and $|b|$ is at least 0 and $|a|+|b| \leq 10$, then $|a|$ is at most 10 and $|b|$ is at most 10 .
We count the number of possible pairs $(a, b)$ by working through the possible values of $|a|$ from 0 to 10 .
Suppose that $|a|=0$. This means that $a=0$. There is 1 possible value of $a$ in this case. Since $|a|=0$ and $|a|+|b| \leq 10$, then $|b| \leq 10$ which means that the possible values for $b$ are $-10,-9,-8, \ldots,-1,0,1, \ldots, 8,9$, or 10 . There are 21 possible values of $b$ in this case.
Since there is 1 possible value for $a$ and there are 21 possible values for $b$, then overall there are $1 \times 21=21$ pairs $(a, b)$ when $|a|=0$.
Suppose that $|a|=1$. This means that $a=1$ or $a=-1$. There are 2 possible values of $a$ in this case.
Since $|a|=1$ and $|a|+|b| \leq 10$, then $|b| \leq 9$ which means that the possible values of $b$ are $-9,-8,-7, \ldots,-1,0,1, \ldots, 7,8$, or 9 . There are 19 possible values of $b$ in this case.
Since there are 2 possible values for $a$ and 19 possible values for $b$, then overall there are $2 \times 19=38$ pairs $(a, b)$ when $|a|=1$.
Suppose that $|a|=2$. This means that $a=2$ or $a=-2$. There are 2 possible values of $a$ in this case.
Here, $|b| \leq 8$ which means that $b$ could equal $-8,-7,-6, \ldots,-1,0,1, \ldots, 6,7$, or 8 . There are 17 possible values of $b$ in this case.
Overall, there are $2 \times 17=34$ pairs $(a, b)$ when $|a|=2$.
As $|a|$ increases from 2 to 9 , at each step, the largest possible value of $|b|$ will decrease by 1 , which means that there will be 2 fewer possible values of $b$ from each step to the next. Since there are 2 possible values for $a$ at each step, this means that there will be $2 \times 2=4$ fewer pairs $(a, b)$ at each step.
We check the final case $|a|=10$ to verify that nothing different happens in the last case. Suppose that $|a|=10$. This means that $a=10$ or $a=-10$. There are 2 possible values of $a$ in this case.
Here, $|b| \leq 0$ which means that $b$ can only equal 0 . There is 1 possible value of $b$ in this case. Overall, there are $2 \times 1=2$ pairs $(a, b)$ when $|a|=10$.
In total, this means that there are
$$
21+38+34+30+26+22+18+14+10+6+2
$$
pairs $(a, b)$ with $|a|+|b| \leq 10$.
Grouping the last 10 numbers in pairs from the outside towards the middle we obtain
$$
21+(38+2)+(34+6)+(30+10)+(26+14)+(22+18)
$$
which equals $21+5 \times 40$ or 221 .
Thus, there are 221 pairs.
(This problem can also be solved using a neat result called Pick's Theorem. We encourage you to look this up and think about how you might apply it here.)
ANSWER: 221