$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?

Question

$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?

My approach:

For $x=1$, $z$ is not divisible by $6$.

For $x=2$, $z$ is divisible by $6$.

For $x=3$, $z$ is not divisible by $6$.

For $x=4$, $z$ is divisible by $6$.

For $x=5$, $z$ is not divisible by $6$.

For $x=6$, $z$ is not divisible by $6$.

For $x=7$, $z$ is divisible by $6$.

For $x=8$, $z$ is divisible by $6$.

I could not identify the pattern in these questions. Also can this problem be solved with a better approach?

Answer 1

Just a different way to highlight what has already been correctly noted, and to explain why the answer could be $66$, as stated in one of the initial comments by the author of the OP. To avoid confusion, I interpreted the problem as the question of how many values of $x$ lead to a positive $z$ that is divisible by $6$.

Let us start from $z=(100-x)(100+x)$, where the two factors $100-x$ and $100+x$ clearly have the same parity. First, we can observe that the only possibility that $z$ is divisible by $6$ occurs when at least one of these two factors is divisible by $6$. In fact, if any of the two factors is divisible by $3$ and not by $2$, the other factor is also not divisible by $2$, and so $z$ cannot be divisible by $6$.

Then, we can note that $100-x$ is divisible by $6$ only when $x=6j+4$ (with $j$ integer), whereas $100+x$ is divisible by $6$ only when $x=6j+2$.

If we limit the problem to solutions that give positive values of $z$ (although this is not specified in the OP), then $|x|$ can only assume integer values $<100$. It is not difficult to note that there are $16$ positive integers $<100$ of the form $x=6j+4$ (they are $4,10,16,22...94$) and $17$ positive integers $<100$ of the form $x=6j+2$ (they are $2,8,14,20...98$). So, there are $33$ positive integer values of $x$ that lead to a $z$ value that is divisible by $6$.

Lastly, because each of these solutions with a given positive $x$ corresponds to another symmetric solution with $-x$, we have to count other $33$ negative values of $x$. This finally leads to a total of $66$ solutions.

Answer 2

HINT:

$$z=100^2-x^2=(100-x)(100+x)$$

As $(100-x)+(100+x)=200,100\pm x$ have same parity and

if one is divisible by $3,$the other is not

So if $2|z,100\pm x$ must be even

If $3|z,3|(100-x)(100+x)\implies$

either $3|(100-x)\iff x\equiv1\pmod3$

or $3|(100+x)\iff x\equiv-1\pmod3$

Answer 3

I think a good idea for the problem is using modulos. So, the problem can be written like this $$ z=4-x^2 \text{ }mod(6), $$ We know that $100=96+4=4\quad mod(6)$. So $100^2=(100)*(100)=4*4=16=12+4=4\quad mod(6)$. If $z$ is divisible for $6$, then $z=0\quad mod(6)$. For all of this we have to find $x$ such that $$ x^2=4\quad mod(6). $$ We have to make a table where we can analize $x^2\quad mod(6)$.

For $x=0$, $x^2=0$ and $x^2=0\quad mod(6)$.

For $x=1$, $x^2=1$ and $x^2=1\quad mod(6)$.

For $x=2$, $x^2=4$ and $x^2=4\quad mod(6)$.

For $x=3$, $x^2=9$ and $x^2=3\quad mod(6)$.

For $x=4$, $x^2=16$ and $x^2=4\quad mod(6)$.

For $x=5$, $x^2=25$ and $x^2=1\quad mod(6)$.

So, the last proposition implies that $x=2 \quad mod(6)$ or $x=4 \quad mod(6)$.

In summary, if $x=2+6m$ or $x=4+6n$ for $m,n\in\mathbb{Z}$. Then, $z$ is divisible by $6$. There are infinity solutions.