If $x \equiv 23 \pmod{36}$, what is $x \bmod 9$ and why?

Question

If $x \equiv 23 \pmod{36}$, what is $x \bmod 9$ and why? I got this question in school today. I know the answer is $5,$ but I don't know how to prove it.

Answer 1

====== Preface ======

In response to a comment by Bill Dubuque, I feel I must make it clear that $x \mod 9$ (or $x \pmod {9}$) is not a value as result of an operation, but a statement about how $x$ is related to some other number in terms of equivalence modulo $9$.

"What is $x \mod 9$" doesn't actually make any sense (although it is clear what you meant). $x \equiv 23 \pmod{36}$ means that $x$ and $23$ are related in a certain way. (Namely that $36| x- 23$). It doesn't meant that there is some number equal to $23 \mod 36$ or a number equal to $x \mod 9$.

It is similar to saying $x < 23$ describes a relationship between $x$ and $23$ (namely that $x$ is less than $23$). It doesn't mean that there is a number equal to $< 23$ and $x$ is that number.

So in a way "What is $x \mod 9$?" is a bit like saying "What is $x>$?"

What it should be is what number or numbers (there are more than one) is $x$ equivalent to modulo $9$.

Sorry for the pendantry. But it is aan important (albeit subtle) distinction.

And now back to the answer:

=====================

Just do it:

$x \equiv 23 \pmod{36}$ means that there is an integer $k$ so that $x = 23 + 36k$

So $x = 23 + 9*(4k)$. Which means $x \equiv 23 \pmod 9$. And that can be (is) our final answer.

We can reduce that further to $x \equiv 23 \equiv 5 \pmod 9$, if the question was going for an answer where $0 \le x < 9$ but the question doesn't state that that is a requirement. So any $x = 5 + 9j$ for an integer $j$ will be equally correct. So it's perfectly acceptable to say $x \equiv 23 \pmod 9$.

(When we noted $x = 23 + 9*(4k)$ we could take that further $x = 5 + 18 + 9*(4k) = 5 + 9*(4k + 2)$ so $x \equiv 5 \pmod 9$.

......

That's the answer... but there's more....

It should be clear:

Lemma: If $x \equiv a \pmod n$ and $m|n$, then $x \equiv a \pmod m$.

Pf: $x\equiv a \pmod n \iff n|x-a$. So if $m|n$ and $n|x-a$ then $m|n-a$ so $x \equiv a \pmod m$. (But I think that proof is a little "cold" and distant and won't be as convincing to students and the argument above. If $x \equiv a \pmod n$ then there is a $k$ so $x = a+k*n$. But $m|n$ so there is a $j$ so that $n = m*j$ os $x = a + (jk)*m$. So $x \equiv a \pmod m$.

......

Its important to notice it does not go in the reverse order:

If $x \equiv 5 \pmod 9$ it does not follow that $x \equiv 5 \pmod{36}$. (After all $23 \equiv 5 \pmod 9$ but $23 \not \equiv 5 \pmod{36}\,.$)

That is if $x \equiv a \pmod m$ and $m|n$ it does not follow that $x\equiv a \pmod n$.

If $x \equiv a \pmod m$ then there is a $j$ so that $x = a + j*m$ and if $m|n$ there there is a $k$ so that $n = km$ so $x = a + j*\frac nk= a + \frac jk*n$ but we don't know that $\frac jk$ is an integer.

However we do know that $x \equiv a + j*m \pmod n$ for some integer $j$.

So if $x \equiv 5 \pmod 9$ then $x \equiv 5 + 9j \pmod{36}$ for some $j$.

Example: $5 \equiv 5 \pmod 9$ and $5 \equiv 5 + 9*0 \pmod{36}$.

And $14 \equiv 5 \pmod 9$ and $14 \equiv 5 + 9*1 \pmod{36}$.

And $23 \equiv 5 \pmod 9$ and $23 \equiv 5 + 9*2 \pmod{36}$.

Answer 2

Hint: $x \equiv 23 \pmod{36}$ means that $(x-23)$ is a multiple of $36$.

Answer 3

If $x \equiv 23 \pmod{36}$ then $x = 36q+23 = 9(4q+2)+5 \equiv 5 \pmod{9}.$

Answer 4

Always use the definitions. What does $x\equiv 23 \mod 36$ mean? It means that $(x-23)/36$ is an integer. So give that a name: $(x-23)/36=n\in \Bbb Z.$ So $x-23=36 n,$ so $x=23 +36 n=(9)(2+4n)+5.$ And $2+4n\in \Bbb Z$ because $n\in \Bbb Z$. Therefore $x$ is $5$ more than an integer-multiple of $9,$ so $x\equiv 5 \mod 9.$

BTW.If you asked me the Q in your title I could logically answer "It's $x$" because as stated it does't actually ask for a value from $0$ to $8$ even though we can presume that was the intent.