The sum of digits of $3(3x+3)$ is always $9$ for any $x$ between $1$ and $9$

Question

Given the following 'joke' I stumbled across today

This calculation will tell you who you really are:

• Pick any number between 1 to 9.
• Now multiply it with 3.
• Add 3.
• Now again multiply it with 3.
• You will get 2 digits, now add them.

Now this number will tell you what you really are...

The rest is not really relevant, but it's easy enough to figure out that the answer is always 9. However when I tried to 'prove' this for friends in an easy manner I stumbled upon two problems.

First of all I had to add the different parts, which I wasn't able to do with 'simple' maths, so I used $\lfloor \frac{x}{10} \rfloor + x \mod 10$. Due to the jump in mathematical functions I had no idea how to continue with 'pure mathematics' so I wanted to get WolframAlpha to simply do the work for me and spit out the number 9, but I wasn't able to get it to work only within $1\leq x \leq 9,\ x \in \mathbb{Z} $.

Either way, practically speaking I have

$\lfloor \frac{(x \times 3 + 3) \times 3}{10} \rfloor + ((x \times 3 + 3) \times 3 \mod 10)\ \ \ where\ 1\leq x \leq 9,\ x \in \mathbb{Z}$

How would one solve this, either using normal mathematics or how to get this inputted in WolframAlpha in a way it will be understood (here is the WA link without the restriction on $x$)?

Answer 1

For any integer $k$, $3(3k+3)=9(k+1)$ is a multiple of $9$, and one can prove that such a number has digital root $9$.

Answer 2

You can show them that they are adding the digits of multiples of $9$ between $18$ and $90$ and then just list them.

$18 \implies 1+8=9$

$27\implies 2+7=9$

$36\implies 3+6=9$

$45\implies 4+5=9$

$54\implies 5+4=9$

$63\implies 6+3=9$

$72\implies 7+2=9$

$81\implies 8+1=9$

$90\implies 9+0=9$

While this approach is not sophisticated, it might be fit-for-purpose when discussing with those who are not mathematically inclined.