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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 $x$)?

rschwieb
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David Mulder
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2 Answers2

9

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.

Mark Viola
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  • It definitely makes the brute force approach far more comprehensible, though it definitely feels quite... ungeneric. – David Mulder Jun 21 '15 at 16:37
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    @DavidMulder Well, for those less mathematically inclined, sometimes a more general approach doesn't connect. We are therefore left with keeping things simple and more specific. – Mark Viola Jun 21 '15 at 16:40
  • Yeah, definitely true, I guess I was chasing two different birds and trying to hit them with the same stone: Not gonna work. – David Mulder Jun 21 '15 at 16:42
  • Would the downvoter care to comment? – Mark Viola Oct 07 '22 at 14:32
8

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 $9$.

please delete me
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  • This was I guess the answer I was looking for, even if it helps me less in communicating it. In the end this is the same thing as taking a good look at the puzzle and just seeing that the answer is always 9. I guess it's just the programmer in me that wishes to have a model like that, and then running the calculations till I get the answer I am looking for. – David Mulder Jun 21 '15 at 16:38
  • @David See casting out nines for the arithmetical essence of the matter. – Bill Dubuque Jun 21 '15 at 16:50
  • @DavidMulder Since you wrote $\mod{10}$ I suppose that you are familiar with modular arithmetic. Given a natural number $n$ and the fact that $10 \equiv 1 \pmod{9}$ you can prove (in a fairly straightforward way) that $n$ is congruent to the sum of its digits, modulo $9$. Since here $n = 9(x+1)$ has two digits it doesn't take much to spell the proof out for your friends that don't know about modular arithmetic. – A.P. Jun 21 '15 at 16:54