Number trick

[Mark44]

I posted this in one of the math technical forum sections a few weeks ago, but didn't get much in the way of responses. In hopes of getting more involvement, I thought I would give it another shot here in General Discussion.

Here goes:

1. Write down a 3-digit positive integer. The only restriction is that the first and third digits should be different.
2. Reverse the digits of the number you started with. (If the first number were, say 340, the new number would be 43.) Edit: if you end up with a 2-digit number, write this number with a leading 0.
3. Subtract the smaller of the two numbers from the larger.
   For example, if you started with 340, the new number would be 43. Subtracting results in 340 - 43 = 297.
4. Now take the number you ended up with (297 in the example above) and reverse its digits.
5. Add the new number (792 in this example) to the number you ended with in step 3. If your arithmetic is correct, you'll always end up with exactly the same number. What number do you get?

I came across this "trick" in a book I came across whose name was "Number 9" or something similar. I can't recommend the book as it was full of a bunch of New Age woo and numerology. The only bit that was interesting to me was this trick. The book provided no explanation of how or why it worked, so I devoted about three pages of scratch paper to determine what was going on and prove to myself that you always do get the same result.

I've showed this trick to my 8-year-old grandson (who had a little help from his grandmother). The trick absolutely blew his mind, so much so that he drew several hearts on the paper he was using. I also showed it to my grandniece, who also loved it. I'm confident that she will show this trick to her friends at school.

 

[Hill]

One can see what is going on quite fast, e.g., like this:

After the subtraction on step 3, the middle digit disappears from the game. So, it is sufficient to consider just a number of the form $100a+b$, with the digits $a$ and $b$. After reversing it becomes $100b+a$, and after subtracting, $99|a-b|$.

That is, one of the nine numbers: 99, 198, 297, 396, 495, 594, 693, 792, 891. There is always 9 in the middle and the sum of the other two digits is always 9. So, after reversing any of these numbers and adding, one always gets 1089.

P.S. Starting with a 2-digit number, one always gets 99.

 

[mcastillo356]

If I have understood, your example ends up with $999$

Counterexample

1- $495$
2- $594$
3- $594-495=99$
4- $99$
5- $99+99=198$

$999\neq{198}$

 

[Hill]

If I have understood, your example ends up with $999$

Counterexample

1- $495$
2- $594$
3- $594-495=99$
4- $99$
5- $99+99=198$

$999\neq{198}$

It's $099+990=1089$.

 

[Gavran]

I posted this in one of the math technical forum sections a few weeks ago, but didn't get much in the way of responses. In hopes of getting more involvement, I thought I would give it another shot here in General Discussion.

Here goes:

1. Write down a 3-digit positive integer. The only restriction is that the first and third digits should be different.
2. Reverse the digits of the number you started with. (If the first number were, say 340, the new number would be 43.) Edit: if you end up with a 2-digit number, write this number with a leading 0.
3. Subtract the smaller of the two numbers from the larger.
   For example, if you started with 340, the new number would be 43. Subtracting results in 340 - 43 = 297.
4. Now take the number you ended up with (297 in the example above) and reverse its digits.
5. Add the new number (792 in this example) to the number you ended with in step 3. If your arithmetic is correct, you'll always end up with exactly the same number. What number do you get?

I get $1089$.
There is a number $a=100x+10y+z$ and there is a number $b=100z+10y+x$ where $a\gt b$.
$c=a-b=99(x-z)=9\cdot11\cdot(x-z)\implies c\equiv0\mod11$
$(c=100x’+10y’+z’\wedge c\equiv0\mod11)\implies x’+z’=y’$
$c=99x’+11y’=11(9x’+y’)$

$c=9\cdot11\cdot(x-z)\implies c\equiv0\mod9$
$(c=11(9x’+y’)\wedge c\equiv0\mod9)\implies y’=9$

$d=100z’+10y’+x’$
$c+d=100x’+10y’+z’+100z’+10y’+x’=121y’=1089$

 

[mcastillo356]

It's $099+990=1089$.

How does my arithmetic end up with $1089$?. I don't understand. Maybe missing something of the statement?.
Attempt
The statement imposes to arrange the digits so they must be always three.

 

[anuttarasammyak]

Interesting.

Spoiler

1. 100a+10b+c
2. 100c+10b+a
3. (100-1)|a-c|= 100(|a-c|-1)+90+(10-|a-c|)
4. 100 (10-|a-c|)+90+(|a-c|-1)
5. 900+90+90+9=1089

In general case of base-d system, the result value is
$d^3+d^2-d-1=(d-1)(d+1)^2$
For 2-digit it is
$(d-1)(d+1)$

 

[Mark44]

I get 1089.

Right.

How does my arithmetic end up with 1089?.

In your step 3, you ended up with 99, which is the same as 099. Reversing the digits of the latter 3-digit number results in 990 so that 099 + 990 = 1089.

When I wrote the problem I first stated that if you ended up with a 3-digit number in step 2, you need to tack on a leading 0. The example I chose to elucidate the process didn't need that additional leading 0, so I revised what I wrote. Obviously, that was the wrong call. I have revised that step in post #1.

 

[Hill]

Note that each of these is a multiple of 99.

Yes, that's what I said:

and after subtracting, 99|a−b|.

 

[Mark44]

Some of the things about this problem are interesting to me because of the number theoretic concepts involved, notably that the result of the subtraction in step 3 is divisible by 9 as well as 11 and that the final result from step 5 is always the same multiple of 99.

Suppose you start with a 3-digit number in step one, say abc, where a, b, and c represent the digits of this number. It will fall into one of 9 possible equivalence classes modulo 9. That is, $abc \equiv n \mod 9$.
I'll state without proof (but the proof is easy) that reversing the digits doesn't change the equivalence class modulo 9. I.e., $cba \equiv n \mod 9$.

Assuming without loss of generality that a > c, the result abc - cba will be divisible by 9. That is, $abc - cba \equiv 0 \mod 9$.

Furthermore, the only possible 3-digit results can be only 099, 198, 297, 396, 495, 594, 693, 792, and 891, all of which have 9 as their middle digit and all of which are multiples of 99. This means that in addition to being divisible by 9, they are all divisible by 11. A somewhat arcane rule about divisibility by 11 is that if the even-index digits of an integer have the same sum as the odd-index digits, the number must be divisible by 11. By virtue of this rule one can see at a glance that 101200 is divisible by 11.

If we extend this list of the previous paragraph by one more value, 990, this will include all possible reversed values of the original list. Let $L = \{n_0, n_1, n_2, \dots, n_9 \}, k = 0, \dots, 9$ be the numbers 099, 198, and so on, with $n_9$ being 990. In the subtraction process of step 3, if you end up with $n_k$ on this list, reversing the digits in step 4 results in $n_{9 - k}$. Adding these two numbers always results in a sum of 1089.

 

[Gavran]

In general case of base-d system, the result value is
$d^3+d^2-d-1=(d-1)(d+1)^2$

Great.
The formula can also be written in the form of $(10_d-1_d)\cdot11_d\cdot11_d$.