Recently I discovered this video

In the video the man asks you to think of a three digit number, then through a series of steps, he makes your number into \(1089\) every single time.

The purpose of this note is to explain how he does it.

First let's list the steps.

- Think of a three digit number with unique digits (all of them must be different)
- Find the reverse of that number (swap the first and last digits)
- Take the largest of the two numbers and subtract the other one
- Find the reverse of that number
- Sum the reverse and the number you just worked out, you should get \(1089\) as the answer

Now let's translate each step into an algebraic formula.

Step 1: A three digit number with unique digits is represented by the following notation:

\[\overline{ABC} = 100A + 10B + C\]

With each letter representing a different digit.

Step 2: The reverse of that first number can be represented using the same notation just with the letters flipped horizontally.

\[\overline{CBA} = 100C + 10B + A\]

Again with each letter representing a different digit.

Step 3: This step depends on which of the two numbers (\(\overline{ABC}\) or \(\overline{CBA}\)) is larger. Either way will work but I'll show both anyway.

If \(\overline{ABC} > \overline{CBA}\) (This also implies \(A > C\))

\[\overline{ABC} - \overline{CBA} = 99(A - C)\]

If \(\overline{CBA} > \overline{ABC}\) (This also implies \(A < C\))

\[\overline{CBA} - \overline{ABC} = 99(C - A)\]

To make things easier we'l make \(x = A - C\), this makes the equations read:

\[\overline{ABC} - \overline{CBA} = 99x\]

\[\overline{CBA} - \overline{ABC} = -99x\]

The second equation comes about because \(C - A = -(A - C) = -x\). I'll use the first equation from now on and prove that the second equation works later.

We'll say for now that this new number we've gotten is \(\overline{DEF}\), I'll show what each of those letters equals later on, we're just listing the steps as formula for now.

Step 4: The new number we have is equal to \(\overline{DEF}\) so it's reverse is \(\overline{FED}\).

Step 5: The final step can be represented by the equation below:

\[\overline{DEF} + \overline{FED} = 1089\]

Now that we have the steps sorted out, we find we only have two unknowns (technically three) out of all of the steps. The two unknowns are \(\overline{DEF}\) and \(\overline{FED}\), however we only need to find out one to figure out the other. It's not as easy as you might think.

We know \(\overline{DEF} = 99x\) but that doesn't help us find the three digits we need in order to calculate \(\overline{FED}\).

Let's start by taking a step back, what's \(99x\) equal to?

\[99x = \overline{ABC} - \overline{CBA}\]

\[99x = (100A + 10B + C) - (100C + 10B + A)\]

We can use this to find the separate digits, since each digit in \(\overline{DEF}\) is equal in some way to the corresponding digits in the two previous numbers. In other words:

\[D = A - C\]

\[E = B - B = 0\]

\[F = C - A\]

So now \(\overline{DEF} = 100(A-C) + 10(0) + (C-A)\)

We already said that \(x = A - C\) so we'll substitute that in

\[\overline{DEF} = 100x - x = 99x\]

We've now just proved that we still have the same number, so we'll stop using \(x\) for now and switch back to using the three digits.

\[\overline{DEF} = 100(A - C) + 10(0) + (C - A)\]

Now we have a problem, \(A > C\) so \(C - A < 0\), this doesn't make any sense since a digit can't be negative. To fix this problem we'll take a \(10\) from the second digit and add it to the third digit.

\[\overline{DEF} = 100(A - C) + 10(-1) + (10 + (C - A))\]

We've encountered the same problem again so we'll take a \(100\) from the first digit and add it to the second.

\[\overline{DEF} = 100(A - C - 1) + 10(10 - 1) + (10 + (C - A))\]

Now that all the digits are positive and below \(10\), we can use the notation.

\[\overline{DEF} = \overline{(A - C - 1)9(10 + (C - A))}\]

Each set of brackets represent a digit in this case. We'll switch back to using \(x\) to make it easier to calculate.

\[\overline{DEF} = \overline{(x - 1)9(10 - x)}\]

We can now calculate \(\overline{FED}\)

\[\overline{FED} = \overline{(10 - x)9(x - 1)}\]

We know \(\overline{DEF} = 99x\) so we just need to calculate \(\overline{FED}\) now

\[\overline{FED} = 100(10 - x) + 10(9) + (x - 1)\]

\[\overline{FED} = 1000 - 100x + 90 + x - 1\]

\[\overline{FED} = 1089 - 99x\]

Well we found the \(1089\).

The last step involves summing the two opposites so we'll do that.

\[\overline{DEF} + \overline{FED} = (99x) + (1089 - 99x)\]

\[\overline{DEF} + \overline{FED} = 1089 + (99x - 99x)\]

\[\overline{DEF} + \overline{FED} = 1089\]

That shows what happens when \(\overline{ABC} > \overline{CBA}\), so what happens when \(\overline{CBA} > \overline{ABC}\).

The exact same thing happens, the only difference to the equations is that \(x = C - A\).

Hope this wasn't too long for a note.

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## Comments

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TopNewestYou have the general idea, though step 5 can be simplified much more. For example, there are only 10 possibilities, and it would be easier to check all the cases than work through your algebra. Think about how we can improve this proof further.

Can you also add this to Mind reading? Thanks!

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Does it work for the number 132 ??

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\(231 - 132 = 099\)

\(099 + 990 = 1089\)

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LOL thanx.No wonder i was getting it wrong.....

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