Maths trick

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 10891089 every single time.

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

First let's list the steps.

  1. Think of a three digit number with unique digits (all of them must be different)
  2. Find the reverse of that number (swap the first and last digits)
  3. Take the largest of the two numbers and subtract the other one
  4. Find the reverse of that number
  5. Sum the reverse and the number you just worked out, you should get 10891089 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:

ABC=100A+10B+C\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.

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

Again with each letter representing a different digit.

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

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

ABCCBA=99(AC)\overline{ABC} - \overline{CBA} = 99(A - C)

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

CBAABC=99(CA)\overline{CBA} - \overline{ABC} = 99(C - A)

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

ABCCBA=99x\overline{ABC} - \overline{CBA} = 99x

CBAABC=99x\overline{CBA} - \overline{ABC} = -99x

The second equation comes about because CA=(AC)=xC - 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 DEF\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 DEF\overline{DEF} so it's reverse is FED\overline{FED}.

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

DEF+FED=1089\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 DEF\overline{DEF} and FED\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 DEF=99x\overline{DEF} = 99x but that doesn't help us find the three digits we need in order to calculate FED\overline{FED}.

Let's start by taking a step back, what's 99x99x equal to?

99x=ABCCBA99x = \overline{ABC} - \overline{CBA}

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

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

D=ACD = A - C

E=BB=0E = B - B = 0

F=CAF = C - A

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

We already said that x=ACx = A - C so we'll substitute that in

DEF=100xx=99x\overline{DEF} = 100x - x = 99x

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

DEF=100(AC)+10(0)+(CA)\overline{DEF} = 100(A - C) + 10(0) + (C - A)

Now we have a problem, A>CA > C so CA<0C - A < 0, this doesn't make any sense since a digit can't be negative. To fix this problem we'll take a 1010 from the second digit and add it to the third digit.

DEF=100(AC)+10(1)+(10+(CA))\overline{DEF} = 100(A - C) + 10(-1) + (10 + (C - A))

We've encountered the same problem again so we'll take a 100100 from the first digit and add it to the second.

DEF=100(AC1)+10(101)+(10+(CA))\overline{DEF} = 100(A - C - 1) + 10(10 - 1) + (10 + (C - A))

Now that all the digits are positive and below 1010, we can use the notation.

DEF=(AC1)9(10+(CA))\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 xx to make it easier to calculate.

DEF=(x1)9(10x)\overline{DEF} = \overline{(x - 1)9(10 - x)}

We can now calculate FED\overline{FED}

FED=(10x)9(x1)\overline{FED} = \overline{(10 - x)9(x - 1)}

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

FED=100(10x)+10(9)+(x1)\overline{FED} = 100(10 - x) + 10(9) + (x - 1)

FED=1000100x+90+x1\overline{FED} = 1000 - 100x + 90 + x - 1

FED=108999x\overline{FED} = 1089 - 99x

Well we found the 10891089.

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

DEF+FED=(99x)+(108999x)\overline{DEF} + \overline{FED} = (99x) + (1089 - 99x)

DEF+FED=1089+(99x99x)\overline{DEF} + \overline{FED} = 1089 + (99x - 99x)

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

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

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

Hope this wasn't too long for a note.

#Algebra #NumberTheory #Trick #Digits

Note by Jack Rawlin
5 years, 8 months ago

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Comments

You 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!

Calvin Lin Staff - 5 years, 7 months ago

Does it work for the number 132 ??

Athiyaman Nallathambi - 5 years, 7 months ago

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231132=099231 - 132 = 099

099+990=1089099 + 990 = 1089

Jack Rawlin - 5 years, 7 months ago

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

Athiyaman Nallathambi - 5 years, 7 months ago
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