## The 1089 Trick

Those who like to read books about puzzles or recreational mathematics would likely be familiar with the trick that involves the number 1089. The trick begins by asking someone to select a three-digit number that has a different first and last digits, reverse it to create a new number, and subtract the smaller number from the larger one. For instance, someone selects 457:

754 – 457 = 297

Afterward, ask him to add the answer that he got to the reverse of itself:

297 + 792 = 1089

No matter which three-digit number he selects, as long as he correctly followed all the instructions, the answer will always be 1089.

## Why Does It Work?

To explain why this trick works every time, we need to use simple algebra. The proof is based on the proof provided in the book *Recreations in the Theory of Numbers* by Albert Beiler but I further simplified the algebra used in the proof for easier understanding. For a more detailed proof, you can refer to that book.

Let a be the initial hundreds digit, b the initial tens digit and C the initial ones digit. Thus, the three-digit number chosen by someone is represented by abc, then abc is subtracted to cba, its reversal. The initial three-digit number can then be written as:

(a – 1) + (b + 9) + (c + 10)

Now, deduct c + b + a from (a – 1) + (b + 9) + (c + 10):

Hundreds Tens Ones

a – 1 b + 9 c + 10

__– c b a__

a – 1 – c 9 c + 10 – a

Finally, add the answer to the reverse of itself:

a – 1 – c 9 c + 10 – a

__+ c + 10 – a 9 a – 1 – c__

9 18 9

Thus, the answer will always be 1089.

## Other Properties of 1089

Aside from the property mentioned above, 1089 has more properties which may not be that well-known:

- If you multiply 1089 by 9, the answer is 9801, which is the reversal of 1089.
- Also, 108910891089 × 9 = 980198019801.
- 1089 = 33
^{2}and 9801 = 99^{2}. These can be continued:

33^{2} = 1089

333^{2} = 110889

3333^{2} = 11108889

33333^{2} = 1111088889

etc.

99^{2} = 9801

999^{2} = 998001

9999^{2} = 99980001

99999^{2} = 9999800001

etc.

1089 × 1 = 1089 9801 = 1089 × 9

1089 × 2 = 2178 8712 = 1089 × 8

1089 × 3 = 3267 7623 = 1089 × 7

1089 × 4 = 4356 6534 = 1089 × 6

1089 × 5 = 5445

- 1089 × 9801 = 10673289. Note that the concatenation of the first two and the last two digits of the product is 1089 and the square root of 10673289 is equal to 3267
^{2}. Interestingly, the digits of 3267 are the cyclic permutation of the middle four digits of 3267^{2 }(10673289). Lastly, also note that 3267 is equal to 1089 × 3.

**Magic Squares**

The number 1089 can also be connected to magic squares. A magic square is an arrangement of numbers from 1 to *n ^{2} *in an

*n*matrix in which each number should only occur once. Every row, column, and diagonal of a magic square have the same

^{2}**magic sum**

*which is similar to an addition magic pentagram that I’ve discussed before where each sum of the vertices is the same. The formula for determining the magic sum*

**,***S*of any

*n*order magic square is:

The smallest possible magic square is order 3, which is also called the Luo Shu square. Using the formula above, we can easily figure that the magic sum for an order 3 magic square is 15:

Here is an order 3 magic square with a magic sum of 15:

There is only a single magic square of order 3 excluding rotations and reflections. On the other hand, there are 880 solutions for order 4 magic squares and 275305224 solutions for order 5 (A006052)! The number of order 6 magic squares and up is still unknown.

## 1089 Magic Squares

This section shows how the number 1089 is connected with order 3 magic squares.

First, multiply each of the numbers in the order 3 magic square above by** **1089 as illustrated:

For convenience, the new magic square on the right will be referred as the 1089 magic square.

The magic sum of the 1089 magic square is 16335.

It is no surprise that the products would produce a magic square since all the numbers in the magic square were multiplied by the same number (1089). However, what’s surprising is from this magic square, we can form several other magic squares. What this means is, for instance, if we only use the** last **3 digits of the numbers in the 1089 magic square, we would form another magic square with a magic sum of 1335:

If we instead use the **last **2 digits of the numbers in the 1089 magic square, the new magic square produced would have a magic sum* *of 135:

We an even just write the **first **3 digits of the numbers in the 1089 magic square to form a magic square with* *a magic sum of 1632 or the **first **2 digits of the numbers in the 1089 magic square to form yet another magic square with** **a magic sum of

**162**, as shown below:

Note that the last digits of the numbers in the 1089 magic square can form a rotated version of the order** **3** **magic square above:

Furthermore, the magic sums of the 1089 magic squares also have patterns:

1089 magic square S = 16335

Last 3 digits S = 1335

Last 2 digits S = 135

First 3 digits S = 1632

First 2 digits S = 162

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Am going to enjoy showing this magic. You are an absolute genius. Thank you for explaining it so well.

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Thanks for reading and I’m glad that you liked it.

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Way cool Ed can’t wait to show the guys at the office this one lol your my go to for the cool stuff

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Thanks! 🙂

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I studied Maths in college, we all discussed a lot about these magic 1089 & magic squares, thanks for taking me to those days !

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Interesting. From what I know, recreational mathematics is barely talked about in higher math classes in Hong Kong – even in courses like number theory and combinatorics.

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Yes, but it was discussed within ourselves (myself & my friends) not in the presence of our lecturer.

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Oh, I see.

I know some lecturers who like to incorporate recreational math in their lectures, even in graduate level courses. Though there are only a small number of them.

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That is totally amazing. I like that games 🙂

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Thanks. I’m glad that you liked it.

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Amazing explanation! Now I know why some people seem so intellectual in math when they solve puzzles in seconds!! Thanks😊

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Anyone who is familiar about the algoritm of a puzzle can quickly solve it (e.g. Rubik’s Cube). Mathematical prowess isn’t really necessary though it helps in certain instances. 🙂

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*slaps forehead* aargh I can never complete Rubik’s cube!! Duffer me😊

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The beginner method is actually easy. I have taught several people to solve the Rubik’s cube using the beginner’s method without any problem.

I was pretty good at it before. My average was around 13 sec. which was still impressive back then (but it’s slow by today’s standards).

I just tested it now. It took me 35 sec. to finish it lol.

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You are really fun! 35 secs compared to 13 secs! How come? 🤔keep practising or else “no olympic gold medal” – 😂😂 But 35 secs is impressive compared to mine it took me months and I have given up!!

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Once I figured out that I can’t go faster without learning hundreds of algorithms more (and committing them to muscle memory), I quickly lost interest lol. And it has been many year since I practiced. 🙂

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