*I believe that mathematical reality lies outside of us. Our function is to discover, or observe it, and that the theorems which we describe grandiloquently as our “creations” are simply notes on our observations.***Godfrey H. Hardy**

Numbers have a stunning beauty and there is a whole world of wonder to explore within them. This article focuses on a specific number that holds a special place in the name of this website. Let’s delve into the mathematical wonders of this number.

1089 is a unique number that possesses several fascinating mathematical properties. In this article, we will delve into the unique qualities of the number 1089 and examine its various properties.

1089 is the integer after 1088 and before 1090. It is a perfect square number and 33^{2} = 1089.

When translated into binary form, 1089 is represented as 100010001. In Roman numerals, it is denoted as MLXXXIX.

**Using the digits 1 to 9 in ascending and descending order**

1089 = 12 × 3

^{4}+ 5 × 6 + 78 + 91089 = 987 + 65 + 4 + 32 + 1

**Using the digits 1, 7, 2 and 9**

1089 = [17 + 2 × (9 + 17) × 2] × 9

**Using the digits 9, 8, 7, 6 and 5**

1089 = 9 × (8 × 7 + 65)

**Using the digits 4, 3, 2, 1 and 0**

1089 = (4! + 3

^{2})^{1 + 0!}

**Single digit representations**

1089 is a frequently used number in magic tricks because it can be obtained from any two three-digit numbers. This feature makes it the basis for a magician’s choice. Start with a three-digit number and perform some basic arithmetic operations, the result is always 1089!

**Magic.** **The steps to obtain 1089 from any two three-digit numbers are as follows:**

Example 1. Consider the number731(difference between the first and last digits is 6 > 2).Reversed number is

137.Their difference is 731 – 137 =

594.The reverse of the number 594 is

495.The sum of the last two numbers is 594 + 495

= 1089.

Example 2. Consider the number684(difference between the first and last digits is 2).Reversed number is

486.Their difference is 684 – 486 =

198.The reverse of the number 198 is

891.The sum of the last two numbers is 198 + 891

= 1089.

Example 3. Consider the number685(difference between the first and last digits is 1 < 2).Reversed number is

586.Their difference is 685 – 586 =

99.The reverse of the number 99 is

99.The sum of the last two numbers is 99 + 99

= 198.

**An algebraic proof of the trick**

Let one of the 3-digit numbers be represented as *ABC*. The reversed number will then be *CAB*. Without loss of generality, let *ABC* be the larger of the two numbers, so that *A* > *C* (note that we did not allow *A* = *C* in the problem statement). We are considering numbers in base 10.

Represent *ABC* and *CAB* as 100*A* + 10*B* + *C* and 100*C* + 10*B* + *A*, respectively. Then, *ABC* – *CAB* is

ABC–CBA= (100

A+ 10B+C) – (100C+ 10B+A)= 99

A– 99C= 99(

A–C).

Since the first and last digits in *ABC* differ by at least 2, (*A* – *C*) must be any of 2, 3, 4, 5, 6, 7, 8, or 9. Therefore, 99(*A* – *C*) is one of the following:

- 99 × 2 = 198
- 99 × 3 = 297
- 99 × 4 = 396
- 99 × 5 = 495
- 99 × 6 = 594
- 99 × 7 = 693
- 99 × 8 = 792
- 99 × 9 = 891.

We want to add this number to its reverse. Let’s call the number *XYZ*, which is 100*X* + 10*Y* + *Z*, and *ZYX* is 100*Z* + 10*Y* + *X*. We want to add *XYZ* to *ZYX*.

Looking at the possibilities for *XYZ* above, we notice that the middle number *Y* is always 9. Also note that the first and third digits always add up to 9, i.e., *X* + *Z* = 9. Therefore,

XYZ+ZYX= (100

X+ 10Y+Z) + (100Z+ 10Y+X)= 101

X+ 101Z+ 20Y= 101(

X+Z) + 20Y= 101(9) + 20(9)

= 909 + 180

= 1089, as expected.

The products formed by multiplying **1089** by integers from **1 to 9** exhibit a unique symmetry in the digits of the products. Some of the observations of this symmetry include:

- The thousand’s digits of the products form an increasing sequence from 1 to 9;
- The unit’s digits of the products form a decreasing sequence from 9 to 1, which is the reverse of the thousand’s digits;
- The sum of the digits in the unit’s place and the thousand’s place is always equal to 10;
- The hundred’s digits of the products form an increasing sequence from 0 to 8;
- The ten’s digits of the products form a decreasing sequence from 8 to 0, which is the reverse of the hundred’s digits.
- The sum of the digits in the tens’ place and the hundred’s place is always equal to 8.
- The sum of the digits in each product is always equal to 18, which is the
**sum of the digits in****1089**.

When **1089** is multiplied by integers **from 1 to 9**, the products formed by multipliers that *add up to *10 are *digit reversals* of each other. For example:

This pattern holds true for all multipliers that add up to 10, making 1089 a special number with unique properties. In particular, the multiplication of 1089 by 9 is of special interest. The digits are reversed to the original number as a result. In other words, 1089 turned into 9801.

The number **1089 **has yet another cute property that depends upon its digits. The process is as follows:

- Consider a number;
- Calculate the sum of the squares of its digits;
- If it is
**1 or 89**, stop; otherwise - Repeat the process
**until it reaches either 1 or 89**.

Here are a few examples:

**Example 1**. Consider *n* = 1.

1 → 1

^{2}=1→ 1^{2}=1→ . . .

It’s clear that once **1** is reached for a second time, the process enters a repeating loop and will always result in **1**.

**Example 2**. Consider *n* = 28.

28 → 2

^{2}+ 8^{2}= 68 → 6^{2}+ 8^{2}= 100 → 1^{2}+ 0^{2}+ 0^{2}=1→ . . .

**Example 3**. Consider *n* = 89.

89 → 8

^{2}+ 9^{2}= 145 → 1^{2}+ 4^{2}+ 5^{2}= 42 → 4^{2}+ 2^{2}= 20 → 2^{2}+ 0^{2}= 4 → 4^{2}= 16 → 1^{2}+ 6^{2}= 37 → 3^{2}+ 7^{2}= 58 → 5^{2 }+ 8^{2}=89→ 8^{2}+ 9^{2}= 145 → 1^{2}+ 4^{2}+ 5^{2}= 42 → 4^{2}+ 2^{2}= 20 → 2^{2}+ 0^{2}= 4 → 4^{2}= 16 → 1^{2}+ 6^{2}= 37 → 3^{2}+ 7^{2}= 58 → 5^{2 }+ 8^{2}=89→ . . .

It’s clear that once **89** is reached for a second time, the process enters a repeating loop and will always result in **89**.

**Example 4**. Consider *n* = 30.

30 → 3

^{2}+ 0^{2}= 9 → 9^{2}= 81 → 8^{2}+ 1^{2}= 65 → 6^{2}+ 5^{2}= 61 → 6^{2}+ 1^{2}= 37 → 3^{2}+ 7^{2}= 58 → 5^{2}+ 8^{2}=89→ . . .

A two-digit number *ab* with *a* > *b* can be used to form a square using the following process:

**The only time this difference is a square is when the result is 1089.**

**In other words, 1089 is the only square that can be obtained from the difference between the squares of two two-digit numbers formed by reversing the digits of each other.**

** Proof**. The algebraic explanation behind the above process is straightforward. The difference between the squares of two two-digit numbers

*ab*= 10

*a*+

*b*and

*ba*= 10

*b*+

*a*(obviously, 10 >

*a*>

*b*> 0) is

(10

a+b)^{2}− (10b+a)^{2}= 100

a^{2}+ 20ab+b^{2}– 100b^{2}– 20ab–a^{2}= 99

a^{2}– 99b^{2}= 11(

a+b) ⋅ 9(a−b).

Clearly, this expression will be a square if and only if 11(*a* + *b*)(*a* − *b*) is a square. For this to happen, we must have 11 | (*a* + *b*)(*a* − *b*).

Since *a* > *b*, we get 11 ∤ (*a* − *b*). The only possibility is 11 ∣ (*a* + *b*). As 11 is a prime and 1 ≤ *a* + *b* ≤ 17, it follows that ** a + b = 11**.

Thus, *a* – *b* should be a perfect square. We can write, *a* – *b* = (*a* + *b*) – 2*b*, so that when *a* – *b* is even (or odd), *a* + *b* is even (or odd). Therefore, *a* + *b* and *a* – *b* are of the same parity. Only two possibilities are thus *a* − *b* = 1^{2} or *a* − *b* = 3^{2}.

If we take *a* − *b* = 3^{2}, then *a* = *b* + 9 giving the minimum value of *a* as 10, which is not possible. Thus, *a* = *b* + 1 giving ** a – b = 1**.

Solving the two equations, we get *a* = 6 and *b* = 5.

More significantly, we have

65^{2}– 56^{2}= 11

^{2}⋅ 3^{2}= 33

^{2}=

1089.

**An important observation:**

Here is a magic square with magic sum **1089**.

Your suggestions are eagerly and respectfully welcome! See you soon with a new mathematics blog that you and I call **“****Math1089 – Mathematics for All!**“.