Comment by (Keith) HUNG Kut Man:

The same thing happens for the cube of a sequence of consecutive numbers too. The recurring pattern of the last digit of the product is 0,1,8,7,4,5,6,3,2,9.

0 x 0 x 0 = 0

1 x 1 x 1 = 1

2 x 2 x 2 = 8

3 x 3 x 3 = 27

4 x 4 x 4 = 64

5 x 5 x 5 = 125

6 x 6 x 6 = 216

7 x 7 x 7 = 343

8 x 8 x 8 = 512

9 x 9 x 9 = 729

10 x 10 x 10 = 1000

11 x 11 x 11 = 1331

12 x 12 x 12 = 1728

13 x 13 x 13 = 2197

14 x 14 x 14 = 2744

15 x 15 x 15 = 3375

16 x 16 x 16 = 4096

17 x 17 x 17 = 4913

18 x 18 x 18 = 5832

19 x 19 x 19 = 6859

In fact, it is true for the nth power of a sequence of consecutive numbers, where n=1,2,3,...,infinity. The proof of it is left to the other students.

The following table shows the recurring pattern with various nth powers:

If we look at the above table carefully, we can find that these patterns are also under a recurring manner of

I have been thinking there may be some kinds of relationship between the above table and the magic square. A magic square is a square of numbers (n x n). The numbers in each row, column or diagonal are under the same kind of mathematical relationship, eg.

e.g. n1+n2+n3 = n4+n5+n6 = n7+n8+n9 = n1+n4+n7 = n2+n5+n8 = n3+n6+n9 = n1+n5+n9 = n3+n5+n7

If we consider only the natural numbers, zero will be taken away, and the above table becomes:

Figure 1

Number 5 occurs in each row and is the midpoint of it. I think of putting it in the middle of the magic square.

Since the odd numbers and the even numbers are in alternate positions, so I also put them in alternate positions around 5 in the magic square. Number 1 also occurs in each row of figure 1. There may be some kind of importance for this number. In this way, I put it in the middle of the column or row of the magic square.

or

If we group the numbers in figure 1 in the manner of (first number, last number), (second number, second last number),..., we will have the following results:

Row 2 and row 4 are neglected since they are of the same numbers within a group. The combinations become (1,9), (2,8), (3,7) and (4,6). If we put these pairs in the magic square, we would construct the following magic squares. (remember to make use of the criteria: 1. To have the number 5 as their middle number;  2. The odd numbers and the even numbers are in alternate positions, as shown in figure 1.)  Since the odd and even numbers are in alternate positions (in figure 1), so I also put them in alternate positions around 5 in the magic square.

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be 5.

In each row, column or diagonal, if we add up all the digits, the result will always be 15.

written around March 1999.

I created the following generalized creation method of a 3 x 3 magic square from a sequence of any 9 consecutive integers, where the starting number may be 0, 1, 2, 3 or....

1.  put the middle number of the sequence in the middle of the magic square;

2.  group the numbers of the sequence in the manner of (first number, last number), (second number, second last number),....

3.  put the first number of the sequence in the middle of the column or row of the magic square;

or

4. Since we have fixed the location of the first number, we can put the counterpart number of its group (first number, last number), i.e. the last number, in its opposite position:

or

5.i.  get a total by adding up the three numbers.  Let's call it x.

   ii. get a number y = 1st no. + last no. - middle no.  And you will find that y is always equal to the middle no.

6.  put in the next group of numbers, i.e. (second number, second last number). (remember to make use of the criteria: 1. To have the middle number of the sequence as their middle number;  2. The odd numbers and the even numbers are in alternate positions around the centre of the magic square.)

7.  put in the next group of numbers.  This time, make sure (a) the sum of the three numbers in a row or column or diagonal is equal to x, or (b) 1st no. + last no. - middle no. = y  (i.e. the number at the centre of the magic square)

8.  follow step 7 until you use up all groups of numbers.  And you will create 2 magic squares, one follows the criterion of 7(a) i.e. the sum of the three numbers in a row or column or diagonal is equal to x, and the other follows 7(b) 1st no. + last no. - middle no. = y  (i.e. the number at the centre of the magic square)

For example:

1.  From a sequence of 9 consecutive integers, 0, 1, 2, 3, 4, 5, 6, 7, 8: 

Its middle no. = 4, and its groups of numbers are

(0, 8)

(1, 7)

(2, 6)

(3, 5)

Two magic squares are created:

In each row, column or diagonal, if we add up all the digits, the result will always be 12.

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e.4.

2.  From a sequence of 9 consecutive integers, 111, 112, 113, 114, 115, 116, 117, 118, 119:

Its middle no. = 115, and its groups of numbers are

(111, 119)

(112, 118)

(113, 117)

(114, 116)

Two magic squares are created:

In each row, column or diagonal, if we add up all the digits, the result will always be 345.

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e. 115.

The above method is also true for a sequence of 9 consecutive even or odd integers.

For example:

1.  From a sequence of 9 consecutive even integers, 110, 112, 114, 116, 118, 120, 122, 124, 126: 

Its middle no. = 118, and its groups of numbers are

(110, 126)

(112, 124)

(114, 122)

(116, 120)

Two magic squares are created:

In each row, column or diagonal, if we add up all the digits, the result will always be 354.

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e. 118.

2.  From a sequence of 9 consecutive odd integers, 1113, 1115, 1117, 1119, 1121, 1123, 1125, 1127, 1129: 

Its middle no. = 1121, and its groups of numbers are

(1113, 1129)

(1115, 1127)

(1117, 1125)

(1119, 1123)

Two magic squares are created:

In each row, column or diagonal, if we add up all the digits, the result will always be 3363.

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e. 1121.

Updated on June 17, 1999.

Further, the above creation method is also true for the following numbers and sequence:

1.  negative number;

2.  decimals;

3.  n₁+c, n₂+c, n₃+c, n₄+c, n₅+c, n₆+c, n₇+c, n₈+c, n₉+c, where n₁, n₂, n₃,..., n₉ is a feasible sequence in forming a magic square, and c is a constant which can be a positive or negative number or a decimal.

4.  n₁ x c,  n₂ x c,  n₃ x c,  n₄ x c,  n₅ x c,  n₆ x c,  n₇ x c,  n₈ x c,  n₉ x c, where n₁, n₂, n₃,..., n₉ is a feasible sequence in forming a magic square, and c is a constant which can be a positive or negative number or a decimal.

For example:

1.  From a sequence of 9 consecutive integers, -3, -2, -1, 0, 1, 2, 3, 4, 5: 

Its middle no. = 1, and its groups of numbers are

(-3, 5)

(-2, 4)

(-1, 3)

(0, 2)

Two magic squares are created:

In each row, column or diagonal, if we add up all the digits, the result will always be 3.

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e. 1.

2.  From a sequence of 9 consecutive negative integers, -7777, -7776, -7775, -7774, -7773, -7772, -7771, -7770, -7769:

Its middle no. = -7773, and its groups of numbers are

(-7777, -7769)

(-7776, -7770)

(-7775, -7771)

(-7774, -7772)

Two magic squares are created:

In each row, column or diagonal, if we add up all the digits, the result will always be -23319.

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e. 7773.

3.  From a sequence of 9 consecutive decimals, 1234.77, 1234.78, 1234.79, 1234.80, 1234.81, 1234.82, 1234.83, 1234.84, 1234.85:

Its middle no. = 1234.81, and its groups of numbers are

(1234.77, 1234.85)

(1234.78, 1234.84)

(1234.79, 1234.83)

(1234.80, 1234.82)

Two magic squares are created:

In each row, column or diagonal, if we add up all the digits, the result will always be 3704.43.

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e. 1234.81.

4.  From a sequence of 9 consecutive negative decimals, -8.3, -8.2, -8.1, -8.0, -7.9, -7.8, -7.7, -7.6, -7.5:

Its middle no. = -7.9, and its groups of numbers are

(-8.3, -7.5)

(-8.2, -7.6)

(-8.1, -7.7)

(-8.0, -7.8)

Two magic squares are created:

In each row, column or diagonal, if we add up all the digits, the result will always be -23.7

.

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e. -7.9.

5.  Illustration of "n₁+c, n₂+c, n₃+c, n₄+c, n₅+c, n₆+c, n₇+c, n₈+c, n₉+c, where n₁, n₂, n₃,..., n₉ is a feasible sequence in forming a magic square, and c is a constant which can be a positive or negative number or a decimal.":

Let the above example (example 4) be the feasible sequence, n₁, n₂, n₃,..., n₉ and let c = 4.

The magic square in example 4,

becomes

=

In each row, column or diagonal, if we add up all the digits, the result will always be -11.7.

The magic square in example 4,

becomes

=

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e. -3.9.

6.  Illustration of "n₁ x c,  n₂ x c,  n₃ x c,  n₄ x c,  n₅ x c,  n₆ x c,  n₇ x c,  n₈ x c,  n₉ x c, where n₁, n₂, n₃,..., n₉ is a feasible sequence in forming a magic square, and c is a constant which can be a positive or negative number or a decimal."

Illustration 1:

Let the above example (example 4) be the feasible sequence, n₁, n₂, n₃,..., n₉ and let c = 0.4

The magic square in example 4,

becomes

=

In each row, column or diagonal, if we add up all the digits, the result will always be -9.48.

The magic square in example 4,

becomes

=

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e. -3.16.

7.  Illustration 2:

Let the above example (example 4) be the feasible sequence, n₁, n₂, n₃,..., n₉ and let c = -0.477

The magic square in example 4,

becomes

=

In each row, column or diagonal, if we add up all the digits, the result will always be 11.3049.

The magic square in example 4,

becomes

=

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, i.e. 3.7683.

written on June 18, 1999.

In mathematical presentation:

Two 3 x 3 magic squares can be created by the following sequence of 9 consecutive numbers:

C₁(n - 4? + C₂, C₁(n - 3? + C₂, C₁(n - 2? + C₂, C₁(n - ? + C₂, C₁n + C₂, C₁(n + ? + C₂, C₁(n + 2? + C₂, C₁(n + 3? + C₂, C₁(n + 4? + C₂

where C₁, C₂, n, and ?#060;/font> are any real, imaginary or complex numbers, and C₁?is the difference between any two consecutive numbers. 

The two 3 x 3 magic squares are:

C1(n + 3? + C2	C1(n - 4? + C2	C1(n + ? + C2
C1(n - 2? + C2	C1n + C2	C1(n + 2? + C2
C1(n - ? + C2	C1(n + 4? + C2	C1(n - 3? + C2

In each row, column or diagonal, if we add up all the digits, the result will always be 3(C₁n + C₂).

C1(n - 3? + C2	C1(n - 4? + C2	C1(n - ? + C2
C1(n - 2? + C2	C1n + C2	C1(n + 2? + C2
C1(n + ? + C2	C1(n + 4? + C2	C1(n + 3? + C2

In each row, column or diagonal, if we add the first and the last digits together and then minus the middle one, the result will always be the number at the centre of the magic square, C₁n + C₂.

written on June 25, 1999.

A magic square can be constructed by Excel. If you want to know how, please click

.

P.S. The square of a sequence of consecutive numbers (0, 1, 2, 3,...,infinity) also shows the following recurring patterns:

The last 2 digits: A recurring pattern occurs every 50 consecutive numbers.

The last 3 digits: A recurring pattern occurs every 1,000 consecutive numbers.

The last 4 digits: A recurring pattern occurs every 10,000 consecutive numbers.

The last 5 digits: A recurring pattern occurs every 100,000 consecutive numbers.

       .                                           .

       .                                           .

       .                                           .

The last N digits: A recurring pattern occurs every 1 x 10^N consecutive numbers, where N=3, 4, 5,..., infinity.

A more generalized description is:

The last N digits: A recurring pattern occurs every 1 x 10^N consecutive numbers (where N=1, 2, 3,..., infinity) of the nth power of a sequence of consecutive integers, where n=1, 2, 3,..., infinity.

The last 2 digits: A recurring pattern occurs every 50 consecutive numbers of the nth power of a sequence of consecutive integers, where n=even number.

written around June 10, 1999.

For High School Students or Above

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