*Calendars and clocks exist to measure time, but that signifies little because we all know that an hour can seem as eternity or pass in a flash, according to how we spend it.***Michael Ende**

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A calendar is a tabular register which displays the date and the day of the week, and often the whole of a particular year divided up into months, weeks, and days.

Today is **23rd January**, **2023**. There are **seven columns** and **five rows** in the calendar for this month. The presence of mathematics is due to the presence of numbers. In this blog post, let us consider the calendar of January 2023 and try to discuss various mathematical facts.

### Various mathematical facts that can be observed in a Calendar

In this calendar, the first 31 natural numbers are present. The numbers 1, 3, 5, 7, . . . , 31 are odd numbers, whereas the numbers 2, 4, 6, 8, . . . , 30 are even.

The numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29 are primes, and there are five twin prime number pairs between 1 and 31, which are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31).

There are two perfect numbers, namely 6 and 28; five square numbers, namely 1, 4, 9, 16, and 25; three square numbers, namely 1, 8, and 27.

**(a)** **The difference in the sequence of numbers with the same colour is always 6**. In other words, if we start from an element in a row and move diagonally to the left, the difference in numbers is 6. For example,

- 2 ⟶ 8
- 3 ⟶ 9 ⟶ 15
- 4 ⟶ 10 ⟶ 16 ⟶ 22
- 5 ⟶ 11 ⟶ 17 ⟶ 23 ⟶ 29
- 6 ⟶ 12 ⟶ 18 ⟶ 24 ⟶ 30
**[this sequence produces the multiples of 6]** - 7 ⟶ 13 ⟶ 19 ⟶ 25 ⟶ 31
- 14 ⟶ 20 ⟶ 26
- 21 ⟶ 27

**(b) The difference in the sequence of numbers with the same colour is always 7**. In other words, if we start from an element in a row and go down, the difference in numbers is 7. For example,

- 1 ⟶ 8 ⟶ 15 ⟶ 22 ⟶ 29
- 2 ⟶ 9 ⟶ 16 ⟶ 23 ⟶ 30
- 3 ⟶ 10 ⟶ 17 ⟶ 24 ⟶ 31
- 4 ⟶ 11 ⟶ 18 ⟶ 25
- 5 ⟶ 12 ⟶ 19 ⟶ 26
- 6 ⟶ 13 ⟶ 20 ⟶ 27
- 7 ⟶ 14 ⟶ 21 ⟶ 28
**[this sequence produces the multiples of 7]**

**(c) The difference in the sequence of numbers with the same colour is always 8**. In other words, if we start from an element in a row and move diagonally to the right, the difference in numbers is 8. For example,

- 6 ⟶ 14
- 5 ⟶ 13 ⟶ 21
- 4 ⟶ 12 ⟶ 20 ⟶ 28
- 3 ⟶ 11 ⟶ 19 ⟶ 27
- 2 ⟶ 10 ⟶ 18 ⟶ 26
- 1 ⟶ 9 ⟶ 17 ⟶ 25
- 8 ⟶ 16 ⟶ 24
**[this sequence produces the multiples of 8]** - 15 ⟶ 23 ⟶ 31
- 22 ⟶ 30

(**d**) **If we divide this calendar into 2 × 2 smaller matrices, the sum of numbers with the same colour is always same**. For example,

- 4 + 12 =
**16**= 5 + 11 - 12 + 20 =
**32**= 13 + 19 - 20 + 8 =
**48**= 21 + 27 - 1 + 31 =
**32**= 3 + 29 - 11 + 28 =
**39**= 14 + 25 - 10 + 27 =
**37**= 13 + 24

(**e**) Divide the calendar into 3 × 3 smaller matrices (**rows and columns are consecutive**). The following facts are worth pointing out:

**The sum of numbers with the same colour is always same**. For example,

4 + 20 = **24** = 6 + 18

5 + 19 = **24** = 11 + 13

**The average of 4 and 20 is 12 (in the center), as is the average of 6 and 18**. Similarly, the averages of 5, 19 and 11, 13 (which also equals 12) are the same.**The sum of the elements along any row equals 36**.**The sum of the elements along any column equals 36**.**The sum of the elements along any diagonal equals 36**.**Also, the average of all nine numbers is 12**.**Certainly, the sum of the numbers is 9 × 12 = 108**.

Similar results can be obtained with other smaller matrices of order 3 × 3 (**rows and columns are consecutive**). A few examples of 3 × 3 matrices are

If we allow the skipping of rows or columns, a few of the above-mentioned results may or may not happen. The fact becomes clear from the following examples:

(**f**) Divide the calendar into 4 × 4 smaller matrices. Then

**The sum of the numbers along the diagonals are same**. For example

- 4 + 12 + 20 + 28 = 64
**[diagonal sum]** - 7 + 13 + 19 + 25 = 64
**[diagonal sum]**

**The sum of the numbers with the same colour shades is the same**. For example

- 4 + 7 + 28 + 25 = 64 [
**green colour**] - 5 + 6 + 26 + 27 = 64 [
**yellow colour**] - 11 + 18 + 14 + 21 = 64 [
**blue colour**] - 12 + 13 + 20 + 19 = 64 [
**tan colour**]

The following sums are also equal:

4 + 7 = 11 = 5 + 6 11 + 14 = 25 = 12 + 1318 + 21 = 39 = 19 + 20 25 + 28 = 53 = 26 + 27 | 4 + 25 = 29 = 11 + 18 5 + 26 = 31 = 12 + 19 6 + 27 = 33 = 13 + 20 7 + 28 = 35 = 14 + 21 |

Certainly, the sum of the numbers is 32 × 8 = 256. Otherwise, the sum

= (4 + 28) + (7 + 25) + (5 + 27) + (6 + 26) + (11 + 21) + (14 + 18) + (12 + 20) + (13 + 19)

= 32 + 32 + 32 + 32 + 32 + 32 + 32 + 32

= 32 × 8

= 256, as expected.

Similar results can be obtained with other smaller matrices of order 4 × 4. A few examples of 4 × 4 matrices are

(**g**) **A game with any 4 × 4 matrix. Perform the following steps: **

(i)Choose a number fromany row(or column) and circle it;

(ii)Determinewhich row and columnthat number belongs in;

(iii)Cross out all the numberslying in that particular row (or column);

(iv)Repeat the procedurewith uncovered numbers until all the elements are covered;

(v)The sum of the circled numbersmust beequal to the sum of the four corner numbers.

Here is an example.

**(i)** Choose 2 from the *first column* and circle it;

**(ii)** Clearly, 2 lies in the **first column** and **first row**;

**(iii)** *Cross out all the numbers* lying in the *first column* and *first row*;

**(iv)** Next, choose 12 from the *second row* and cross out the numbers lying in the *second row* and *fourth column*. Now we choose 17 in the *third row,* and we cross out the numbers lying in the *third row* and *second column*. Finally, take 25 from the *fourth row* and cross out the corresponding rows and columns.

**(v)** We have the **sum of the circled numbers: **2 + 12 + 17 + 25 = **56** and the **sum of the corner numbers:** 2 + 5 + 26 + 23 = **56** same, as required.

The above ideas can easily be extended to matrices of other orders. For example, let us consider the following matrices of order 4 × 7 and 5 × 3. What number relationships can you discover? If something different comes out, do let us know by writing in the comments.

Finally, we want to find the sum of all the numbers from 1 to 31. From the first calculation, can we see that the average of the numbers from 1 to 31 is 16? Clearly, the sum is

1 + 2 + 3 + 4 +··· + 15 + 16 + 17 + ··· + 29 + 30 + 31

= (1 + 31) + (2 + 30) + (3 + 29) + ··· + (15 + 17) + 16

= 32 + 32 + ··· + 32 + 16

= 15 × 32 + 16

= 480 + 16

= 496.

**Alternatively**, the sum

= (31/2)[1 + 31]

= (31/2) × 32

= 31 × 16

= 496, as before.

We now turn our attention from number relations to geometry. There are many Pythagorean triplets within this range. They are as follows: (3, 4, 5), (6, 8, 10), (5, 12, 13), (9, 12, 15), (8, 15, 17), (12, 16, 20), (15, 20, 25), (20, 21, 29), and (24, 30).

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