*In the presence of so many beautiful creations of his thought the mathematician lives long and lives young: he rejoices in the grandeur of the height to which his controlled imagination attains.***Robert D. Carmichael**

A clock is a device used to measure and indicate time. Normally, it displays the time in hours, minutes, and often seconds during a 12- or 24-hour period. In this blog post, we will consider several mathematical aspects of a clock.

**The 12-hour Clock**

The 12-hour clock is a time convention in which the 24 hours of the day are divided into two periods: a.m. and p.m. Each period consists of 12 hours numbered 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 (with 12 acting as 0). In general, 12 p.m. indicates 12 o’clock noon, while 12 a.m. means 12 o’clock midnight.

Considering a clock showing numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12, we can easily observe the following **six recreational facts**:

**Unlocking the Sum of 12 Through Addition **

We can always find a sum of 12 by adding the selected numbers as shown below:

- 12 = 12
- 12 = 11 + 1
- 12 = 10 + 2
- 12 = 9 + 3
- 12 = 8 + 4
- 12 = 7 + 5
- 12 = 6 + 6

**Unlocking the Mystery of Achieving 13 Through Addition**

We can always find a sum of 13 by adding the selected numbers as shown below:

- 13 = 12 + 1
- 13 = 11 + 2
- 13 = 10 + 3
- 13 = 9 + 4
- 13 = 8 + 5
- 13 = 7 + 6

**Achieving an Even Sum Through Addition**

We will find the sums as 18, 16, 14, 12, 10, and 8 by adding the numbers lying on the opposite faces of a clock, as shown below:

- 18 = 12 + 6
- 16 = 11 + 5
- 14 = 10 + 4
- 12 = 9 + 3
- 10 = 8 + 2
- 8 = 7 + 1

**Attaining a Subtraction Margin of 6** **Through Subtraction**

We can always find a difference of 6 by subtracting the chosen numbers, as demonstrated below:

- 6 = 7 ‒ 1
- 6 = 8 ‒ 2
- 6 = 9 ‒ 3
- 6 = 10 ‒ 4
- 6 = 11 ‒ 5
- 6 = 12 ‒ 6

**Achieving Even Numbers from 12 to 2 Through **Subtraction

We will find the differences of 12, 10, 8, 6, 4, 2, and 0 by subtracting the numbers located on the opposite faces of a clock, as depicted below:

- 6 = 7 ‒ 1
- 6 = 8 ‒ 2
- 6 = 9 ‒ 3
- 6 = 10 ‒ 4
- 6 = 11 ‒ 5
- 6 = 12 ‒ 6

**Achieving Odd Numbers from 11 to 1 Through **Subtraction

We will find the differences of 11, 9, 7, 5, 3, and 1 by subtracting the numbers located on the opposite faces of a clock, as depicted below:

- 11 = 12 ‒ 1
- 9 = 11 ‒ 2
- 7 = 10 ‒ 3
- 5 = 9 ‒ 4
- 3 = 8 ‒ 5
- 1 = 7 ‒ 6

So far, we have discussed about various number symmetries in a clock and we have **considered only 12 hour clock for our discussion**. In this case, *an hour* after 11 o’clock it is 12 o’clock; *two hours* after 11 o’clock is 1 o’clock; *ten hours* after 11 o’clock is 9 o’clock; *twenty hours* after 11 o’clock is 7 o’clock etc. We now take our turn into **clock arithmetic**. We all know how to add numbers, but did you ever think about how strangely we add the hours on a clock?

Suppose, it is 5 o’clock and we add 3 hours to the time that will put us at (5 + 3) = 8 o’clock, so we could write 5 + 3 = 8. But if it is 10 o’clock and we add 6 hours the time will be 4 o’clock, so we should write 10 + 6 = 4! We all know that 10 + 6 is actually 16, but there is no number 16 in a usual clock. In fact, when we go past 12 on the clock we start counting the hours at 1 again. If we add numbers the way we add hours on the clock, we say that we are doing clock arithmetic. So, in clock arithmetic 7 + 7 = 2, because 7 hours after 7 o’clock is 2 o’clock.

**Test Yourself**

**5 + 12 = ?****7 + 6 = ?****12 + 9 = ?****9 + 14 = ?**

As we did the addition in a clock, can we do subtraction also? Yes, we can. On a usual clock with 12 hours, 2 ‒ 3 = 11 because if we start at 2 o’clock and move backward 3 hours, we would be at 11 o’clock. Draw your own picture to see that this is true.

**Example**. Suppose the current time is 5 o’clock. Find the time 32 hours back.

*Solution*. First, subtract 32 from 5:

5 ‒ 32 = ‒27, a negative value.

If we encounter a negative result, our next step is to identify the nearest integer that is a multiple of 12. In this instance, we need to find the next integer greater than 27 that is divisible by 12, which happens to be 36.

Now, let’s express ‒27 in terms of ‒36. We can represent ‒27 as ‒36 + 9.

Therefore, the time 32 hours back was 9 hours.

**Test Yourself**

**5 ‒ 7 = ?****4 ‒ 6 = ?**

You may have noticed that when you go all the way around the clock exactly once you end up just where you started. In clock arithmetic, going around the clock a whole number of times has the same effect as doing nothing! So, if we had a 12 hour clock, adding or subtracting 12 is the same as adding or subtracting 0. Hence, we can use modular arithmetic in any standard clock.

Given an integer *n* > 1. Two integers *a* and *b* are said to be congruent modulo *n* if *n* is a divisor of their difference. In other words, there is an integer *k* such that *a* − *b* = *kn*. We denote this fact by *a* ≡ *b* (mod *n*).

For example, 24 ≡ 8 (mod 4) and 13 ≡ 1 (12). Congruence modulo *n* is an *equivalence relation*. In fact, it satisfies

*Reflexivity*:*a*≡*a*(mod*n*)*Symmetry*:*a*≡*b*(mod*n*) if*b*≡*a*(mod*n*).*Transitivity*: If*a*≡*b*(mod*n*) and*b*≡*c*(mod*n*), then*a*≡*c*(mod*n*).

In the present discussion, we only consider **mod 12**.

**Example**. What time is it in 79 hours if it is 4 o’clock now?

Solution. We have 4 + 79 = 83 and if we divide it by 12, we get 11 as the remainder.

In other words, (4 + 79) ≡ 11 (mod 12), so that the time will be 11 o’clock.

**Definition**. A finite family of non-empty subsets *S*_{1}, *S*_{2}, . . . , *S _{n}* of a non-empty set

*S*is said to form a partition of

*S*if

(i)

S_{1}∪S_{2}∪ · · · ∪S=_{n}S(ii)

S∩_{i}S= Ø for_{j}i≠j.

We know that every congruence relation on the set ℤ determines a partition of ℤ.

Amazing

Thank you