Clock Arithmetic

Section 3.5 Clock Arithmetic

In the following we give applications of the combination of the operation

mod

and the addition of integers called clock arithmetic.

🔗

Recall that the operation

mod

yields the remainder of integer division. That is, for an integer

a

and a natural number

b

the number

r = a mod b

is the number such that

a = (q \cdot b) + r

for some integer

q

and

r

is a non-negative integer and

r < b .

🔗

Applications of the operation

mod

can be found in arithmetic with hours, days of the week, and months. We start with examples of adding hours and then relate this to using addition and the operation

mod .

(a) We conduct the familiar addition of hours and replace 12 on the clock face by 0.

🔗

Example 3.62. 12 hour clock.

🔗

When using the 12 hour clock we have:

An hour after 11 o’clock it is 12 o’clock.
Two hours after 11 o’clock it is 1 o’clock.
10 hours after 11 o’clock it is 9 o’clock.
20 hours after 11 o’clock it is 7 o’clock.
25 hours after 11 o’clock it is 12 o’clock.

🔗

These operations can be considered as adding hours to a time. To compute these additions we add the hours and then subtract

12

as many times as necessary to obtain a number between

1

and

12 .

With a similar method we had computed the remainder in Algorithm 3.6.

🔗

The main difference between the two approaches is that using the

12

hour clock we obtain numbers between

1

and

12

and when computing remainders we obtain numbers between

0

and

11 .

That is, we replace

12

0,

as shown in Figure 3.61.(a). We call this arithmetic modulo

12 .

Figure 3.61.(b) illustrates how the number line wraps around the clock face in arithmetic modulo

12 .

The remainder modulo

12

of two numbers is the same if they differ by a multiple of

12 .

🔗

Example 3.63. Addition $mod 12$ .

🔗

We formulate the computations from Example 3.62 using remainders. Recall that we denoted the remainder of the division of

a

12

a mod 12 .

$(11 + 1) mod 12 = 12 mod 12 = 0$
$(11 + 2) mod 12 = 13 mod 12 = 1$
$(11 + 10) mod 12 = 21 mod 12 = 9$
$(11 + 20) mod 12 = 31 mod 12 = 7$
$(11 + 25) mod 12 = 36 mod 12 = 0$

🔗

When the numbers are larger than

12

we can simplify our computations by applying Theorem 3.46.

🔗

Example 3.64. Addition $mod 12$ .

🔗

By Theorem 3.46 that we can add first or take the remainder first, and we will get the same answer, so in Item 4 and Item 5 above, we could have done the following:

$(11 + 20) mod 12$ $= ((11 mod 12) + (20 mod 12)) mod 12$ $= (11 + 8) mod 12 = 19 mod 12 = 7$
$(11 + 25) mod 12$ $= ((11 mod 12) + (25 mod 12)) mod 12$ $= (11 + 1) mod 12 = 12 mod 12 = 0$

🔗

The second computation appears to have more steps, but the arithmetic can be much simpler. In practice we combine both approaches.

🔗

In the remainder of the section we solve every day problems with clock arithmetic.

🔗

Problem 3.65. $79$ hours from $4$ o’clock.

🔗

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

Solution.

We have

(4 + 79) mod 12 = 83 mod 12 = 11 .

Thus 79 hours from now, it is 11 o’clock.

The same result can also be obtained by first computing

79 mod 12 = 7

and then

(4 + 7) mod 12 = 11 mod 12 = 11 .

🔗

Try for yourself in Checkpoint 3.66.

Checkpoint 3.66. Hours from now with 12 hour clock.

We use the 12 hour clock. Assume it is 8 o’clock. What time will it be 49 hours from now ?

select
12 o’clock
1 o’clock
2 o’clock
3 o’clock
4 o’clock
5 o’clock
6 o’clock
7 o’clock
8 o’clock
9 o’clock
10 o’clock
11 o’clock

🔗

Everything we have done for hours above also works for other counts that wrap around. For example, when answering questions about weekdays we compute

mod 7 .

🔗

Problem 3.67. $110$ days from Friday.

🔗

Which day of the week is it in 110 days from today if today is Friday ?

Solution.

The days of the week wrap around after seven days. When adding days and we want the result as a weekday, any multiples of 7 do not change the day of the week. Instead of adding 110 days we add