Clock arithmetic

What happens if we change the rules and make the usual straight number line into a circle?
Age
11 to 18
Challenge level
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Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving
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Problem

You may be familar with a straight number line, but what happens if we change the rules and make our number line into a circle?

We are used to using this sort of arithmetic when we use clocks.  

The numbers go from 1 to 12, but when you get to "13 o'clock", it actually becomes 1 o'clock again (think of how the 24 hour clock numbering works). So 13 becomes 1, 14 becomes 2, and so on.

Clock arithmetic
Clock arithmetic

 

This can keep going, so when you get to "25 o'clock'', you are actually back round to where 1 o'clock is on the clock face (and also where 13 o'clock was too).

Click on "Task 1" to find the first set of questions.  Once you have tried these you can check your solutions.

Can you answer these questions?  Don't worry about whether the time is am or pm.
  • What time is 5 hours after 10 o'clock?
  • What time is 9 hours after 4 o'clock?
  • What time is 13 hours after 6 o'clock?
  • What time is 20 hours after 5 o'clock?
  • What time is 36 hours after 2 o'clock?
  • 3 o'clock
  • 1 o'clock
  • 7 o'clock
  • 1 o'clock
  • 2 o'clock

You may have noticed that everytime you add $12$ hours you end back where you were, so instead of adding on 30 hours you can use the fact that $30=12+12+6$ and so just add on $6$ hours instead.

We write $30 \equiv 6 \text{ mod } 12$ which reads as "$30$ is congruent to $6$ mod (or modulo) $12$".

 

Instead of using the 12 numbers of a clock, we could instead take the portion of the number line between 0 and 7 inclusive and wrap this into a circle so that the 0 and 7 are in the same place (and we write 0 in this position).

Clock arithmetic

You can still use this curved number line to calculate sums like "3+2" but instead of "+2" meaning move two places to the right, it means moving 2 places "clockwise".   When working with this clock you are working in modulo 7 (there are 7 numbers, including 0, around the clock), so you would write $4+5 \equiv 2 \text{ mod } 7$.

Can you work out the answers to these sums using our circular number line?
  • $2 + 3 \equiv \text{ ? mod }7$
  • $5 + 2 \equiv \text{ ? mod }7$
  • $6 + 6 \equiv \text{ ? mod }7$
  • $4 \times 2 \equiv \text{ ? mod }7$
  • $3 \times 3 \equiv \text{ ? mod }7$
  • $\text{ 5 mod }7$
  • $\text{ 0 mod }7$
  • $\text{ 5 mod }7$
  • $\text{ 1 mod }7$
  • $\text{ 2 mod }7$

Subtracting can be thought of as a move anticlockwise around the circle.  Can you work out the answers to these?

  • $2 - 5 \equiv \text{ ? mod }7$
  • $1-2 \equiv \text{ ? mod }7$
  • $\text{ 4 mod }7$
  • $\text{ 6 mod }7$

When using clock arithmetic, subtracting is not really necessary, as instead of moving round anticlockwise we can move clockwise and end up in the same position. We have $1 - 2 \equiv 6$ mod $7$ but we also have $1 + 5 \equiv 6$ mod $7$. This means that the operation "$-2$" is equivalent to "$+5$" when we are working modulo $7$.

What additions could be used instead of these subtractions when working modulo $7$?
  • $-1$
  • $-3$
  • $-5$
  • $-6$
  • $+6$
  • $+4$
  • $+2$
  • $+1$

We can also keep travelling round our modulo $7$ clock. If we wanted to represent $16$ we would start at $0$ and move $16$ places clockwise until we ended on $2$. We would write $16 \equiv 2 \text{ mod }7$.

Find what these numbers would be on the modulo 7 clock.
  • $20 \equiv \text{ ? mod }7$
  • $32 \equiv \text{ ? mod }7$
  • $50 \equiv \text{ ? mod }7$
  • $91 \equiv\text{ ? mod }7$
If I know that $x \equiv 3 \text{ mod } 7$ then what values could $x$ be?

Can you find a general expression for $x$?

If I know that $x \equiv a \text{ mod } 7$ then can you find a general expression for $x$?

 
  • $20 \equiv \text{ 6 mod }7$
  • $32 \equiv \text{ 4 mod }7$
  • $50 \equiv \text{ 1 mod }7$
  • $91 \equiv\text{ 0 mod }7$

If $x \equiv 3 \text{ mod } 7$ then $x$ could be $... 10, 17, 24, 31... 73... 1403... 777780...$

In general $7n+3$.

If $x \equiv a \text{ mod } 7$ then $x=7n+a$.

You can use some of these ideas to solve problems about days of the week. 

Can you use your knowledge of arithmetic modulo 7 to solve this final set of problems?

  • If today is a Monday, what day will it be in $31$ days' time?
  • If April $1^{\text{st}}$ is a Thursday, what day will Christmas day be on?
  • If Jan $1^{\text{st}}$ is a Friday, and it is a leap year, what day will Jan $1^{\text{st}}$ be on next year?
  • Can you use today's date to work out what day of the week your birthday will be on?

If you have enjoyed working on this problem, then you may enjoy More Adventures with Modular Arithmetic and Clock Squares 



We are very grateful to the Heilbronn Institute for Mathematical Research for their generous support for the development of this resource.