AUMAT 229 Activities

Example 3.1.2. Addition modulo $12$.

Suppose it is $7$ o'clock now.

1. What time will it be in $9$ hours?

2. What time will it be in $19$ hours?

3. What time will it be in $109$ hours?

For the first question, we can break up the $9$ hours into two time periods: first the $5$ hours necessary to get to $12$ o'clock, and then the remaining $4$ hours takes us to $4$ o'clock. But it will be tedious to answer the other two questions this way, so let's redo this first question using some clock arithmetic instead. We have

but our clock doesn't have a $16$ on it, so we recognize that $16$ is $4$ past $12\text{,}$ and we again arrive at the same answer: the time $9$ hours past $7$ o'clock is $4$ o'clock.

Let's tackle the second question now. We have

The clock “resets” every time it hits $12\text{,}$ and it will hit $12$ twice in $26$ hours, the second time at $24\text{.}$ So the time $19$ hours past $7$ o'clock is $2$ o'clock.

Finally, the third question. We have

Since

over $116$ hours the last clock “reset” occurs after $108$ hours, and $116$ is $8$ past that. So the time $109$ hours past $7$ o'clock is $8$ o'clock.

Notice in these last two parts, we could also “reset” our numbers to proper “o'-the-clock” numbers before adding, and we would still come to the same answer:

Example 3.1.3. Multiplication modulo $12$.

You need to schedule $8$ consecutive $10$-hour work shifts for your employees, with the first shift beginning at $12$ o'clock. What time will it be when the last shift ends?

Just as in the addition examples, if we recognize that the clock “resets” at multiples of $12\text{,}$ we can calculate modulo $12$:

That is, it will be $8$ o'clock at the end of the last of the $8$ shifts.