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#### The ` % ` remainder function

To understand ` (% 1 length) `, we need to understand ` % `. According to its documentation (which I just found by typing C-h f % RET), the ` % ` function returns the remainder of its first argument divided by its second argument. For example, the remainder of 5 divided by 2 is 1. (2 goes into 5 twice with a remainder of 1.)

What surprises people who don't often do arithmetic is that a smaller number can be divided by a larger number and have a remainder. In the example we just used, 5 was divided by 2. We can reverse that and ask, what is the result of dividing 2 by 5? If you can use fractions, the answer is obviously 2/5 or .4; but if, as here, you can only use whole numbers, the result has to be something different. Clearly, 5 can go into 2 zero times, but what of the remainder? To see what the answer is, consider a case that has to be familiar from childhood:

• 5 divided by 5 is 1 with a remainder of 0;

• 6 divided by 5 is 1 with a remainder of 1;

• 7 divided by 5 is 1 with a remainder of 2.

• Similarly, 10 divided by 5 is 2 with a remainder of 0;

• 11 divided by 5 is 2 with a remainder of 1;

• 12 divided by 5 is 1 with a remainder of 2.

By considering the cases as parallel, we can see that

• zero divided by 5 must be zero with a remainder of zero;

• 1 divided by 5 must be zero with a remainder of 1;

• 2 divided by 5 must be zero with a remainder of 2;

and so on.

So, in this code, if the value of ` length ` is 5, then the result of evaluating

```(% 1 5)
```

is 1. (I just checked this by placing the cursor after the expression and typing C-x C-e. Indeed, 1 is printed in the echo area.)

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