Why the digit-sum test works

To explain a divisibility shortcut, you first need a general way to write a number. A five-digit number edcba means e × 10000 + d × 1000 + c × 100 + b × 10 + a. In general: … + 1000d + 100c + 10b + a.

That makes the rule for 10 obvious. Every place value except the units is already a multiple of 10, so the whole number is divisible by 10 exactly when the units digit a is 0. The rules for 5, 2, 4 and 8 come out the same way. (For 4, check the last two digits; for 8, the last three.)

Nine is harder, because you can't read it off the last digit: 99 and 109 both end in 9, and only one is a multiple of 9.

Start with something easier. Is 10 divisible by 9? No: it leaves remainder 1. What about 20, 30, 40? Remainders 2, 3, 4. For any number of tens, the remainder on division by 9 is just the number of tens. The same goes for hundreds: 400 leaves remainder 4.

So take 427. It has 4 hundreds (remainder 4), 2 tens (remainder 2) and 7 units (remainder 7). Add the leftovers: 4 + 2 + 7 = 13. That's one more group of 9, plus 4, so 427 ÷ 9 leaves remainder 4.

Why does it work at every place? Because every place value is a multiple of 9 plus exactly 1:

Watch it for 7309: 7 × 1000 + 3 × 100 + 0 × 10 + 9 = 7 × (999 + 1) + 3 × (99 + 1) + 0 × (9 + 1) + 9 × 1 = (7 × 999 + 3 × 99 + 0 × 9) + (7 + 3 + 0 + 9). The first bracket is a multiple of 9 and can be thrown away; what's left is the digit sum, 19. Add again: 1 + 9 = 10, and once more: 1 + 0 = 1. So 7309 leaves remainder 1. (A shortcut: any 9s among the digits can simply be crossed out before you add.)

The rule: a number is divisible by 9 if and only if the sum of its digits is divisible by 9, and adding the digits again and again down to one digit gives the remainder. The test for 3 works the same way, because the powers of 10 are also one more than a multiple of 3. As the book says, learning maths isn't about knowing shortcuts, but about understanding why they work.