## Monday Math 61

Divisibility Tests
Part 1: Factors of powers of 10

A common topic in mid-level aritmetic are tests to determine whether a number is divisible by a given small integer without needing to perform a full division; for example, that a number is divisible by nine if and only if the sum of its digits is also a multiple of nine. Proving these rules, however, can involve a bit of higher level math, such as modular arithmetic or other number-theoretic reasoning. Today, I will consider the simplest of divisibility tests; those for factors of powers of ten, which are simple due to our use of the base-ten number system.

First, divisiblity by ten is simply detemined by whether or not the ones digit is zero; divisiblity by 100; by whether the ones and tens digits are both zero; and so on. The reason is obvious.
Next, we see that divisibilty by the factors of 10, 2 and 5, also depend only on the ones digit: a number is divisible by 2 (even) if and only if the ones digit is even, and a number is divisible by five if and only if the ones digit is 0 or 5.

Those numbers which divide 100 but not 10 divide a number only if they divide the last two digits. This gives us the rules for 4, 20, 25, and 50.

For numbers dividing 1000 but not 100, we test the last three digits. This gives the tests for 8, 40, 125, 200, 250, and 500.

And so on. These, as we can see, are simple, and are obvious consequences of our base-ten system.

### One Response to “Monday Math 61”

1. meichenl Says:

For $2^n$, here’s an explicit algorithm using the last $n$ digits.

Divide the units digit by two. If this is not an integer, stop. It’s not divisible by $2^n$
Add the next digit (one power of ten greater) and divide by two. If this is not an integer, stop (test fails). If it is an integer and you have used all $n$ digits already, stop (success).
repeat previous

So to test 8904 for 2^4 = 16, you’d do:

4/2 = 2
2+0 = 2. 2/2 = 1.
1+9 = 10. 10/2 = 5.
5+8 = 13. 13/2 = 6.5

So 8904 is not a multiple of 16, but it is a multiple of 8.