Divisibility Rules in Different Bases

Just like we have divisibility rules in base 10, we can have rules in any base.

Divisors of the Base

If a number divides the base, this is similar to our divisibility by 2 and 5 rules in base-10.

Just look at the last digit!

$101_2$ is not divisible by $2$ since it ends in a $1$

$3F94_{16}$ is divisible by $2$ and $4$, but not $8$ or $16$ since it ends in a $4$.

$A3B5_{15}$ is divisible by $5$ since it ends in a $5$, but it is not divisible by $3$ since its last digit is $5$ (not divisible by 3).

Check your understanding: Use a random number generator to generate a number between 2 and 10, then generate another number between 1 and 10,000. Using the first number as the base, can you determine if the second number has any easy-to-find divisors?

Base - 1

We were able to determine a rule for divisibility by $9$ in base $10$. We can similarly find that for any base $b$, a number is divisible by $b-1$ if the digits (in that base) sum to a number that is divisible by $b-1$.

For example:

$2433_7$

$2 + 4 + 3 + 3 = 12$

This number is divisible by 6!

In fact, if we convert it,

$2 \cdot 7^3 + 4 \cdot 7^2 + 3 \cdot 7^1 + 3 \cdot 7^0 = 906$

$906$ is divisible by $2$ and $3$, therefore it is divisible by $6$!

So, $2433_7$ is divisible by $6$

Check your understanding: Use a random number generator to generate a number between 2 and 10, then generate another number between 1 and 10,000. Using the first number as the base, can you determine if the second number is divisible by the base minus $1$?

Base + 1

Think about it: Can you come up with a divisibility rule for the base plus 1? Hint: divisibility by 11 would be an example here. What would happen in general?

Multiplication in Different Bases

We know that 10 times any number in base 10 just moves all the digits one place value.

Think about it: What happens if I multiply any number by $(10)_b$? Hint: Remember that this is just $b$ itself.