Divisibility Algebra

The Logic of Place Value

Any number $N$ can be written using powers of 10. Example: $4725 = 4(1000) + 7(100) + 2(10) + 5$.

To check divisibility, we replace $10$ with remainders.

Divisibility by 9

Key Fact: $10 \div 9$ leaves a remainder of $1$. Therefore, $100, 1000, 10^n$ all leave a remainder of $1$ when divided by $9$.

Proof: Consider a 3-digit number $abc = 100a + 10b + c$.

The first part $(99a + 9b)$ is clearly divisible by 9. Therefore, $N$ is divisible by 9 if and only if $(a + b + c)$ is divisible by 9.

Divisibility by 3

Key Fact: $10 \div 3$ leaves a remainder of $1$. The logic is identical to 9.

Since $(99a + 9b)$ is divisible by 3, $N$ depends entirely on the sum of digits $(a+b+c)$.

Divisibility by 11

Key Fact: Powers of 10 alternate remainders when divided by 11.

• $1 \equiv 1$
• $10 = 11 - 1 \equiv -1$
• $100 = 99 + 1 \equiv 1$
• $1000 = 1001 - 1 \equiv -1$

Proof: Consider $abcd = 1000a + 100b + 10c + d$.

Composite Rules

• Divisibility by 6: Must be divisible by 2 AND 3.
• Divisibility by 24: Checking 4 and 6 is NOT enough (e.g., 12 is divisible by 4 and 6 but not 24). You must check 3 and 8 (since 3 and 8 are coprime).