Divisibility rules for 2 and 4

Divisibility by 2

A number is divisible by $2$ iff its last digit is even (0, 2, 4, 6, 8).

Divisibility by 4

A number is divisible by $4$ iff the number formed by its last two digits is divisible by $4$.

Why the “last few digits” rules work: $10^k = 2^k \cdot 5^k$

Write a positive integer in base 10:

$$N = d_0 + d_1\cdot 10 + d_2\cdot 10^2 + \cdots + d_n\cdot 10^n.$$

Since $10 = 2 \cdot 5$, we have $10^k = 2^k \cdot 5^k$. So $2^a \mid 10^k$ exactly when $k \ge a$ (the factor $2^a$ appears in $10^k$). So modulo $2^a$:

• $10^0 = 1$ is not divisible by $2$ (only $k \ge 1$ gives $2 \mid 10^k$).
• $10^0, 10^1$ are not both divisible by $4$; $10^2, 10^3, \ldots$ are divisible by $4$.
• $10^3, 10^4, \ldots$ are divisible by $8$; only the last three digits matter.
• $10^4, 10^5, \ldots$ are divisible by $16$; only the last four digits matter.

So:

Examples

• $18\,572$ is divisible by $2$ (last digit $2$).
• $7\,236$ is divisible by $4$ (last two digits $36$).
• $2\,136$ is divisible by $8$ (last three digits $136 = 8 \times 17$).
• $12\,048$ is divisible by $16$ (last four digits $2048 = 16 \times 128$).