Divisibility rules — 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 (with proofs where possible)

Core Idea

Think of a number like a recipe: $n = d_k \cdot 10^k + d_{k-1} \cdot 10^{k-1} + \ldots + d_1 \cdot 10 + d_0$. Divisibility rules translate "Does this recipe divide evenly?" into "Do these ingredients (digits) satisfy a simpler check?"

Foundation: Place Value Representation

Any positive integer can be written as: $n = d_k \cdot 10^k + d_{k-1} \cdot 10^{k-1} + \ldots + d_1 \cdot 10 + d_0$

where $d_i$ are the digits (0–9). For example, $n = 1234 = 1 \cdot 10^3 + 2 \cdot 10^2 + 3 \cdot 10 + 4$.

Key insight: If we want to test divisibility by $m$, we ask whether $n ≡ 0 \pmod{m}$. We can reduce each power of 10 modulo $m$ and work with the digits alone.

The Rules & Proofs

Rule 1: Divisibility by 2

Why? $n = \underbrace{d_k \cdot 10^k + \ldots + d_1 \cdot 10}_{\text{all multiples of 10, hence even}} + d_0$

Since $10 ≡ 0 \pmod{2}$, every power of 10 is divisible by 2. So: $n ≡ d_0 \pmod{2}$

Example: $n = 1238$. Last digit is 8 (even) → divisible by 2. Check: $1238 = 2 \times 619$. ✓

Rule 2: Divisibility by 3

Proof from first principles: $n = d_k \cdot 10^k + d_{k-1} \cdot 10^{k-1} + \ldots + d_1 \cdot 10 + d_0$

Observe that $10 ≡ 1 \pmod{3}$ (since $10 = 3 \times 3 + 1$). Therefore: $10^2 ≡ 1^2 = 1 \pmod{3}, \quad 10^3 ≡ 1 \pmod{3}, \quad \text{etc.}$

So: $n ≡ d_k \cdot 1 + d_{k-1} \cdot 1 + \ldots + d_1 \cdot 1 + d_0 \pmod{3}$ $n ≡ d_k + d_{k-1} + \ldots + d_1 + d_0 \pmod{3}$

Why this step? We replaced every $10^i$ with 1 because they're all congruent to 1 mod 3.

Rule 3: Divisibility by 4

Proof: $n = \underbrace{d_k \cdot 10^k + \ldots + d_2 \cdot 10^2}_{\text{multiples of 100}} + d_1 \cdot 10 + d_0$

Since $100 = 4 \times 25$, we have $100 ≡ 0 \pmod{4}$, and so all terms with $10^i$ for $i ≥ 2$ vanish: $n ≡ d_1 \cdot 10 + d_0 \pmod{4}$

Why this step? We isolated the last two digits because higher powers of 10 are all multiples of 100, which is divisible by 4.

Rule 4: Divisibility by 5

Proof: $n ≡ d_0 \pmod{5}$ because $10 ≡ 0 \pmod{5}$. All higher powers vanish.

Example: $n = 1235$. Last digit is 5 → divisible by 5. Check: $1235 = 5 \times 247$. ✓

Rule 5: Divisibility by 6

Why? $6 = 2 \times 3$ and $\gcd(2,3) = 1$. If $n ≡ 0 \pmod{2}$ and $n ≡ 0 \pmod{3}$, then by the Chinese Remainder Theorem, $n ≡ 0 \pmod{6}$.

Rule 6: Divisibility by 7

Proof sketch: Let $n = 10a + b$ where $b$ is the last digit and $a$ is the remaining part.

We want to find a linear combination that eliminates the factor involving $b$ modulo 7. Note that $10 ≡ 3 \pmod{7}$, so: $n ≡ 3a + b \pmod{7}$

We want to construct a new number $a - kb$ that preserves divisibility. If we choose $k = 2$: $a - 2b ≡ a - 2b \pmod{7}$

We need to verify: $n ≡ 0 \pmod{7}iff a - 2b ≡ 0 \pmod{7}$.

Why this step? From $n = 10a + b$: $n ≡ 0 \pmod{7} \iff 10a + b ≡ 0 \pmod{7} \iff 3a + b ≡ 0 \pmod{7}$

Multiply both sides by 5(since $5 \times 3 = 15 ≡ 1 \pmod{7}$, so 5 is the modular inverse of 3): $5(3a + b) ≡ 0 \pmod{7} \iff 15a + 5b ≡ 0 \pmod{7} \iff a + 5b ≡ 0 \pmod{7}$

But $5b ≡ -2b \pmod{7}$ (since $5 ≡ -2 \pmod{7}$), so: $a - 2b ≡ 0 \pmod{7}$

Another: $n = 1057$

• $a = 105$, $b = 7$
• $a - 2b = 105 - 14 = 91$
• $a = 9$, $b = 1$: $9 - 2 = 7$ ✓
• So $1057$ is divisible by 7. Check: $1057 = 7 \times 151$. ✓

Rule 7: Divisibility by 8

Proof: $n ≡ d_2 \cdot 10^2 + d_1 \cdot 10 + d_0 \pmod{8}$ because $1000 = 8 \times 125 ≡ 0 \pmod{8}$.

Why this step? $10^3$ and higher powers are all divisible by 8.

Rule 8: Divisibility by 9

Proof: Identical to the divisibility by 3 proof. $10 ≡ 1 \pmod{9} \implies n ≡ \sum d_i \pmod{9}$

Rule 9: Divisibility by 10

Proof: $n ≡ d_0 \pmod{10}$, so $n ≡ 0 \pmod{10} \iff d_0 = 0$.

Example: $n = 1230$. Last digit is 0 → divisible by 10. ✓

Rule 10: Divisibility by 11

Proof: $n = d_k \cdot 10^k + d_{k-1} \cdot 10^{k-1} + \ldots + d_1 \cdot 10 + d_0$

Observe that $10 ≡ -1 \pmod{11}$. Therefore: $10^0 ≡ 1, \quad 10^1 ≡ -1, \quad 10^2 ≡ 1, \quad 10^3 ≡ -1, \quad \ldots$

So: $n ≡ d_k \cdot (-1)^k + d_{k-1} \cdot (-1)^{k-1} + \ldots - d_1 + d_0 \pmod{11}$

Why this step? We replaced each power of 10 with $(-1)^i$, creating an alternating pattern.

For practical use, compute: $(d_0 - d_1 + d_2 - d_3 + \ldots)$ or equivalently (sum of digits at odd positions) − (sum of digits at even positions).

Visual Summary

Common Mistakes

Why it's wrong: $6 = 2 \times 3$. The number must be divisible by BOTH. Example: 15 is divisible by 3 but not by 2, so not by 6.

The fix: Check BOTH conditions. Even last digit AND sum divisible by 3.

Why it's wrong: The key is $10 ≡ -1 \pmod{11}$, not $10 ≡ 1$. This creates an alternating pattern, not a plain sum.

The fix: Alternate signs: $(d_0 - d_1 + d_2 - d_3 + \ldots)$. Pattern matters!

Why it's wrong: The algebraic proof specifically requires subtraction to preserve the modular relationship.

The fix: Remember: "7 is heaven, so SUBTRACT to get there." Always $a - 2b$.

Active Recall

Instead of doing long division every time, there are shortcuts!

• Divisibility by 2: Just check the last digit. If it's even (0, 2, 4, 6, 8), you can divide by 2. Why? Because everything else in the number is already a multiple of 10, which is divisible by 2.

• Divisibility by 3 or 9: Add up all the digits. If that sum can be divided by 3 (or 9), the whole number can! This works because of a cool pattern: 10, 100, 1000… all leave remainder of 1 when divided by 3 or 9.

• Divisibility by 5: Last digit is 0 or 5? You're good! Just like money (₹5, ₹10).

• Divisibility by 11: This one's fun! Alternate adding and subtracting the digits (start from the right). If you get 0 or a multiple of 11, the number is divisible by 11.

These rules save time and make you look like a math wizard!

Mnemonic

4, 8: Look at last 2 or 3 digits

• 4: last 2 digits
• 8: last 3 digits

3, 9: Sum all digits

• Pattern: powers of 10 are ≡ 1

11: Alternating sum (because $10 ≡ -1 \pmod{11}$)

6: Must pass both 2 and 3

7: The tricky one—subtract twice the last digit

"2-5-10 look at the end, 4-8 need two or three friends, 3-9 sum to transcend, 11 alternates without end, 6 needs two rules to defend, 7's the one where you subtract and descend."

Connections Modular Arithmetic — divisibility rules are applications of congruence

• Chinese Remainder Theorem — why divisibility by 6 = divisibility by 2 AND 3
• Place Value System — foundation for understanding digit manipulation
• GCD and LCM — composite divisors like 6, 10, 12rely on prime factorization
• Casting Out Nines — digit sum techniques used for error checking
• Fermat's Little Theorem — deeper number-theoretic patterns involving modular arithmetic

Summary Table

| Divisor | Rule | Key Modular Fact | |---------|---------------| | 2 | Last digit even | $10 ≡ 0 \pmod{2}$ | | 3 | Sum of digits divisible by 3 | $10 ≡ 1 \pmod{3}$ | | 4 | Last2 digits divisible by 4 | $100 ≡ 0 \pmod{4}$ | | 5 | Last digit 0 or 5 | $10 ≡ 0 \pmod{5}$ | | 6 | Divisible by both 2 and 3 | $\gcd(2,3)=1$ | | 7 | $a - 2b$ divisible by 7 | $10 ≡ 3 \pmod{7}$ | | 8 | Last 3 digits divisible by 8 | $1000 ≡ 0 \pmod{8}$ | | 9 | Sum of digits divisible by 9 | $10 ≡ 1 \pmod{9}$ | | 10 | Last digit is 0 | $n ≡ d_0 \pmod{10}$ | | 11 | Alternating digit sum divisible by 11 | $10 ≡ -1 \pmod{11}$ |

#flashcards/maths

What is the divisibility rule for 2? :: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8).

Is 2728 divisible by 11? :: Yes. Alternating sum: $8 - 2 + 7 - 2 = 11$, which is divisible by 11.

Concept Map

Hinglish (regional understanding)