Divisibility Rules Practice & Tester
Test any number against all divisibility rules from 2 to 13 simultaneously and view step-by-step mathematical proofs.
Comprehensive Guide: Divisibility Rules, Modular Arithmetic & Mental Tests
Master fast mental division shortcuts for numbers 2 through 12, the underlying modular proofs, and alternating sum theorems.
1. What Are Divisibility Rules & Why Do They Work?
A divisibility rule is a mental arithmetic algorithm that determines whether a given integer is divisible by a fixed divisor without performing long division. These rules are grounded in modular arithmetic and our base-10 positional decimal system.
Because every decimal number can be expanded as a polynomial in powers of 10: N = dₙ10ⁿ + ... + d₁10¹ + d₀10⁰, inspecting the behavior of powers of 10 modulo k reveals patterns in the digits.
2. Complete Reference Table: Divisibility Rules (2 through 12)
| Divisor | Rule Criterion | Quick Example |
|---|---|---|
| 2 | Last digit is even (0, 2, 4, 6, 8) | 7,498 (Ends in 8 ✓) |
| 3 | Sum of all digits is divisible by 3 | 483 → 4+8+3=15 (15÷3=5 ✓) |
| 4 | Last two digits form a number divisible by 4 | 19,316 → 16÷4=4 ✓ |
| 5 | Last digit is 0 or 5 | 82,915 (Ends in 5 ✓) |
| 6 | Divisible by both 2 AND 3 simultaneously | 792 (Even, and 7+9+2=18 ✓) |
| 7 | Double last digit and subtract from remaining. Result divisible by 7 | 672 → 67 − 2(2) = 63 (63÷7=9 ✓) |
| 8 | Last three digits form a number divisible by 8 | 54,128 → 128÷8=16 ✓ |
| 9 | Sum of all digits is divisible by 9 | 2,871 → 2+8+7+1=18 (18÷9=2 ✓) |
| 10 | Last digit is 0 | 4,930 (Ends in 0 ✓) |
| 11 | Alternating sum of digits is divisible by 11 | 9,482 → 9−4+8−2 = 11 (11÷11=1 ✓) |
| 12 | Divisible by both 3 AND 4 simultaneously | 648 (Ends in 48 [÷4], 6+4+8=18 [÷3] ✓) |
3. Mathematical Proof: Why the Sum of Digits Works for 3 and 9
Why does summing digits work for 3 and 9? Consider that 10 ≡ 1 (mod 9) and 10 ≡ 1 (mod 3). Therefore, any power of 10 satisfies 10ⁿ ≡ 1ⁿ ≡ 1 (mod 9).
Expanding an arbitrary 4-digit number abcd:
N = 1000a + 100b + 10c + d N = (999+1)a + (99+1)b + (9+1)c + d N = [999a + 99b + 9c] + [a + b + c + d]
The bracketed quantity [999a + 99b + 9c] is clearly an exact multiple of 9 (and 3). Therefore, N is divisible by 9 (or 3) if and only if the remainder term [a + b + c + d] is divisible by 9 (or 3)!
4. Composite Divisibility: The Coprime Factor Rule
To test divisibility by any composite number M = p × q, you can test divisibility by p and q separately if and only if p and q are coprime (gcd(p, q) = 1):
- Divisibility by 6: Test 2 and 3 (gcd(2, 3) = 1) ✓
- Divisibility by 12: Test 3 and 4 (gcd(3, 4) = 1) ✓ (Testing 2 and 6 fails because gcd(2, 6) = 2!)
- Divisibility by 15: Test 3 and 5 (gcd(3, 5) = 1) ✓
- Divisibility by 18: Test 2 and 9 (gcd(2, 9) = 1) ✓
5. Mental Math Pro-Tips & Casting Out Nines
When summing digits for 3 or 9 in large numbers, you can ignore any digits that are already 9 (or 3, 6, 9 for the rule of 3). For example, to check 959,949 for divisibility by 9: disregard all the 9s, leaving only 5 + 4 = 9. Immediate conclusion: it is divisible by 9!
6. Frequently Asked Questions (FAQ)
Is 0 divisible by all numbers?
Yes, zero is divisible by every non-zero integer because 0 ÷ k = 0 with remainder 0. However, division by zero itself is mathematically undefined.
Why is there no simple single-digit rule for 7?
Because 10 is not adjacent to a multiple of 7 (10 ≡ 3 mod 7, 100 ≡ 2 mod 7, 1000 ≡ 6 ≡ −1 mod 7). The cycle of remainders (1, 3, 2, 6, 4, 5) has length 6, meaning simple digit addition does not work without modular weights.