Divisibility tricks for 7, 11, 13, 17, 19 & 23 — (easy rules + worked examples)

Introduction

When you want to check whether a large number is divisible by a given small prime (or composite) without long division, divisibility tricks are handy. Many rules are iterative: they transform the number into a smaller one that has the same remainder mod the divisor. Below are simple, reliable rules for 7, 11, 13, 17, 19, and 23 — each with clear examples you can easily understand.

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1) Divisibility by 7

Rule (one common version): Remove the last digit, double it, and subtract that from the remaining truncated number. Repeat until you get a small number. If the result is divisible by 7, the original number is divisible by 7.

Why (brief): If number = 10×(rest) + lastDigit. Since 10 ≡ 3 (mod 7), you can transform to an equivalent smaller expression; the “subtract double the last digit” method works by producing an equivalent remainder modulo 7.

Example: Test 20,203

1. Split: rest = 2020, last = 3. Compute 2020 − 2×3 = 2020 − 6 = 2014.

2. Repeat: rest = 201, last = 4 → 201 − 2×4 = 201 − 8 = 193.

3. Repeat: rest = 19, last = 3 → 19 − 2×3 = 19 − 6 = 13.
   13 is not divisible by 7 → 20,203 is not divisible by 7.

Another quick example: 203 → 20 − 2×3 = 20 − 6 = 14 → 14 is divisible by 7 → 203 is divisible by 7.

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2) Divisibility by 11

Rule (standard): Compute the alternating sum of digits (sum of digits in odd positions − sum of digits in even positions). If that result is a multiple of 11 (including 0 or ±11, ±22, …), the original number is divisible by 11.

Example: Test 27,528
Digits from left: 2 7 5 2 8
Alternating sum = (2 − 7 + 5 − 2 + 8) = 6 → 6 is not a multiple of 11 → 27,528 is not divisible by 11.

Example 2: 34,287 → (3 − 4 + 2 − 8 + 7) = 0 → 0 is divisible by 11 → 34,287 is divisible by 11.

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3) Divisibility by 13

Rule (useful iterative rule): Remove the last digit, multiply it by 4, and add that to the truncated number. Repeat until small. If the result is divisible by 13, the original number is divisible by 13.

Example: Test 351

1. rest = 35, last = 1 → 35 + 4×1 = 35 + 4 = 39.

2. 39 is divisible by 13 → 351 is divisible by 13 (indeed 13×27 = 351).

Another example: Test 2,226

1. rest = 222, last = 6 → 222 + 4×6 = 222 + 24 = 246.

2. rest = 24, last = 6 → 24 + 4×6 = 24 + 24 = 48.

3. 48 is not divisible by 13 → 2,226 is not divisible by 13.

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4) Divisibility by 17

Rule (one easy form): Remove the last digit, multiply it by 5, and subtract from the truncated number. Repeat. If the final number is divisible by 17, the original number is too.

(We use: new = rest − 5×last)

Example: Test 289

1. rest = 28, last = 9 → 28 − 5×9 = 28 − 45 = −17.
   −17 is divisible by 17 → 289 is divisible by 17 (indeed 17×17 = 289).

Another quick one: 1,533 → rest 153, last 3 → 153 − 15 = 138 → 13 − 5×8? (we continue) rest 13, last 8 → 13 − 40 = −27 (not divisible by 17) → 1,533 is not divisible by 17.

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5) Divisibility by 19

Rule (simple): Remove the last digit, multiply it by 2, and add to the truncated number. Repeat. If the final result is divisible by 19, so is the original number.

(new = rest + 2×last)

Example: Test 133
rest = 13, last = 3 → 13 + 2×3 = 13 + 6 = 19 → divisible → 133 is divisible by 19.

Another: 1,902
rest = 190, last = 2 → 190 + 4 = 194.
rest = 19, last = 4 → 19 + 8 = 27 → not divisible by 19 → 1,902 not divisible by 19.

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6) Divisibility by 23

Rule (one version): Remove the last digit, multiply it by 7, and add to the truncated number. Repeat. If the result is divisible by 23, the original number is divisible by 23.

(new = rest + 7×last)

Example: Test 115
rest = 11, last = 5 → 11 + 7×5 = 11 + 35 = 46 → 46 = 2×23 → 115 is divisible by 23.

Another: Test 5,842
rest = 584, last = 2 → 584 + 14 = 598.
rest = 59, last = 8 → 59 + 56 = 115 → 115 is divisible by 23 → 5,842 is divisible by 23.

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Practice problems

Try these quickly (work with the given rules); answers below.

1. Is 1,234 divisible by 7?

2. Is 11,891 divisible by 11?

3. Is 1,014 divisible by 13?

4. Is 5,102 divisible by 17?

5. Is 2,285 divisible by 19?

6. Is 4,602 divisible by 23?

Answers (work shown briefly)

1. 1,234 → (for 7) 123 − 2×4 = 123 − 8 = 115 → 11 − 2×5 = 11 − 10 = 1 → not divisible → No.

2. 11,891 → alternating sum = 1 − 1 + 8 − 9 + 1 = 0 → divisible → Yes.

3. 1,014 → for 13: 101 + 4×4 = 101 + 16 = 117 → 11 + 4×7 = 11 + 28 = 39 → 39 = 3×13 → Yes.

4. 5,102 → for 17: 510 − 5×2 = 510 − 10 = 500 → 50 − 5×0 = 50 → 5 − 5×0 = 5 → not divisible → No.

5. 2,285 → for 19: 228 + 2×5 = 228 + 10 = 238 → 23 + 2×8 = 23 + 16 = 39 → 39 not multiple of 19 → No. (39 = 19×2 + 1)

6. 4,602 → for 23: 460 + 7×2 = 460 + 14 = 474 → 47 + 7×4 = 47 + 28 = 75 → not divisible → No.

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Quick tips & final notes

• Apply the operation repeatedly until the number is small enough to check by eye. You may need to work with negatives — if the small final result is a multiple of the divisor, the original number is divisible.

• These rules arise because of how 10 behaves modulo the divisor; each rule transforms 10×rest + last into an equivalent (mod divisor) smaller expression.

• If you prefer, you can always do regular division or use a calculator — but these tricks are fast in head-calculation and great for contests and interviews.

Special thanks to:

• Classic references from Elementary Number Theory (David Burton)

• Open educational resources (NCERT & CBSE mathematics)

• Inspiration from various math forums and educational websites

Thank You

Alok Pradhan

02.09.2025