Divisibility Rule of 23

Last Updated : 23 Jul, 2025

Divisibility rules are simple shortcuts that help determine if one number is divisible by another without performing full division. The divisibility rule for 23 helps us quickly determine if a number is divisible by 23 without needing to perform long division.

Divisibility Rule of 23:

  • Take the number's last digit (unit digit) and multiply it by 7.
  • Add this value to the rest of the digits in the number.
  • Check the result: If the number is divisible by 23 (or is 0), then the original number is divisible by 23.

Mathematical Proof

A general number N can be written as:

N = 10^n a_n + 10^{n-1} a_{n-1} + \cdots + 10 a_1 + a_0

Here, a_n, a_{n-1}, \dots, a_1, a_0​ are the digits of the number. We want to show that N is divisible by 23, i.e., N = 23k for some integer k.

We can factor out 10 from all terms except the last one, giving:

N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + \cdots + 10 a_2 + a_1 \right) + a_0

Now, to introduce the rule of subtracting twice the last digit, we add and subtract 70 a_0

N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + \cdots + 10 a_2 + a_1 \right) + 70 a_0 - 70 a_0 + a_0

This simplifies to:

N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + \cdots + 10 a_2 + a_1 + 7 a_0 \right) - 69 a_0

Now, notice that:

  • 69 \equiv 0 \mod 23 so the term 69 a_0 contributes nothing to the remainder modulo 7.
  • We only need to check whether 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + \cdots + 10 a_2 + a_1 + 7 a_0 \right) \equiv 0 \mod 23

10 (\overline{a_na_{n-1}......a_2a_1} + 7a_0) \equiv 0 (mod 23)

since 10 ≡ 10 (mod 23), for N to be divisible by 7, \overline{a_na_{n-1}......a_2a_1} + 7a_0) \equiv 0 (mod 7)

This leads to a rule for checking if a three-digit number is divisible by 23:

  1. Remove the last digit c.
  2. Multiply c by 7.
  3. Subtract this result from the number formed by the remaining two digits.

If the result is divisible by 23, then the original number N, is also divisible by 23.

More Examples of Divisibility by 23 Rule

Here are a few examples of numbers divisible by 23, applying the divisibility rule:

For 2829:

  • Take the last digit (9) and multiply it by 7 to get 63.
  • Add it to the remaining number (282): 282 + 63 = 345.
  • Since 345 is still large, apply the rule again:
    • Take the last digit (5) of 345, and multiply it by 7 to get 35.
    • Add it to the remaining number (34): 34 + 35 = 69.
  • Divide 69 by 23: 69 ÷ 23 = 3.

Since 69 is divisible by 23, 2829 is also divisible by 23.

For 4761:

  • Take the last digit (1) and multiply it by 7 to get 7.
  • Add it to the remaining number (476): 476 + 7 = 483.
  • Since 483 is still large, apply the rule again
    • Take the last digit (3) of 483, and multiply it by 7 to get 21.
    • Add it to the remaining number (48): 48 + 21 = 69.
  • Divide 69 by 23: 69 ÷ 23 = 3.
  • Since 69 is divisible by 23, 4761 is also divisible by 23.

Divisibility Rule of 23 Solved Questions

Question 1: Check if the given number is divisible by 23 or not: 28395041

Solution: 

Check 28395041 is divisible by 23 

  • 28395041 ⇒ 2839504 + 1 × 7 = 2839511
  • 2839511 ⇒ 283951 + 1 × 7 = 283958
  • 283958 ⇒ 28395 + 8 × 7 = 28451
  • 28451 ⇒ 2845 + 1 × 7 = 2852
  • 2852 ⇒ 285 + 2 × 7 = 299
  • 299 ⇒ 29 + 9 × 7 = 92

Since 92 is divisible by 23 (92 ÷ 23 = 4)

28395041 is also divisible by 23.

Question 2: Check if the given number is divisible by 23 or not: 142857

Solution: 

Check 142857 is divisible by 23

  • 142857 ⇒ 14285 + 7 × 7 = 14334
  • 14334 ⇒ 1433 + 4 × 7 = 1461
  • 1461 ⇒ 146 + 1 × 7 = 153
  • 153 ⇒ 15 + 3 × 7 = 36
  • 36 ⇒ 3 + 6 × 7 = 45

Since 45 is not divisible by 23 (45 ÷ 23 ≈ 1.956),

142857 is not divisible by 23.

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