Divisibility Rules

1. Basic Divisibility Rules

• Idea: Checking whether a number is divisible by common numbers such as 2, 3, 4, 5, 8, 9, 10 and 11.

• Formula / Rule:

  • Divisible by 2 → last digit is even.
  • Divisible by 3 → sum of digits is divisible by 3.
  • Divisible by 4 → last 2 digits are divisible by 4.
  • Divisible by 5 → last digit is 0 or 5.
  • Divisible by 8 → last 3 digits are divisible by 8.
  • Divisible by 9 → sum of digits is divisible by 9.
  • Divisible by 10 → last digit is 0.
  • Divisible by 11 → difference between sums of alternate digits is 0 or a multiple of 11.

• Example: Is 5832 divisible by 8?

  Solution: Last three digits are 832.

  832 ÷ 8 = 104

  Therefore, 5832 is divisible by 8.

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2. Missing Digit / Missing Number Divisibility

• Idea: Finding an unknown digit so that a number becomes divisible by a given number.

• Formula / Rule: Apply the divisibility rule of the required divisor.

• Example: Find the digit (x) so that (54x32) is divisible by 9.

  Solution:

  5 + 4 + x + 3 + 2 = 14 + x

  For divisibility by 9:

  14 + x = 18
  
  x = 4

  Answer: (x=4)

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3. Divisibility by Composite Numbers

• Idea: Breaking a number into smaller factors and checking divisibility by each factor.

• Formula / Rule: If (a) and (b) are coprime,

$$n \text{ is divisible by } ab \iff n \text{ is divisible by both }a\text{ and }b$$

• Example: Check whether 9765 is divisible by 15.

  Solution:

  15 = 3 × 5

  Digit sum:

  9 + 7 + 6 + 5 = 27

  So it is divisible by 3.

  Last digit is 5, so it is divisible by 5.

  Therefore, it is divisible by 15.

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4. Divisibility by 6, 12, 15, 18, 24, 36 and Similar Numbers

• Idea: Combining basic divisibility rules to handle common composite numbers.

• Formula / Rule:

  6  = 2 × 3
  12 = 3 × 4
  15 = 3 × 5
  18 = 2 × 9
  24 = 3 × 8
  36 = 4 × 9

  Check the required factors separately.

• Example: Is 7380 divisible by 36?

  Solution:

  For 4:

  Last two digits = 80
  80 is divisible by 4

  For 9:

  7 + 3 + 8 + 0 = 18
  18 is divisible by 9

  Therefore, 7380 is divisible by 36.

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5. Algebraic Divisibility

• Idea: Using standard algebraic divisibility patterns instead of calculating large powers.

• Formula / Rule:

$$x^n-y^n \text{ is divisible by }x-y$$ $$x^n+y^n \text{ is divisible by }x+y \quad\text{when }n\text{ is odd}$$

• Example: Is (7^{15}+3^{15}) divisible by 10?

  Solution:

  Since 15 is odd,

$$x^{15}+y^{15}$$

is divisible by (x+y).

Here,

$$x+y=7+3=10$$

Therefore,

$$7^{15}+3^{15}$$

is divisible by 10.

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6. Remainder-Based Divisibility

• Idea: Using the remainder of a number or expression to determine divisibility without calculating the complete value.

• Formula / Rule:

  If

$$N\equiv r\pmod d$$

then (N) is divisible by (d) when (r=0).

• Example: Find the remainder when (2^{10}) is divided by 7.

  Solution:

  2³ = 8 ≡ 1 (mod 7)

  Therefore,

  2¹⁰ = 2⁹ × 2
       ≡ 1³ × 2
       ≡ 2 (mod 7)

  Remainder = 2

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7. Cyclicity of Powers

• Idea: Finding the last digit or remainder of very large powers using repeating patterns.

• Formula / Rule: Powers of a number often repeat their last digits in a fixed cycle.

• Example: Find the last digit of (7^{103}).

  Solution:

  Powers of 7 have the cycle:

  7, 9, 3, 1

  Cycle length = 4.

  103 ÷ 4 → remainder 3

  Take the 3rd number in the cycle:

  3

  Answer: 3

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8. Digital Root

• Idea: Quickly reducing a number to a single digit to test divisibility patterns or compare numerical properties.

• Formula / Rule:

$$\text{Digital root of }N ======================== 1+(,N-1\pmod 9,)$$

• Example: Find the digital root of 987654.

  Solution:

  9 + 8 + 7 + 6 + 5 + 4 = 39
  3 + 9 = 12
  1 + 2 = 3

  Answer: 3

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9. Factorial Divisibility

• Idea: Finding how many times a prime number divides a factorial such as (10!), (50!), or (100!).

• Formula / Rule:

$$\text{Power of }p\text{ in }n! ============================== \left\lfloor\frac np\right\rfloor+ \left\lfloor\frac n{p^2}\right\rfloor+ \left\lfloor\frac n{p^3}\right\rfloor+\cdots$$

• Example: Find the highest power of 5 in (100!).

  Solution:

$$\left\lfloor\frac{100}{5}\right\rfloor+ \left\lfloor\frac{100}{25}\right\rfloor$$ $$=20+4=24$$

Therefore, (5^{24}) divides (100!), but (5^{25}) does not.

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10. Number of Zeros in a Factorial

• Idea: Finding trailing zeros in (n!).

• Formula / Rule:

$$\text{Trailing zeros in }n! =========================== \left\lfloor\frac n5\right\rfloor+ \left\lfloor\frac n{25}\right\rfloor+ \left\lfloor\frac n{125}\right\rfloor+\cdots$$

• Example: Find the number of trailing zeros in (100!).

  Solution:

$$\left\lfloor\frac{100}{5}\right\rfloor+ \left\lfloor\frac{100}{25}\right\rfloor$$ $$=20+4=24$$

Answer: 24

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Advanced

11. Highest Power of a Composite Number in (n!)

• Idea: Finding the highest power of a composite number such as 12, 18 or 24 that divides (n!).

• Formula / Rule: Factor the divisor into primes and compare the available powers in (n!).

• Example: Find the highest power of 12 dividing (50!).

  Solution:

  12 = 2² × 3

  In (50!):

  Power of 2 = 25 + 12 + 6 + 3 + 1 = 47
  Power of 3 = 16 + 5 + 1 = 22

  Each factor of 12 needs 2 twos and 1 three.

  47 ÷ 2 = 23
  22 ÷ 1 = 22

  Therefore, the limiting factor is 3.

  Answer: (12^{22}) divides (50!).

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12. Advanced Missing-Digit Divisibility

• Idea: Finding one or more missing digits when several divisibility conditions must hold simultaneously.

• Formula / Rule: Apply all required divisibility conditions together.

• Example: Find (x) such that (35x4) is divisible by both 3 and 4.

  Solution:

  Divisibility by 4:

  Last two digits = x4

  Possible values:

  04, 24, 44, 64, 84

  Divisibility by 3:

$$3+5+x+4=12+x$$

So (x) must be divisible by 3.

From the possible digits:

x = 0 or 6

Therefore:

x = 0 or 6