Highest Power in Factorial: Formula, Methods and Examples

Highest Power in Factorial means finding the greatest exponent of a number that divides a factorial exactly. For a prime p, use Legendre’s formula by adding the quotients obtained from n divided by p, p², p³ and so on. The same method extends to composite numbers through prime factorisation and minimum exponents.

On this page

What Is the Highest Power in a Factorial?

The highest power of a prime p in n! is the greatest exponent e for which p^e divides n!. It is denoted by v_p(n!) and is calculated using Legendre’s formula.

For n!, count the multiples of p, p², p³ and higher powers of p among the numbers from 1 to n. Thus, v_p(n!) = floor(n/p) + floor(n/p²) + floor(n/p³) + ... . Stop when p^k > n. For example, v_3(10!) = floor(10/3) + floor(10/9) = 3 + 1 = 4, so the highest power of 3 dividing 10! is 3^4.

Highest Power in Factorial Formula & Tricks

Important Formulas

Legendre’s Formula
v_p(n!) = ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + ...

Here p is prime, n! = 1 × 2 × 3 × ... × n, and terms are included only while p^k ≤ n.

Highest Power of a Prime
Highest power of p in n! = p^{v_p(n!)}

First calculate v_p(n!), then raise p to that exponent.

Highest Power of a Composite Number
If a = p₁^{e₁}p₂^{e₂}...p_r^{e_r}, then the highest power of a in n! is a^k, where k = min(⌊v_{p₁}(n!)/e₁⌋, ..., ⌊v_{p_r}(n!)/e_r⌋).

Factor the base a and find how many complete copies of every prime factor are available in n!.

Quick Tricks

Write the successive powers first

For a prime p, list p, p², p³ and stop after the first power greater than n. This prevents missing a term in Legendre’s sum.

Example: For 50!, the powers of 2 needed are 2, 4, 8, 16 and 32. Therefore v₂(50!) = 25 + 12 + 6 + 3 + 1 = 47.
Use the minimum for composite bases

After finding the prime exponents in n!, divide each by the corresponding exponent in the composite base. The smallest quotient determines the answer.

Example: For 20!, v₂(20!) = 10 + 5 + 2 + 1 = 18 and v₃(20!) = 6 + 2 = 8. Since 12 = 2² × 3, k = min(18/2, 8) = 8, so 12^8 is the highest power of 12 in 20!.
Ignore powers greater than n

If p^j > n, then floor(n/p^j) = 0, so no further terms are needed.

Example: For 15!, the terms for p = 5 are floor(15/5) = 3; 25 > 15, so v₅(15!) = 3.

Highest Power in Factorial Concepts

Legendre’s Method for a Prime Factor

To find the exponent of a prime p in n!, add the integer parts of n/p, n/p², n/p³ and all later quotients that are non-zero.

The first quotient counts multiples of p, the second counts an additional factor of p in multiples of p², and so on. For 25!, v_5(25!) = floor(25/5) + floor(25/25) = 5 + 1 = 6. Hence 5^6 divides 25!, but 5^7 does not.

Example: v_2(10!) = floor(10/2) + floor(10/4) + floor(10/8) = 5 + 2 + 1 = 8. Therefore the highest power of 2 in 10! is 2^8.

Counting Prime Factors by Multiples

Each multiple of p contributes at least one factor p, each multiple of p² contributes one additional factor, and each multiple of p³ contributes another.

This layered counting explains why simply counting multiples of p is insufficient. For 16!, the multiples of 2 contribute 8, those of 4 add 4, those of 8 add 2, and 16 adds 1. Therefore v₂(16!) = 8 + 4 + 2 + 1 = 15.

Example: The number 8 contributes three factors of 2, while 12 contributes two. Legendre’s separate terms count these extra factors automatically.

Highest Power of a Composite Number

For a composite base, factor it into primes and use the smallest available complete-group count.

Suppose a = 2^2 × 3. If n! contains 2^u and 3^v, then it contains 12^k only when 2k ≤ u and k ≤ v. Hence k = min(floor(u/2), v). For 25!, v₂ = 12 + 6 + 3 + 1 = 22 and v₃ = 8 + 2 = 10, so the highest power of 12 is 12^10.

Example: For 10!, v₂(10!) = 8 and v₃(10!) = 4. Since 18 = 2 × 3², k = min(floor(8/1), floor(4/2)) = 2. Thus the highest power of 18 dividing 10! is 18².

Highest Power of a Product or Factorial Expression

When the required divisor is written as a product of prime powers, calculate the exponent of each prime separately before taking the minimum ratio.

For a divisor such as 72 = 2^3 × 3², determine v₂(n!) and v₃(n!), then calculate min(floor(v₂/3), floor(v₃/2)). The prime with the smaller ratio limits the power of 72 that can divide n!.

Example: For 15!, v₂ = 7 + 3 + 1 = 11 and v₃ = 5 + 1 = 6. For 72 = 2³ × 3², k = min(floor(11/3), floor(6/2)) = min(3, 3) = 3. Therefore 72³ is the highest power of 72 in 15!.

Highest Power in Factorial Video Lessons

Watch short topic-wise lessons for quick revision.

14 Lessons
Lesson 1 of 14 Quick Revision

Factorials: Trailing Zeros and Prime Powers

Learn how to find trailing zeros in factorials and determine the highest power of a prime that divides a factorial using standard number system methods.

Continue with more lessons and practice in PrepShots.Watch More in App - Start ₹1 Trial →
More Highest Power in Factorial Lessons Scroll to explore →

Practice Highest Power in Factorial Questions

Practise published questions related to this topic.

1The LCM of the numbers 5.4 and 0.384 is:→ 2The traffic lights at three different road crossings change after every 90 seconds, 120 seconds and 150 seconds respectively. If they all change simultaneously at 5:20 p.m., then find the time (in p.m.) at which they will change simultaneously again?→ 3The LCM of the numbers 1.3, 7.6 and 0.13 is:→ 4198 chips and 330 toffees are to be distributed among maximum number of boys. Each boy gets the same number of chips. Each boy get the same number of toffees. How many chips does each boy get?→ 5Find the HCF of the numbers 2520, 3600, 5880.→ 6How many two-digit numbers satisfy this property? The last digit (unit's digit) of the square of the two-digit number is 8.→ 7Two numbers are such that their HCF is 18 and their LCM is 540. If the sum of the two numbers is 198, find the numbers.→ 8What is the least number that must be added to 1789, so that the resulting sum is completely divisible by 29 ?→ 9A record entry gives √32 . Simplify.→ 10What is the Highest Common Factor of 15, 25 and 35?→

Highest Power in Factorial Quick Quiz

Attempt 5 questions and check your score instantly.

Quick Revision Notes

Highest Power in Factorial: Quick Revision

Use these rules to calculate prime and composite powers in factorials.

  • For prime p, v_p(n!) = floor(n/p) + floor(n/p²) + floor(n/p³) + ... .
  • Stop the sum when p^k > n.
  • The highest power of p dividing n! is p^{v_p(n!)}.
  • For a = p₁^{e₁}p₂^{e₂}..., the exponent of a is the minimum of floor(v_{p_i}(n!)/e_i).
  • Counting only multiples of p misses the extra factors contributed by multiples of p², p³ and higher powers.
  • If p > n, then p does not divide n!, so v_p(n!) = 0.

Highest Power in Factorial FAQs

What is the formula for the highest power of a prime p in n!?

Use v_p(n!) = floor(n/p) + floor(n/p²) + floor(n/p³) + ... . The highest power is then p^{v_p(n!)}.

What is the highest power of 5 in 100!?

v₅(100!) = floor(100/5) + floor(100/25) = 20 + 4 = 24. Therefore, the highest power of 5 is 5^24.

What is the highest power of 2 in 25!?

v₂(25!) = 12 + 6 + 3 + 1 = 22. Hence 2^22 is the highest power of 2 dividing 25!.

How do you find the highest power of 12 in 25!?

Since 12 = 2² × 3, calculate v₂(25!) = 22 and v₃(25!) = 10. Thus k = min(floor(22/2), 10) = 10, so the answer is 12^10.

Why are powers such as p² and p³ included in Legendre’s formula?

A multiple of p² contains at least two factors of p, and a multiple of p³ contains at least three. The additional terms count these extra factors after the first factor has been counted.

What happens when the prime p is greater than n?

No number from 1 to n is divisible by p, so v_p(n!) = 0. Therefore p does not divide n!.

Continue learning Highest Power in Factorial on PrepShots

Continue on PrepShots