Trailing Zeros in Factorial: Formula, Tricks and Examples

Trailing Zeros in Factorial means the zeroes at the end of the decimal form of n!. The count depends on the number of pairs of factors 2 and 5 in the factorial. Since factors of 2 are more frequent, the answer is found by counting factors of 5 using successive powers of 5.

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What Are Trailing Zeros in a Factorial?

The number of trailing zeros in n! is the exponent of 10 in its prime factorisation. It is calculated by counting the factors of 5 in n!, because factors of 2 occur more frequently.

Each trailing zero requires one factor pair of 2 × 5 = 10. For a positive integer n, the count is obtained by adding the multiples of 5, 25, 125 and higher powers of 5 that are not greater than n. For example, 25! has 5 + 1 = 6 trailing zeros.

Trailing Zeros in Factorial Formula & Tricks

Important Formulas

Trailing zero formula
Number of trailing zeros in n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...

Continue adding terms while 5^k ≤ n. Each term counts the contribution of factors of 5 from multiples of that power.

Prime-factor form
Number of trailing zeros in n! = min(v₂(n!), v₅(n!)) = v₅(n!)

A zero requires one factor 2 and one factor 5. Since v₂(n!) is greater than v₅(n!), the number of zeros equals the total exponent of 5.

Quick Tricks

Count multiples of 5, 25 and higher powers

Count every multiple of 5 once, every multiple of 25 an additional time, every multiple of 125 another time, and so on.

Example: For 100!, the count is ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24.
Do not stop at the first division by 5

Numbers such as 25, 125 and 625 contain more than one factor of 5. Their extra factors must be counted through successive powers of 5.

Example: For 125!, the count is 25 + 5 + 1 = 31, not just 25.

Trailing Zeros in Factorial Concepts

Why factors of 5 determine the zeros

A trailing zero is produced by a factor of 10, and each factor of 10 is formed from one factor of 2 and one factor of 5.

In n!, factors of 2 are more numerous than factors of 5 because every second number contributes a factor of 2, while only every fifth number contributes a factor of 5. Therefore, the smaller exponent, v₅(n!), determines the number of trailing zeros.

Example: In 10!, there are eight factors of 2 and two factors of 5, so 10! ends in two zeros.

Legendre method for n!

Use the sum of floor divisions by successive powers of 5 to find the number of trailing zeros in n!.

The required calculation is ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ... . Stop when the next power of 5 is greater than n. The first term counts multiples of 5, while later terms count extra factors from multiples of 25, 125 and higher powers.

Example: For 50!, the count is ⌊50/5⌋ + ⌊50/25⌋ = 10 + 2 = 12.

Extra factors from powers of 5

A multiple of 25 contributes two factors of 5, a multiple of 125 contributes three, and a multiple of 625 contributes four.

The first division by 5 counts such numbers once. Dividing by 25, 125 and higher powers counts their additional factors of 5. This prevents undercounting in factorials containing powers of 5.

Example: In 125!, the multiples of 5 contribute 25 factors, multiples of 25 contribute 5 extra factors, and 125 contributes 1 more factor. Thus, the total is 31.

Finding the smallest n for a required number of zeros

To find the smallest n for which n! has at least k trailing zeros, evaluate the formula for increasing candidates or use binary search.

The function Z(n) = ⌊n/5⌋ + ⌊n/25⌋ + ... is non-decreasing. For example, 24! has 4 zeros, while 25! has 6 zeros, so 25 is the smallest n for at least 6 trailing zeros.

Example: Z(24) = 4 and Z(25) = 5 + 1 = 6. Therefore, the smallest n such that n! has at least 6 zeros is 25.

Trailing Zeros in Factorial Video Lessons

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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.

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Practice Trailing Zeros in Factorial Questions

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1What is the largest five-digit number that is divisible by 9, 15, 16, 21, and 5?→ 2The LCM of the numbers 7, 12.4 and 1 is:→ 3Two ropes are of different lengths. The HCF of their lengths is 2 cm and the product of their lengths is 1848 cm². Find the ratio of their LCM to HCF.→ 436P2 is divisible by 6. What is the value of P?→ 5√((0.064 × 0.256)/(0.004 × 0.16)) = ?→ 6What is the digit y if the number 6y512 is exactly divisible by both 8 and 9?→ 7The difference between the two numbers is 1365. On dividing the larger by the smaller, the quotient is 6, and the remainder is 15. Find the smaller number.→ 8The ratio of three numbers is 2:5:6, and their least common multiple (LCM) is 180. Find the greatest common factor (HCF) of the numbers.→ 9The largest four-digit number which when divided by 8, 7 and 11 leaves remainder 4 in each case is:→ 10Least Common Multiple of A and B is 20. Least Common Multiple of C and D is 30. What is the Least Common Multiple of A, B, C and D?→

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Quick Revision Notes

Trailing Zeros in Factorial: Quick Revision

Use these rules to calculate trailing zeros in n! quickly and accurately.

  • A trailing zero is formed by one pair of factors 2 and 5.
  • For n!, factors of 2 exceed factors of 5, so count the factors of 5.
  • Formula: Z(n!) = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ... .
  • Continue until the power of 5 becomes greater than n.
  • Multiples of 25, 125 and higher powers contribute extra factors of 5.
  • For 100!, Z(100!) = 20 + 4 = 24.
  • For 125!, Z(125!) = 25 + 5 + 1 = 31.

Trailing Zeros in Factorial FAQs

What is the formula for trailing zeros in n!?

The formula is Z(n!) = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ... . Add terms until 5^k exceeds n.

How many trailing zeros does 100! have?

Z(100!) = ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24. Therefore, 100! has 24 trailing zeros.

How many trailing zeros does 50! have?

Z(50!) = ⌊50/5⌋ + ⌊50/25⌋ = 10 + 2 = 12. Hence, 50! has 12 trailing zeros.

Why are multiples of 25 counted twice?

Every multiple of 25 contains at least two factors of 5. The division by 5 counts one factor, and the division by 25 counts its additional factor.

How many trailing zeros are in 125!?

Z(125!) = ⌊125/5⌋ + ⌊125/25⌋ + ⌊125/125⌋ = 25 + 5 + 1 = 31.

What is the smallest n for which n! has at least 6 trailing zeros?

The smallest value is n = 25. Since Z(24!) = 4 and Z(25!) = 5 + 1 = 6, 25! is the first factorial with at least 6 trailing zeros.

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