Finding the number of trailing zeroes in 100! (read factorial 100) has become the favorite question in quant among aspirants who prepare for CAT Exam. * *In this section, we will be discussing the approaches to solve similar sorts of questions.

## Exponent of any prime number p in n! (factorial n)

The exponent of any prime number p in n! = \(\left[ {\frac{{\rm{n}}}{{{{\rm{p}}^1}}}} \right] + \left[ {\frac{{\rm{n}}}{{{{\rm{p}}^2}}}} \right] + \left[ {\frac{{\rm{n}}}{{{{\rm{p}}^3}}}} \right] + ..\), where, [ ] is the greatest integer function.

Here is the video explanation of the above concept

Example 1: Find the highest power of 3 in 100!

__Solution:__

Highest power of 3 in 100! = \(\left[ {\frac{{100}}{{{3^1}}}} \right]\) + \(\left[ {\frac{{100}}{{{3^2}}}} \right]\) + \(\left[ {\frac{{100}}{{{3^3}}}} \right]\) + \(\left[ {\frac{{100}}{{{3^4}}}} \right]\) = 33 + 11 + 3 + 1 = **48**

*Point to note:*

The above solution implies that when 100! is prime factorized, then the prime number 3 will have exponent or power equal to 48.

Similarly, we can find the power of 2 in 100! or any prime number less than 100. By calculation, an obvious thing to observe is that power of 3 will always be greater than power of 5 but less than power of 2. We will use this observation in the next example.

Example 2: Find the highest power of 21 in 100!

__Solution:__

As 21 = 3 \( \times \) 7

By formula, exponent of 3 and 7 in 100! is 48 and 16 respectively. Therefore, number of pairs of 3 and 7 (i.e. 3\( \times \)7) will be 16 only. So, the highest power of 21 in 100! is 16

### Wrong Approach:

Solving the above problem by using the formula directly like \(\left[ {\frac{{100}}{{{{21}^1}}}} \right]\) + \(\left[ {\frac{{100}}{{{{21}^2}}}} \right]\) +.. is **wrong** as 21 is NOT a prime, the formula is applicable only for prime numbers.

Example 3: Find the highest power of 12 in 100! (read factorial 100).

__Solution:__

\(12 = {2^2} \times 3\)

By formula, exponent of 2 and 3 in 100! is 97 and 48 respectively.

To make one 12, we need two 2’s and one 3. The total number of 2’s is 97, and 3’s is 48. For every one 3 we need two 2’s, hence we will get 48 group of \(\left( {{2^2} \times 3} \right)\) by using 96 2’s and 48 3’s. So the answer to the question is 48.

### Finding Number of Trailing Zeroes in Factorial:

Example 4: Find the number trailing of zeroes at the end of 100!

__Solution:__

The number of trailing zeros is equal to the number of times 100! is successively divisible by 10.

Also, 10 = 2 \( \times \) 5. As power of 5 will always be less than power of 2, the number of 10’s in 100! is equal to Exponent of 5 in 100!

i.e. \(\left[ {\frac{{100}}{5}} \right]\) + \(\left[ {\frac{{100}}{{{5^2}}}} \right]\) = 20 + 4 = 24. Hence there are 24 trailing zeros in 100!

**Read Also:**

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