For a prime , let be the sum of the digits of when written in base- form. Then the number of factors of that divide is

There are trailing zeroes in . Since $100_{\text{ten}}=400_{\text{five}}$, there are factors of in .

However, there are other zeros that occur earlier, making the total :

$933262154439441526816992388562667\color{#C00000}{00}49\color{#C00000}{0}7159682643816214685929638952175999932299156\color{#C00000}{0}894146397615651828625369792\color{#C00000}{0}82722375825118521\color{#C00000}{0}916864\color{#C00000}{000000000000000000000000}$

Using well known approximations for the length and number of trailing zeroes of n!, and making the reasonable assumption that the inside zeros appear with frequency , we get the following approximation of the total number of zeros, t, in n!: · · Which simplifies to: · · This approximation seems to work well for n up to at least 10,000. · 100!, with digit length 158, has less inside zeroes, 6, with 24 trailing, than the normal expectation for a total of 30, with t=36. · 98! is "zero-perfect", i.e. inside zeroes appear with exactly frequency , with actual total zero count 35 and · Other examples of zero-perfect factorials are: 1009!, 1097!, 1112!, 2993!, 6128!, .... · There appears to be a strong correlation of n having only 0-3 prime factors in {2, 3, 5} if n! is zero-perfect. Uneven n is often a prime number if n! is zero-perfect.

The total number of zeros in n! is given by sequence A027869 in the On-line Encyclopedia of Integer Sequences. There really seems to be no way to compute the total number of zeros in n! short of computing n! and counting the number of zeros. With a big int library, this is easy enough. A simple Python example:

import math

def zeros(n): return str(math.factorial(n)).count('0')

So, for example, zeros(100) evaluates to 30. For larger n you might want to skip the relatively expensive conversion to a string and get the 0-count arithmetically by repeatedly dividing by 10.

As you have noted, it is far easier to compute the number of trailing zeros. Your code, in Python, is essentially:

def trailing_zeros(n):
    count = 0
    p = 5
    while p <= n:
        count += n//p
        p *= 5
    return count

As a heuristic way to estimate the total number of zeros, you can first count the number of trailing zeros, subtract that from the number of digits in n!, subtract an additional 2 from this difference (since neither the first digit of n! nor the final digit before the trailing zeros are candidate positions for non-trailing zeros) and guess that 1/10 of these digits will in fact be zeros. You can use Stirling's formula to estimate the number of digits in n!:

def num_digits(n):
    #uses Striling's formula to estimate the number of digits in n!
    #this formula, known as, Kamenetsky's formula, gives the exact count below 5*10^7
    if n == 0:
        return 1
    else:
        return math.ceil(math.log10(2*math.pi*n)/2 + n *(math.log10(n/math.e)))

Hence:

def est_zeros(n):
    #first compute the number of candidate postions for non-trailing zerpos:
    internal_digits = max(0,num_digits(n) - trailing_zeros(n) - 2)
    return trailing_zeros(n) + internal_digits//10 

For example est_zeros(100) evaluates to 37, which isn't very good, but then there is no reason to think that this estimation is any better than asymptotic (though proving that it is asymptotically correct would be very difficult, I don't actually know if it is). For larger numbers it seems to give reasonable results. For example zeros(10000) == 5803 and est_zeros == 5814.

Between 1 and 100, 20 numbers are divisible by 5. This "contributes" 520. 4 of these are divisible by 25, which gets you another 54. No other numbers contribute any factors of 5. So there can be at most 24 zeroes at the end. 50 numbers between 1 and 100 are divisible by 2. This "contributes" 250. 24 of these twos can be grouped with the fives to produce 24 factors of 10. Conclusion: 24 zeroes.

Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow. Fro example, 125000 has 3 trailing zeros (125 000 ); The number of trailing zeros in the decimal representation of n! , the factorial of a non-negative integer n, can be determined with this formula: \frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k} , where k must be chosen such that 5^(k+1)>n It's more simple if you look at an example: How many zeros are in the end (after which no other digits follow) of 32!? \frac{32}{5}+\frac{32}{5^2}=6+1=7 (denominator must be less than 32, 5^2=25 is less) So there are 7 zeros in the end of 32! The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero. BACK TO THE ORIGINAL QUESTION: According to above 100! has \frac{100}{5}+\frac{100}{25}=20+4=24 trailing zeros. Answer: B. For more on this issues check Factorials and Number Theory links in my signature. Hope it helps.

"Given:  100! Concept Used: No. of zeroes = Sum of all quotient Calculation: Divide divisor by 5 and add the respective quotient 100 ÷ 5 = 20 20 ÷ 5 = 4 Adding the respective quotients  20 + 4 = 24 ∴ No. of zeroes in the end is 24. "

One thing is clear. $2 \times 5 = 10$ and there is no other way to get 10 out of 2 prime numbers.

"trailing zeros" are the zeros at the end of the number.
For example: 3200 has 2 trailing zeros. The units and the tenths position.

One other thing is clear.
Multiplying a number by 10 adds a trailing zero to that number.

So in order to find the number of zeros at the tail of a number, you need to split that number into prime factors and see how many pairs (2, 5) you can form.

For example:

300 has 2 trailing zeros. Why?
because $300 = 3 \times 2 ^ 2 \times 5^2$.
So you get 2 pairs of (5, 2).

An other example:

$4000 = 2^5 \times 5 ^3 = 2 ^2 \times (2 \times 5) ^ 3$
Hence, 3 trailing zeros.

EDIT:

For your specific question. 100!.

$100! = 1 \times 2 \times 3 \times .... \times 100$.

Which is equivalent to splitting every number into prime factors like this:

$100! = 1 \times 2 \times 3 \times 2^2 \times 5 \times (3\times 2) .... \times (2 \times 5)^2$

You need to count how many 10's you can make out of these prime numbers.
As explained above, the only way to get a 10 is $2 \times 5$.
So you need to count how many 2's and how many 5s are among the prime factors above.

The 5s can appear only in numbers that are divisible by 5.

5, 10, 20, 25, 30, 35, 40, 45, 50 , 55, 60 , 65, 70, 75, 80, 85, 90, 95, 100

and the number of 5s in the numbers above are

1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2  

Because for example

$15 = 3 \times 5$ so one single 5.
$50 = 2 \times 5 ^2$ so 2 fives.

Add them all up and you get 24 occurrences of 5.

Now you need to see if there are at least 24 occurrences of 2.
And they are because form 1 to 100 you have 50 even numbers (they all contain at least one 2).

This means that 100! can be written as

$2 ^ {50} \times 5 ^ {24} \times$ (some other factors that are never going to multiply to get to a multiple of 10 so we can ignore them).

We can split the line above (ignoring the factors that cannot multiply to reach a multiple of 10) to

$2 ^ {24} \times 2 ^{26} \times 5^{24} = (2 \times 5) ^{24} \times 2 ^{26}$

We can ignore again $2^{26}$ because that never ends with zeros (it ends with a 4 if I'm not mistaken)

Now you can see that 100! is basically a very large number that does not end with a 0 multiplied by $10 ^ {24}$. So you get 24 trailing zeros.

While adding two numbers, the numbers of zeros will depend on the number with lesser number of zeros. For example, 200 + 2000 will have only 2 trailing zeros and the number of zeros is limited by 200 which has 2 only zeros. So, instead of wasting time in finding the number of zeros of 200!, we can simply find the number of zeros in 100! and mark the answer. The number of zeros in 100! = \frac{100}{5} + \frac{20}{5} = 20 + 4 = 24 Hence the correct answer is Option E . Regards, Saquib e-GMAT Quant Expert