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 (125000);

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 121! has \(\frac{121}{5}+\frac{121}{25}=24+4=28\) trailing zeros.

Answer: C.

For more check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.



Hope this helps.
_________________

\(\frac{121}{5} = 24\)
\(\frac{24}{5} = 4\)

So, Total no of trailing zeroes is 28 , answer will be (C) 28
_________________

Thanks VeritasKarishma for your two cents.



To confirm your approach for highlighted text, you did not perform 121! as 121 * 120 * ...1
Instead you found a pattern that just as 10! has more even numbers from 1 to 10
than multiples of 5 from 1 to 10, similarly 121! will have follow same pattern, correct?
_________________

To determine the number of trailing zeros in a number, we need to determine the number of factors of 10, which is equal to the number of 5- and-2 pairs within the prime factorization of that number.

Since we know there are fewer 5s in 121! than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 121!, we can use the following shortcut in which we divide 121 by 5, then divide the quotient of 121/5 by 5 and continue this process until we no longer get a nonzero quotient.

121/5 = 24 (we can ignore the remainder)

24/5 = 4 (we can ignore the remainder)

Since 4/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 121!.

Thus, there are 24 +4 = 28 factors of 5 within 120! and so there are 28 5-and-2 pairs, creating 28 trailing zeros.

Answer: C
_________________

Hi onyx12102,

IMO, we will carry out the operations(successive division & addition) as long as \({Denominator}\leq{Numerator}\) Or,\(5^k\leq{n}\)

Here n=121, So, \(5^k\) is less than 121 when k=2 (\(5^3=125\), DISCARD)

So, \(\frac{121}{5}+\frac{121}{5^2}\)
_________________

to calculating terminating zeroes, divide the number by powers for 5...

121/5 + 121/5^2 = 121/5 + 121/25 = 24 + 4 =28
_________________