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Highest power of a prime $p$ dividing $N!$

How many factors of 10 are there in $100!$ (IIT Question)?

Is it 26,25,24 or any other value

Please tell how you have done it

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marked as duplicate by Marvis, Amr, Douglas S. Stones, Fabian, Stefan Hansen Jan 12 '13 at 9:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Im getting my ans as 24 is it correct – Bharat Khanna Jan 12 '13 at 5:24
Do you mean how many factors of 10 are there in $100!$? 24 is correct. – NeuroFuzzy Jan 12 '13 at 5:25
For ex 6 factorial value is 720 so the power 0f 10 is 1 – Bharat Khanna Jan 12 '13 at 5:25
Yes neuro fuzzy thank you – Bharat Khanna Jan 12 '13 at 5:26
Can yout tell how you have done it becoz my method is destructable – Bharat Khanna Jan 12 '13 at 5:26
up vote 3 down vote accepted

So first, we want to find how many factors of $5$ there are in $100!$. There are $20$ numbers divisible by $5$ from $1$ to $100$, so we start off our count at $20$. Then, we count how many numbers are divisble by $5^2$. There are four: $25, 50, 75, 100$, and so we add four to our count to get $24$ factors of $5$. (Note that we don't add eight fives - if we did so, we would be counting the first factors of five twice!)

Since $5^3>100$, we don't have to worry about third powers of five. There are at least $100/2=50$ factors of $2$ in $100!$, but we're only going to use $24$ of them to get our $24$ multiples of $10$, so we don't have to calculate the exact number of factors of $2$ in $100!$.

So basic method: To find how many factors of $a$ there are in $b!$, first decompose $a$ into its primes $p_n$, and then find out how many factors of each prime $p_n$ are in numbers less than $b$, by using the method I described of checking for divisibility by $p_n$, then $p_n^2$, etc. Then, from this pool of factors, figure out how many you can take. In our examples to make $10^n$ we could take a maximum of $24$ fives and $24$ twos. If we wanted to find how many factors of $40$ (=$2^3 5$) were less than $100$, we would have needed to find out exactly how many factors of $2$ were less than $100$, and then either take $24*3$ twos if there are enough, or less, if there aren't.

See also: youtube Factors of Factorials Part 1

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