Linked Questions
10 questions linked to/from How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
75
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Highest power of a prime $p$ dividing $N!$
How does one find the highest power of a prime $p$ that divides $N!$ and other related products?
Related question: How many zeros are there at the end of $N!$?
This is being done to reduce abstract ...
5
votes
3
answers
3k
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how many zeroes does 2012! have at the end? [duplicate]
Possible Duplicate:
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
How many zeroes does $2012!$ end with?
My idea is:
402 zeroes come from $...
- 2,695
14
votes
2
answers
7k
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Derive a formula to find the number of trailing zeroes in $n!$ [duplicate]
Possible Duplicate:
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
I know that I have to find the number of factors of $5$'s, $25$'s, $125$'s etc....
- 1,047
9
votes
2
answers
3k
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Number of zeros not possible in $n!$ [duplicate]
Possible Duplicate:
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
The number of zeros which are not possible at the end of the $n!$ is:
$...
- 141
6
votes
1
answer
2k
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How to find the highest power of a prime $p$ that divides $\prod \limits_{i=0}^{n} 2i+1$? [duplicate]
Possible Duplicate:
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
Given an odd prime $p$, how does one find the highest power of $p$ that ...
- 12.1k
32
votes
7
answers
3k
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Prove that $\frac{100!}{50!\cdot2^{50}} \in \Bbb{Z}$
I'm trying to prove that :
$$\frac{100!}{50!\cdot2^{50}}$$
is an integer .
For the moment I did the following :
$$\frac{100!}{50!\cdot2^{50}} = \frac{51 \cdot 52 \cdots 99 \cdot 100}{2^{50}}$$
...
- 2,379
0
votes
2
answers
701
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Factorial related problems [duplicate]
How many zeros are there in $ 25!$ ?
My answer was $6$. But i solved it by finding how many numbers are divisible by $5$ and $2$.here i was told to find out the zeros at the last end.
But what is the ...
- 363
3
votes
1
answer
420
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Given $n$, find smallest number $m$ such that $m!$ ends with $n$ zeros
I got this question as a programming exercise. I first thought it was rather trivial, and that $m = 5n$ because the number of trailing zeroes are given by the number of factors of 5 in $m!$ (and ...
- 9,730
3
votes
2
answers
269
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Concecutive last zeroes in expansion of $100!$ [duplicate]
Possible Duplicate:
Highest power of a prime $p$ dividing $N!$
In decimal form, the number $100!$ ends in how many consecutive zeroes. I am thinking of the factorization of $100!$ but I am stuck. ...
- 1,711
0
votes
1
answer
434
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Two questions on finding trailing digits in (large) numbers and one on divisibility
Without using a calculator, how can we solve the following?
How do we find the number of zeros at the end of $600!$
What are the last 3-digits of $171^{172}$?
What is the sum of all positive numbers ...
- 165