Linked Questions
72 questions linked to/from Highest power of a prime $p$ dividing $N!$
44
votes
3answers
2k views
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes? [duplicate]
Possible Duplicate:
Highest power of a prime $p$ dividing $N!$
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
3
votes
2answers
3k views
Prime powers that divide a factorial [duplicate]
If we have some prime $p$ and a natural number $k$, is there a formula for the largest natural number $n_k$ such that $p^{n_k} | k!$.
This came up while doing an unrelated homework problem, but it is ...
6
votes
2answers
4k views
Exponent of Prime in a Factorial [duplicate]
I was just trying to work out the exponent for $7$ in the number $343!$. I think the right technique is $$\frac{343}{7}+\frac{343}{7^2}+\frac{343}{7^3}=57.$$ If this is right, can the technique be ...
0
votes
2answers
8k views
What is the largest power of 2 that divides $200!/100!$. [duplicate]
What is the largest power of 2 that divides $200!/100!$.
No use of calculator is allowed.
I had proceeded in a brute force method which i know regret..
I would like to know your methods.
1
vote
2answers
785 views
Counting the number of zeros [duplicate]
I am stuck at the question:
How many zeros are there when numbers between 1 and 100 are multiplied including 1 and 100, devise some technique for this .
Regards
0
votes
2answers
919 views
the exponent of the highest power of p dividing n! [duplicate]
The formula for the exponent of the highest power of prime $p$ dividing $n!$ is $\sum \frac{n}{p^k}$, but the question is $n=1000!$ (really, it has the factorial) and $p=5$.
When I use Wolfram Alpha ,...
-4
votes
1answer
366 views
How many factors of $10$ in $100!$ [duplicate]
Possible Duplicate:
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
1
vote
0answers
471 views
Determine the number of 0 digits at the end of 100! [duplicate]
I got this question, and I'm totally lost as to how I solve it!
Any help is appreciated :)
When 100! is written out in full, it equals
100! = 9332621...000000.
Without using a calculator, determine ...
3
votes
2answers
211 views
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. ...
0
votes
2answers
93 views
The method of solving for a factor of $90!$ [duplicate]
If $90! = (90)(89)(88)...(2)(1)$, then what is the exponent of the highest power of $2$ which will divide $90!$ ?
How would I apply one of the easiest method from Here?
I need help on applying the ...
2
votes
1answer
1k views
Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number. [duplicate]
Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number.
Actually, I know a way to solve this, but even if it is very large and ...
2
votes
2answers
103 views
What is the logic/theorem/derivation behind finding the exponent of p in n! By [n/p] + [n/p^2] + [n/p^3] + …? [duplicate]
The exponent of prime number of 3 in 100! is 48.
It means 100! is divisible by $3^48$
$$E_3(100!) = \left\lfloor\frac{100}3\right\rfloor + \left\lfloor\frac{100}{3^2}\right\rfloor + \left\lfloor\frac{...
0
votes
1answer
75 views
Dividing $61!$ by $3$ as we can. [duplicate]
Find maximum possible $n$ in the equation $61!=3^n\cdot m$.
Some textbooks gives a solution to this question like this:
\begin{align*}
61&=\textbf{20}\cdot 3+1 \\
20&=\textbf{6}\cdot 3+2 \\
6&...
0
votes
1answer
94 views
Prove the multiplicity property for $n!$ [duplicate]
I was given this hint in a different problem,
Now use that a prime $p$ occurs in $n!$ with multiplicity exactly $\lfloor n/p\rfloor + \lfloor n/p^2\rfloor + \lfloor n/p^3\rfloor + \lfloor n/p^4\...
2
votes
1answer
119 views
Why does $\sum\limits_{k=1}^\infty \lfloor m/(n^k)\rfloor$ give you the number of times that $n$ divides $m!$? [duplicate]
If $n$ is a prime less than $m$, with $n,m \in \mathbb N$, why does $$\sum_{k=1}^\infty \left\lfloor \frac{m}{n^k}\right\rfloor$$
give you the number of times that $n$ divides $m!$?
Examples:
$n=13$
...