All Questions
Tagged with prime-factorization factorial
55
questions
58
votes
2
answers
8k
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Can I search for factors of $\ (11!)!+11!+1\ $ efficiently?
Is the number $$(11!)!+11!+1$$ a prime number ?
I do not expect that a probable-prime-test is feasible, but if someone actually wants to let it run, this would of course be very nice. The main hope ...
- 85.1k
26
votes
1
answer
536
views
Prime factor wanted of the huge number $\sum_{j=1}^{10} j!^{j!}$
What is the smallest prime factor of $$\sum_{j=1}^{10} j!^{j!}$$ ?
Trial :
This number has $23\ 804\ 069$ digits , so if it were prime it would be a record prime. I do not think however that this ...
- 85.1k
8
votes
2
answers
482
views
Find the greatest power of $104$ which divides $10000!$
Find the greatest power of $104$ which divides $10000!$
I thought $$104=2^3\cdot13$$ so I have to find $n$ such that $$(2^3\cdot13)^n\mid 10000!$$ Obviously, we can see that there are fewer factors $...
- 3,789
7
votes
2
answers
7k
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Factorials and Prime Factors
I need to write a program to input a number and output it's factorial in the form:
$4!=(2^3)(3^1)$
$5!=(2^3)(3^1)(5^1)$
I'm now having trouble trying to figure out how could I take a number and get ...
- 71
7
votes
2
answers
187
views
The number of primes in the factorization of $N!$
Is there an approximation to the number of primes in the factorization of $N!$?
For example:
For $N=10$, this number is $15$.
For $N=100$, this number is $239$.
For $N=1000$, this number is $2877$.
...
- 43.2k
6
votes
3
answers
129
views
If $N$ is a multiple of $100$, $N!$ ends with $\left(\frac{N}4-1 \right)$ zeroes.
Did certain questions about factorials, and one of them got a reply very interesting that someone told me that it is possible to show that
If $N$ is a multiple of $100$, $N!$ ends with $\left(\frac{...
- 3,789
6
votes
2
answers
205
views
(1) Sum of two factorials in two ways; (2) Value of $a^{2010}+a^{2010}+1$ given $a^4+a^3+a^2+a+1=0$.
Question $1$:
Does there exist an integer $z$ that can be written in two different ways as $z=x!+y!$,where $x,y\in \mathbb N$ and $x\leq y$?
Answer: $0!=1!$ so $0!+2!=3=1!+2!$
Question $2$:
If $...
- 3,569
6
votes
2
answers
9k
views
Total number of divisors of factorial of a number
I came across a problem of how to calculate total number of divisors of factorial of a number. I know that total number of divisor of a number $n= p_1^a p_2^b p_3^c $ is $(a+1)*(b+1)*(c+1)$ where $a,...
- 298
6
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1
answer
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Conjecture: $\prod\limits_{k=0}^{n}\binom{2n}{k}$ is divisible by $\prod\limits_{k=0}^n\binom{2k}{k}$ only if $n=1,2,5$.
The diagram shows Pascal's triangle down to row $10$.
I noticed that the product of the blue numbers is divisible by the product of the orange numbers. That is (including the bottom centre number $...
- 25.8k
6
votes
1
answer
2k
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Find smallest positive integer $n$ such that $n!$ is divisible by $p^k$ ($p =$ positive prime, $k =$ positive integer)
I have to find smallest positive integer $n$ in such way that $n!$ is divisible by $p^k$ ($p$ is always positive prime and $k$ is always positive integer).
$p$ and $k$ are given, $n$ is (obviously ...
user506862
6
votes
0
answers
100
views
Can we conclude $n=p-1$?
Let $\ n\ $ be a positive integer and $\ p\ $ a prime number such that $$\ p^2\mid (2n)! + n! + 1$$ The only pairs $\ (n,p)\ $ I found so far are
$(1,2)$ , $(2,3)$ , $(10,11)$ , $(106,107)$ , $(4930,...
- 85.1k
5
votes
6
answers
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Product of r consecutive integers is divisible by r! [duplicate]
Well in a book i am reading it is given that you can also prove this by showing that
Every prime factor is contained in $(n+r)!$ as often at least as it is contained in $n!r!$.
How does this prove ...
- 6,922
5
votes
2
answers
191
views
Show that there is no natural number $n$ such that $3^7$ is the largest power of $3$ dividing $n!$
Show that there is no natural number $n$ such that $7$ is the largest power $a$ of $3$ for which $3^a$ divides $n!$
After doing some research, I could not understand how to start or what to do to ...
- 3,789
4
votes
2
answers
2k
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Factoring the factorials
Just for the fun of it, I've started factoring $n!$ into its prime divisors, and this is what I got for $2\leq n\leq20$:
$$\begin{align}
2! &= 2^\color{red}{1} &S_e=1\\
3! &= 2^\color{red}...
- 7,532
4
votes
1
answer
79
views
Solution of $n!=p+1 $ with $p$ is prime number?
One of my friend asked me to solve this equation $n!=p+1 $ with $p$ is prime number and n is positive integer , it's clear that for $p=2$ there is no solutions because : $n! < 3$ for $n=1$ , But ...
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