All Questions
Tagged with prime-factorization factorial
55
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6
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(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 $...
- 3,569
0
votes
1
answer
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How to count Terminal Zeros from subtraction
How do you count the number of zeros that the subtraction 100$^{100}$ - 100! ends in? In particular, I want to know exactly why my approach is wrong, because I know from my source that the answer is ...
4
votes
1
answer
56
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Reasoning about a sequence of consecutive integers and factorials with hope of relating factorials to primorials
I am looking for someone to either point out a mistake or help me to improve the argument in terms of clarity, conciseness, and more standard mathematical argument.
Let $x$ be an integer such that $x,...
- 10.5k
1
vote
2
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Prime factorization of factorials
Is there a way given a sequence of naturals $a_1, a_2, ..., a_k$ to determine whether $c=n!$ for some number $n$ where
$$ c = 2^{a_1} 3^{a_2}5^{a_3}7^{a_4}...$$ (2,3,5,7,... - primes)
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-2
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Divisor Problem [closed]
Express the numbers $11! = 39,916,800$ and the binomial coefficient $\binom {23} {11}$, each as products of their prime factors. Do this without using your calculator in any way. Use this to calculate ...
0
votes
2
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362
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Express $11!$ and $\binom{23}{11}$ as products of their prime factors
I'm a bit stuck on how to figure this question out without a calculator and what kind of working I'm supposed to show. Any help would be appreciated, thank you. $\ddot\smile$
Factorise $11!$ and $\...
- 55
4
votes
2
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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}...
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Number of Divisors of N factorial
Say $d(N) =$ Number of factors of $N!$
Briefly: I wish to know if there is a Recurrence relation for this problem.
Now I wish to Know if there is a way to calculate $d(N)$ in terms of previously ...
- 13
2
votes
1
answer
182
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Total possible ways of representing n! as a sum of two or more consecutive positive integers.
I need to calculate total possible ways of representing $n!$ as a sum of two or more consecutive positive integers.
Example : $3!=1*2*3=6$ and $6=1+2+3$ the only one possible way.
Answer : $1$
The ...
- 298
6
votes
2
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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
0
votes
2
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834
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Find the number of trailing zeroes. [duplicate]
Find the number of trailing zeroes.
$k=1^1\times 2^2\times 3^3\times \cdots \times100^{100}$
It usually involves calculating number of $5$'s in
$5^5\times 10^{10}\times 15^{15}\times \cdots\times ...
- 2,995
3
votes
3
answers
389
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The ultimate formula to factor them all.
Context
I am working on Integer factorization problem, I found a formula for factoring numbers, and I need your help to simplify it. First I will explain how I get there and then I present the ...
- 1,450
2
votes
2
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229
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Integer factorization: What is the meaning of $d^2 - kc = e^2$
I found an interesting behavior when placing the integer factorization problem in to geometry, I call it pyramid factoring.
Lets assume we have $c$ boxes and we want to order them in to rectangle. ...
- 1,450
1
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Prove that if $p\le n$, then $p$ does not divide $n! + 1$
I'm having trouble on how to approach this problem
Prove that if $p\le n$, then $p$ does not divide $n! + 1$ ($p$ is prime and $n$ is an integer).
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If $x$ and $y$ are positive integers then $\frac{(2x)!(2y)!}{x!y!(x+y)!}$ is an integer
If $x$ and $y$ are positive integers then $\frac{(2x)!(2y)!}{x!y!(x+y)!}$ is an integer
I have to show that the proposition above is true for any $x,y\in\mathbb{Z^+}$ by means of Legendre's formula.....
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