0
votes
2answers
79 views

Prove that $6! \mid n(n+1)…(n+5)$ [closed]

Prove that for all $n \in \mathbb{Z}$, $6! \mid n\cdot(n+1)\cdots(n+5)$ using only criteria of divisibility (without using combinatorial arguments).
0
votes
2answers
24 views

Factorial Summation Problem [duplicate]

$$\sum_{j=0}^n j\cdot j!$$ I got $(n+1)!-1$ as the answer but I'm not sure if that's right or how I even got to that answer exactly. (my paper is a mess of random work and I can't make it out). Can ...
1
vote
2answers
39 views

Factorial as a sum. Insight appreciated

I recently posted an answer to a question about ways to express the factorial function as a sum. I posted the following formula, which I discovered several years ago and I haven't seen anywhere else: ...
2
votes
1answer
51 views

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 ...
0
votes
1answer
37 views

Maximum value of function involving factorials

Define $$g_{(k,j)} = \frac{a^{n-k}b^k(k+n)!x^{k+n-j}}{k!(n-k)!(k+n-j)!}$$, where $n,k,j \in \Bbb{N}$ are fixed such that $(0 \leq x \leq a/b ),(b<a),(0 \leq k \leq n ),(2 \leq j \leq 2n),(0 \leq ...
0
votes
0answers
33 views

Reasoning about factorials and powers of a finite set of primes

I am working on an answer to another question: How to prove $k!+(2k)!+\cdots+(nk)!$ has a prime divisor greater than $k!$ I've reduced the question to showing that the following infinite set of ...
5
votes
4answers
201 views

Direct proof that $n!$ divides $(n+1)(n+2)\cdots(2n)$

I've recently run across a false direct proof that $n!$ divides $(n+1)(n+2)\cdots (2n)$ here on math.stackexchange. The proof is here prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer (it ...
3
votes
3answers
96 views

The sum $1!+2!+3!+…+2007!$ is not a perfect square

Today my teacher told me to prove this:- Prove that $1!+2!+3!+...+2007!=\sum_{n=1}^{2007}(n!) $ is neither a perfect square nor a perfect cube. Not getting any idea. Please help. Thanks in advance.
1
vote
3answers
254 views

Prove that no number in this list is prime - Formatting a proof advice

Question: Let $n \in \mathbb{Z}$ where $n \geq 2$, prove no number in the list: $$n! + 2, n! + 3,...,n! + n$$ is prime. I have written my proof exactly as follows: Proof: $P(n) = n! + n = ...
-2
votes
2answers
211 views

Even or Odd for factorial

Moderator Note: This is a current contest question on codechef.com. Given $N$ and $M$ I need to tell whether $\left\lfloor \large\frac{N!}{M} \right\rfloor$ is even or odd.How to do this ...
8
votes
1answer
120 views

What are the conditions for $n^2 \nmid(n-1)!$

Q: What are the conditions for $n^2 \nmid (n-1)!$, given that $2\le n \le 100$ and $n\in \mathbb{N}$? According to me the two conditions must be: 1. $n$ is a prime number (since the factorization ...
3
votes
2answers
687 views

How many perfect squares divide 1!2!3!4!5!6!7!8!9!

What I naturally did was to find the prime factorisation of the product of factorials which is $ 2^{30}3^{13}5^5 7^3 $. Clearly there is 15 unique perfect squares that divide $2^{30}$, 6 unique ...
6
votes
2answers
135 views

Prove that $\sqrt[2012]{2012!}<\sqrt[2013]{2013!}$

I need to prove that $\sqrt[2012]{2012!}<\sqrt[2013]{2013!}$ My attempt: Let $a=\sqrt[2012]{2012!}$ and $b=\sqrt[2013]{2013!}$ Then $\displaystyle\frac{b^{2012}}{a^{2012}}=\frac{2013}{b}$ ...
10
votes
2answers
544 views

What is the remainder when $1! + 2! + 3! +\cdots+ 1000!$ is divided by $12$?

What is the remainder when $$1! + 2! + 3! +\cdots+ 1000!$$ is divided by $12$. I tried to do it using binomial theorem but that doesn't help. How will we do this? Please help.
4
votes
1answer
85 views

Simple method for $\frac{(2n+1)!}{(n!)^{2}}$ divide $lcm(1,2,\ldots,2n+1)$

The question is to prove that $\frac{(2n+1)!}{(n!)^{2}}$ divides $lcm(1,2,\ldots,2n+1)$. This seems like it should be a simple question, but try as I might, I can't seems to find any way that does ...
1
vote
4answers
154 views

Which is larger :: $y!$ or $x^y$, for numbers $x,y$.

This is a generalization of this question :: Which is larger? $20!$ or $2^{40}$?. No explicit general solution was presented there and I'm just curious :D Thank-you. Edit :: I want a most-general ...
4
votes
2answers
87 views

Show that $p!$ and $(p - 1)! - 1$ are relatively prime

If $p$ is prime number, with $p>3$ Show that $p!$ and $(p - 1)! - 1$ are relatively prime. I tried $\text{gcd}\;(p!,(p-1)!-1)=d\Longrightarrow d\mid p!$ e $d\mid(p-1)!-1$ having ...
1
vote
1answer
35 views

show that if $m,n\in\mathbb{N}$ are such that $(m,n)=1$, then $\frac{(m+n-1)!}{m!n!}\in\mathbb{N}.$

show that if $m,n\in\mathbb{N}$ are such that $(m,n)=1$, then $$\frac{(m+n-1)!}{m!n!}\in\mathbb{N}.$$ I have a theorem (shown in my text) that says $$\fbox{If $a_1,...,a_m\in\mathbb{N}^*$ then ...
5
votes
2answers
101 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 ...
4
votes
3answers
76 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 ...
2
votes
1answer
39 views

Find the smallest value of $n$ so that the greater potency of $5$ which divides $n!$ is $5^{84}$. What are the other numbers that enjoy this property?

Find the smallest value of $n$ so that the greater potency of $5$ which divides $n!$ is $5^{84}$. What are the other numbers that enjoy this property? I thought I would put together an equation ...
7
votes
2answers
85 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 ...
2
votes
1answer
99 views

Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number.

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
114 views

Show that $\frac{(2n)!}{(n)!}=2^n(2n-1)!!$

Show that $\frac{(2n)!}{(n)!}=2^n(2n-1)!!$ is the question I am struggling with. I started by saying: $(2n)!=2n(2n-1)(2n-2)(2n-3)...3*2*1$ But then I'm stuck.
4
votes
1answer
58 views

Congruences with prime number and factorial

Prove that if $p\equiv 1 \pmod{4}$ is a prime number and $$x\equiv \pm \left(\frac{p-1}{2}\right)! \pmod{p}$$ then $x^2\equiv -1 \pmod{p}$ I think Wilson's theorem will come in handy here, used ...
2
votes
2answers
1k views

the nth root of n!?

I am playing around with the root/ratio test to practice with series. I just showed that $\sum \frac{1}{n!}$ converges by using the ratio test. I decided to see how things would go with the root test ...
1
vote
3answers
159 views

Integer ordered pairs $(x,y)$ for which $x^2-y!$…

[1] Total no. of Integer ordered pairs $(x,y)$ for which $x^2-y! = 2001$ [2] Total no. of Integer ordered pairs $(x,y)$ for which $x^2-y! = 2013$ My Try:: (1) $x^2-y! = 2001\Rightarrow x^2 = ...
8
votes
1answer
274 views

How is it possible that $\infty!=\sqrt{2\pi}$?

I read from here that: $$\infty!=\sqrt{2\pi}$$ How is this possible ? $$\infty!=1\times2\times3\times4\times5\times\ldots$$ But \begin{align} 1&=1\\ 1\times2&=2\\ 1\times2\times3&=6\\ ...
3
votes
1answer
99 views

The number of zeros in the decimal representation of the factorial of 126

How many zeros are in $126!$ ... the result is $34$. But can I calculate it manually? I have seen How many zeroes are in 100! but I don't think it's helpful.
14
votes
12answers
2k views

Can the factorial function be written as a sum?

I know of the sum of the natural logarithms of the factors of n! , but would like to know if any others exist.
1
vote
3answers
126 views

Product representations of the factorial function?

Is this the only product representation of the factorial function? $$ {n!} =\prod_{k=1}^{n} k $$
2
votes
2answers
703 views

how to find remainder when $20! + 20^{23}$ is divided by $23$?

how to find remainder when $20! + 20^{23}$ is divided by $23$? I am finding it bit difficult to solve. Does any one has a simpler way to solve this problem??
9
votes
2answers
177 views

Polynomials mapping factorials to factorials

I'm looking for all polynomials $P(x)$ with integer coefficients such that for every $n \in \Bbb N$ there is an $m \in \Bbb N$ such that $P(n!)=m$!. The only solutions seem to be the constant ...
1
vote
0answers
66 views

When is $n!+1$ a square? [duplicate]

I'm looking for the solutions $(n,m)$ of the equation $n!+1=m^2$. I have calculated the values of $\sqrt{n!+1}$ for $n \le $ and found only the solutions $(4,5)$, $(5,11)$ and $(7,71)$. Are these ...
1
vote
0answers
65 views

Simplify $\frac{[m+n-1]!}{[m]![n]!}$ where $[k]=x^k-x^{-k}$ and $[k]!=[2][3]…[k]$.

Adopting the notation $[k] = x^k - x^{-k} $ and $[k]! = [2][3]...[k]$ (note that $[1]$ is omitted), and letting $m,n$ be two integers greater than $1$ such that $n>m$ and $gcd(m,n)=1$, would it be ...
2
votes
2answers
213 views

Factorials and Divisibility

I'm having trouble getting started on the following: Given $n_1, n_2, ..., n_k \in$ $\Bbb N$, show that $n_1!\cdot n_2!\cdot\cdot\cdot n_k! |(n_1+n_2+...+ n_k)!$ I thought about a proof by ...
1
vote
1answer
180 views

Understanding the upper and lower bounds of the error estimate in Stirling's Approximation

Based on the Wikipedia article on Stirling Approximation: $n! = \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n e^{\lambda_n}$ where $\frac{1}{12n+1} < \lambda_n < \frac{1}{12n}$ How would this ...
2
votes
2answers
238 views

Analyzing the lower bound of a logarithm of factorials using Stirling's Approximation

I am trying to get the lower bound for: $f(x) = \ln(\lfloor\frac{x}{4}\rfloor!) - \ln(\lfloor\frac{x}{5}\rfloor!) -\ln(\lfloor\frac{x}{20}\rfloor!) - 2(1.03883)(\sqrt{\frac{x}{4}}) - ...
3
votes
1answer
187 views

Reasoning about the Chebyshev functions: How does one check an upper bound based on the second Chebyshev function?

In Ramanujan's proof of Bertrand's Postulate, Ramanujan states: $\log([x]!) - 2\log([\frac{1}{2}x]!) \le \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{1}{3}x)$ where: $\vartheta(x) = \sum_{p \le x} ...
8
votes
0answers
436 views

Understanding Ramanujan's approach in his proof of Bertrand's Postulate

I've been reading through Ramanujan's proof of Betrand's Postulate and I'm not clear why he didn't state his proof in terms of $\varphi(2x) - \varphi(x)$ What would be wrong with this approach for ...
2
votes
3answers
62 views

Constrictions on A.P with factorials.

There are five numbers $(a_1,a_2,a_3,a_4,a_5)$, such that they are in Arithmetic Progression. Given that $a_1$ and $a_2$ are factorials, is there a possibility that either $a_4$ OR $a_5$ is a ...
3
votes
2answers
190 views

Factorials and Arithmetic Progression.

Are there sets of factorials $(a_1!,a_2!,a_3!,\dots,a_n!)$, such that they exist in Arithmetic progression. $n$ is a natural number I don't see any such examples(Except for $n=2$). And I don't see ...
12
votes
1answer
139 views

Solving $n!+m!+k^2=n!m!$ for positive integers $n,m,k$

I have been running in circles with this for a while now. It seems that the only solution is $(n,m,k)=(2,3,2)$ but I don't know how to prove it. Things I have noticed: WLOG $n\geq m$ we see that ...
2
votes
3answers
314 views

Factorial expressed in terms of two other factorials

Can the factorial of $N$ always be expressed by the sum(addition and subtraction) or the product of two other factorials? Do there always exist integer $A$ and $B$ such that $N! = A! + B!$, or $N! = ...
9
votes
2answers
2k views

Number of zero digits in factorials

Here is a riddle someone has been asked in a job interview: How many zero digits are there in $100!$? Well, I found the first $24$ quite fast by counting how many times five divides $100!$ ($5$ ...
10
votes
2answers
253 views

Prove quotient of factorials is integral

If $n$ is an integer $\gt 0$, prove $$\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$$ is also an integer. I understand that a general approach is proving that the power of any prime factor is greater in the ...
2
votes
1answer
229 views

Need help with Factorial Sums! [duplicate]

Possible Duplicate: How to prove that the number $1!+2!+3!+\dots+n!$ is never square? Show that the sum $$\sum_{k=1}^nk!\neq m^2$$for any integer $m$, for $n\geq4$.
2
votes
3answers
263 views

When is a factorial of a number equal to its triangular number?

Consider the set of all natural numbers $n$ for which the following proposition is true. $$\sum_{k=1}^{n} k = \prod_{k=1}^{n} k$$ Here's an example: $$\sum_{k=1}^{3}k = 1+2+3 = 6 = 1\cdot 2\cdot ...
0
votes
1answer
124 views

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 ...
1
vote
3answers
181 views

Number of solutions for $\frac{1}{X} + \frac{1}{Y} = \frac{1}{N!}$ where $1 \leq N \leq 10^6$

Note: this is a programming challenge at this site For this equation $$\frac{1}{X} + \frac{1}{Y} = \frac{1}{N!}\quad ( N \text{ factorial} ),$$ find the number of positive integral solutions for ...