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1

I found that this is the consequence of Fermat's Little theorem. Fermat's little theorem states that if p is a prime number, then for any integer a, the number a p − a is an integer multiple of p. In the notation of modular arithmetic, this is expressed as a^p = a (mod p) This leads to: a^(p-2) = (1/a) (mod p)


1

Hint: There is a theorem that says that there is always a prime between $n$ and $2n$ (and actually, the constant $2$ can be reduced if a lower bound on $n$ is used). So if $m > 2$, you should be able to find a prime strictly in between $n$ and $nm - 2$, at least if $n > 3$, and then your equation can't hold. Also, using the lowered constant, (for $n$ ...


2

If there is a prime greater than $n$ and at most $nm-2$, it divides the left side but not the right side. Bertrand's hypothesis almost gives this to you.


5

It suffices to show that for infinitely many $n$, the largest prime factor of $n^{2015}+1$ is at most $\sqrt{n}$. Indeed, if $n$ is such a large integer and $p$ is a prime, then the largest value of $a$ for which $p^a\mid n^{2015}+1$ is $\leq c \log n$ for some constant $c$, while $n!$ is divisible by $p^a$ with $a\geq \frac{n}{p}-1\geq \sqrt{n}-1>c \log ...


-1

see, the above number can be written like this $(3!)^{5!*7!*9!*\dots}$ first look at the power of 3!, it will be a number ending with a lot of zeroes let that number be n hence $n(\mod 10)=0$ $\implies n=10*k$, $k$ is a positive integer Now, we have to find $6^n (\mod 11)=$? by applying Fermat's little theorem $6^{10}(\mod 11)=1$ $=R[(6^{10})^k/11]$ ...


3

Since $2015 = 5\cdot 13\cdot 31 $, and $n^a + 1| n^{ab}+1 $ if $b$ is odd, a necessary condition for $n^{2015}+1 | n! $ is $n^m+1 | n!$ for every $m$ in $\{5, 13, 31 , 5\cdot 13 , 5\cdot 31 , 13\cdot 31 \} $. Solutions are going to be hard to find. All those expressions of the form $n^j-n^{j-1}+...-n+1 $ for odd $j$ will have to have all prime factors $\le ...


7

Modest progress. There are infinitely many integers $n$ such that $n^3+1\mid n!$. We always have $n^3+1=(n+1)(n^2-n+1)$. Let $n=k^2+1$. Then $$ n^2-n+1=(1+k+k^2)(1-k+k^2). $$ Assume further that $k\equiv1\pmod3$. In that case $1+k+k^2$ and $n+1=2+k^2$ are both divisible by $3$. For all sufficiently large $k\equiv1\pmod3$ we thus have $$ ...


6

$$(n+1)! = (n+1) \times n!$$ and so $$(n+1)! + n! = (n+1) \times n! + n!$$ and $$(n+1) \times n! + n! = n! ((n+1) +1) = n!(n+2)$$ and hence $$\color{blue}{(n+1)! +n! = n!(n+2)}$$


5

HINT: $(n+1)! = (n+1)\cdot n!$. So... $$ (n+1)! + n! = (n+1)\cdot n! + n! = ... $$


2

We wish to show that $n!$ is bounded by the inequalities $$\left(\frac{n+1}{e}\right)^{n}<n!<e\left(\frac{n+1}{e}\right)^{n+1}\tag1$$ We will prove these inequalities using mathematical induction. Let's start with the left-hand side inequality in $(1)$. First, we establish a base case. For $n=2$, we see that $2!>\left(\frac{3}{e}\right)^2$. ...


0

By Stirling formula, there exists a number $0<\theta<1$ such that $$n!=\sqrt{2\pi n}\left(\frac ne\right)^n e^{\frac \theta{12n}}.$$ So we have to prove that $$\Bigl(\frac{n+1}{e}\Bigr)^{n} < \frac{1}{n+1}\sqrt{2\pi (n+1)}\left(\frac {n+1}e\right)^{n+1} e^{\frac \theta{12(n+1)}}< e\Bigl(\cfrac{n+1}{e}\Bigr)^{n+1},$$ that is $$\frac{e}{n+1} ...


2

Both inequalities boil down to showing that for any $m \in \mathbb N$: $$\int_{1}^{m} \ln(x)dx \ge \sum_{k=1}^{m-1} \ln(k)$$ And: $$\int_{1}^{m} \ln(x) dx \le \sum_{k =1}^{m} \ln(k)$$ Both might be done by induction. The first one is easy to establish. For the second one, use: $$\int_{1}^{m} \ln(x) dx = \int_{1}^{m-1} \ln(x) dx + \int_{m-1}^{m} ...


8

Only $1!$. For $n>1$, let $p$ be the greatest prime with $p\le n$. Between $p$ and $2p$ there is another prime, so $2p>n$. Therefore, $p$ occurs only once in the factorization of $n!$ and hence, $n!$ is not a square.


0

Of all $n!$ only $0!$ and $1!$ are perfect squares. http://mathforum.org/library/drmath/view/55881.html To prove that a factorial bigger than 1 can't be a perfect square, first think about breaking down the factorial into prime factors. For example, 4! = 24, and 24 = 2*2*2*3. In order for any number to be a perfect square, it must contain an ...


4

Hint: (for example) $13!, 14!, \dots , 25!$ are all nonsquare numbers because all of them are divisible by $13$ only once. (because $13$ is a prime) Similarly, $17!, 18!, \dots, 33!$ are nonsquare numbers. Go on like this.


2

From Wilson theorem we have, $(p-1)!=-1$ for $p$ prime. Now suppose $(p-2)!=x$, then $(p-1)!=(p-1)(p-2)!=(p-1) x=-1(\mod p)$.This implies $x=1(\mod p)$.


1

Hint: Multiply by the multiplicative inverses of $p-1$ and $p-2$.


0

If you want to apply some of his ideas to a specific $n$, then I would suggest starting with the formula at the beginning of section 2 that gives the expression for $n!$ as a product of $p^{w_p(n)}$ terms (the prime factorisation). The formula for $w_p(n)$ is simply expressing the fact that every $p-th$ number ranging from $1,...,n$ has a factor of $p$ and ...


3

$$1 + 5/1! + 8/2! + \ldots = -1 + \sum_{n =0}^{\infty} \frac{5 + (3n -3)}{n!} = -1 + 5 \sum_{n=0}^{\infty} \frac{1}{n!} = 5e - 1$$


2

Suppose $m=p_1^{e_1}\cdots p_k^{e_k}$ where $p_1,\cdots,p_k$ are distinct primes and $e_1,\cdots,e_k>0$. Then any divisor of $m$ will look like $p_1^{r_1}\cdots p^{r_1}$ with $0\le r_i\le e_i$ for each $i=1,\cdots,k$. With $i=1$, there are a total of $e_1+1$ choices for $r_1$, with $i=2$ there are $e_2+1$ choices for $r_2$, and so on. Therefore the total ...


5

For any odd prime $p$ we have $\left(p-1\right)!\equiv p-1\,\left(\text{mod}\,p\right)$ and $\left(p-2\right)\left(p-3\right)\equiv 2\,\left(\text{mod}\,p\right)$ so $\left(p-1\right)!\equiv \frac{p-1}{2}\,\left(\text{mod}\,p\right)$. The case $p=2003$ gives $\frac{p-1}{2}=1001$.


5

I solved it like this. $2000! \equiv x \pmod {2003} \Rightarrow 2002\cdot 2001\cdot 2000! \equiv 2002\cdot 2001\cdot x \pmod {2003}$ Now, by Wilson's Theorem, and since $2003$ is prime, we know that $$2002! \equiv -1 \pmod {2003}$$ So, $$2002 \cdot 2001 \cdot x \equiv -1 \equiv 2002 \pmod {2003}$$ In other words, $$2001\cdot x \equiv 1 \pmod {2003}$$ ...


32

Wilson's theorem is your friend here. $$(p-1)! \equiv -1 \mod p$$ for prime $p$. Then notice $$-1 \equiv (2003-1)! = 2002 \cdot 2001 \cdot 2000! \equiv (-1) (-2) \cdot 2000! \mod 2003.$$


14

$\mathbb{Z}_{2003}$ is a finite field. The equation $x^2 = 1$ has exactly two roots in that field: $x_1 = 1$ and $x_2 = -1 = 2002$. Thus, every $n \in \mathbb{Z}_{2003}^* \setminus \{ 1, 2002 \}$ has $n^{-1} \neq n$. Hence $$\prod_{n=2}^{2001} n = 1,$$ because we can split the product into $1000$ pairs of form $(n, n^{-1})$ and the product of each pair ...


6

I can give an approximate answer. For large numbers (and $10^{32}$ certainly qualifies), Stirling's approximation holds: $$n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n.$$ Thus, the number of digits of large factorials is approximately $$\text{dig}(n!)\approx\log_{10}(n!)\approx\frac 12\log_{10}(2\pi n)+n\log_{10}(n/e).$$ In the case that $n=1.6\times ...


0

By Stirling's approximation we have $$n!2^{n}<en^{1/2}\left(\frac{2n}{e}\right)^{n}=\left(n+1\right)^{n}\left(\frac{2ne^{1/n}n^{1/\left(2n\right)}}{e\left(n+1\right)}\right)^{n} $$ and $$\left(\frac{2ne^{1/n}n^{1/\left(2n\right)}}{e\left(n+1\right)}\right)^{n}\longrightarrow0 $$ as $n\longrightarrow\infty$.


3

You may use the AM-GM inequality, for which: $$ n! = \prod_{k=1}^{n} k \leq \left(\frac{1}{n}\sum_{k=1}^{n}k\right)^n = \left(\frac{n+1}{2}\right)^n, $$ or prove that: $$\forall n>1,\qquad \frac{(n+1)^{n}}{n^{n-1}}>2n $$ that is equivalent to: $$ \forall n>1,\qquad \left(1+\frac{1}{n}\right)^n > 2 $$ that follows from the binomial theorem: $$ ...


0

Hint:use inequality $\delta _{ n }=\sqrt [ n ]{ { x }_{ 1 }{ x }_{ 2 }...{ x }_{ n } } $ (Geometric mean),${ \xi }_{ n }=\frac { { x }_{ 1 }{ x }_{ 2 }+...+{ x }_{ n } }{ n } $ ( Arithmetic mean)where $${ \xi }_{ n }\ge \delta _{ n }\quad $$, where $x_{ k }=k,\quad k=\overline { 1,n } $


2

Do you mean the number of digits? The number of digits in $n!$ is the smallest integer which is strictly larger than $\log_{10}(n!)$. (This is slightly different from the integer ceiling function, since for instance $\log_{10}(10)=1$ but 10 has two digits.) This logarithm is the same as $\frac{\ln(n!)}{\ln(10)}$. By Stirling's formula this is asymptotically ...


0

Cheeky answer: The best function to "approximate" the factorial is the Gamma function defined by $\displaystyle \Gamma(n+1) = n!$ or $$\Gamma(n) = \int_0^{\infty} x^{n-1}e^{-x}\ ,\mathrm{d}t$$ More seriously: The growth of the factorial function is larger than the growth of any exponential function. To see this intuitively, just look at $$a^n=a\times ...


7

By Stirling's approximation, we have $$ \ln(n!) \approx n \ln(n) - n $$ In particular, the number of digits in $n!$ is given by $\lfloor \log_{10}(n!) \rfloor$, and we have $$ \lfloor \log_{10}(n!) \rfloor = \left\lfloor\frac 1{\ln(10)}\ln(n!) \right\rfloor \approx \frac 1{\ln(10)}\left(n \ln(n) - n\right) = O(n \ln(n)) $$ So, the growth of the number of ...


0

It seems that you need an additional rule to generate these terms: $a_m>a_n$ if $m>n.$ Of course there is a function $f$ which gives the terms, namely $f(n)=a_n.$ Perhaps you are asking about a function generated by some sort of closed-form formula, in which case you'd need to specify what restrictions this formula would be under. This sequence is ...


5

Take natural logs of both and use Stirling. $$\ln(k^m) =m \ln(k) \text{ and } \ln(n!) \approx \frac12 \ln(2 \pi)+(n+\frac12)\ln(n) - n. $$ Comparing these should be no problem.


3

For every number $n\in\mathbb{N}$ that you can think of, I can give you a sequence of $n-1$ consecutive numbers, none of which is prime: $n!+2$ (divisible by $2$) $n!+3$ (divisible by $3$) $\dots$ $n!+n$ (divisible by $n$) BTW, this proves that there is no finite bound on the gap between two consecutive primes.


2

A smaller number that starts $n+1$ consecutive composite numbers is $P(n) =\prod_{p \le n} p $. This is called the primorial of $n$. To show that $P(n)+i$ is composite for $1 \le i \le n$, just note that each $i$ is divisible by a prime $\le n$, and this prime also divides $P(n)$. $P(n)$ is much smaller than $n!$, because $\ln(n!) \approx n \ln n - n$ ...


5

This was actually much easier than I had expected. $$n!+i=i(\frac{n!}{i}+1)$$ Also, when $1<i<n$, both factors are integers greater than $1$. (Credit to vadim123)


0

A power of exponent $0$ and positive base is $1$. This means that a product of no factors is $1$


1

You can think of the definition of the factorial backwards: $$(n-1)!=\frac{n!}{n}.$$ Then you get \begin{align} 2!&=\frac{1\cdot 2\cdot 3}{3}=1\cdot 2,\\ 1!&=\frac{1\cdot 2}{2}=1,\\ 0!&=\frac{1}{1}=1. \end{align} Voilà!


0

We have by definition $(n+1)! = (n+1)\,n!$. Therefore $1! = (0+1)! = (0+1)\,0! = 0!$


1

We have the gamma function that is defined for all complex number $z$, excluding the non-positive integers, specifically $$\Gamma(z) = \int_0^{\infty}t^{z-1}e^{-t} \, \mathrm{d}t$$ We can then define the factorial as $$n! = \Gamma(n+1).$$ For $n=0$, we have $$0! = \Gamma(1) = \int_0^{\infty} e^{-t} \, \mathrm{d}t = 1$$ Which is an improper integral that ...


1

You have written $$6! = 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\\ 5! = 5\cdot 4\cdot 3\cdot 2\cdot 1$$ and so on, but then concluded $$0! = 0 \cdot 0.$$ This doesn't fit the pattern, which should continue as follows: $$4! = 4\cdot 3\cdot 2\cdot 1\\ 3! = 3\cdot 2\cdot 1\\ 2! = 2\cdot 1\\ 1! = 1\\ 0! = $$ At this point you simply need to define what $0!$ should ...


10

The number $n!$ counts the number of bijections from a set of $n$ elements to itself. There is exactly one bijection from the emptyset to itself, the empty function. Thus $0!=1$.


3

Note that with the usual definition, $n!=n(n-1)(n-2)\cdots1$, we have $n!=n(n-1)!$ If we extend this using $n=1$, we get $1!=1\cdot0!$. It is also useful to define a product of $0$ numbers to be $1$. This also works for $0!$ which would be a product of $0$ integers. The Binomial Theorem also works when we define $0!=1$, for example $$ ...


3

You should watch this Numerphile video on Youtube: https://www.youtube.com/watch?v=Mfk_L4Nx2ZI Here the argument is made that $0!$ should be one because then the rule $$ n! = \frac{(n+1)!}{n+1} $$ is true even for $n=0$. One can make other arguments, this is just one.


0

For positive integers, it is common to define $n!=n(n-1)(n-2)\cdots 2\cdot 1$, but this definition does not work for zero, negative integers, real numbers between integers, or complex numbers. For these other numbers, we can be free to define the factorial however we want, but the definition has to be useful for anybody to care. You could say $0!=0$, but ...


3

Following @ZevChonoles, we have $$\prod_{n=1}^{N}n!=\prod_{n=1}^{N}n^{N-(n-1)} \tag 1$$ We can prove this by induction. To that end, let's establish a base case. For $N=2$, we have $$\prod_{n=1}^{2}n!=(1!)\,(2!)=2$$ and $$\prod_{n=1}^{2}n^{2-(n-1)} =(1^2)\,(2^1)=2$$ Now assume that $(1)$ is true for $N=K$. Then, examine $$\begin{align} ...


5

Here is a simple demonstration, laid out pictorally: $$\begin{array}{c|ccccc} N!\strut_\strut & 1 & 2 & \cdots & (N-1) & N\\ (N-1)!\strut_\strut & 1 & 2 & \cdots & (N-1)\\ \vdots\strut_\strut & \vdots& &{\cdot}^{\Large\cdot^{\huge\cdot}} \\ 2!\strut_\strut & 1 & 2\\ 1!\strut_\strut & 1\\\hline ...


6

There are $mn$ people ($m,n$ are nonnegative integers). We want to put them into $n$ groups, each consisting of $m$ people. Let $N$ be the number of ways to do so. If we label the groups, say, groups $1$, $2$, $\ldots$, $n$, there are $n!$ ways for the labeling. Hence, there are $n!\cdot N$ ways to put $mn$ people into $n$ labeled groups. Now, from ...


9

You can see that $n! \cdot 2^n$ is exactly $\prod_{i=1}^n 2i$. Then any of these (even) number appears in the numerator $(2n)! = \prod_{j=1}^{2n}j$. Then the division is an integer. It is exactly $\prod_{i=1}^n (2i-1)$.


3

Induction works. For inductive step: $$\frac{(2n+2)!}{(n+1)!\cdot 2^{n+1}}=\frac{(2n)!}{n!\cdot 2^n}\cdot\frac{(2n+1)(2n+2)}{(n+1)\cdot 2}=\frac{(2n)!}{n!\cdot 2^n}\cdot(2n+1)$$



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