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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 ...


11

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 $0!=1$.


10

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 $$ ...


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 $\prod_{i=1}^n (2i-1)$.


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.


8

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 ...


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 ...


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 ...


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 ...


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... $$ (n+1)! + n! = (n+1)\cdot n! + n! = ... $$


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, $$2001\cdot x \equiv 1 \pmod {2003}$$ ...


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.


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 ...


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)


4

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 ...


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.


3

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


3

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


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} ...


3

For trailing zeroes its easy. A number will end in 0 if it is a multiple of 2 and 5. The multiples of 5 between 0 and 30 are: $5, 10, 15, 20, 25, 30$ so you should expect there to be $7$ zeroes at the end of $30!$. (Notice $25 = 5^2$) For the interior zeroes there's not short cut. You have to multiply out to discover both of them.


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.


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.


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: $$ ...


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 ...


2

Note $$2^n\cdot n! = (2 \times \cdots \times 2) \times (n \times \cdots \times 1) = (2n) \times (2(n - 1)) \times \cdots \times 2.$$ Therefore $$\frac{2^n\cdot n!}{(2n + 1)!} = \frac{(2n) \times (2(n - 1)) \times \cdots \times 2}{(2n + 1) \times (2n) \times \cdots \times 1} = \frac{1}{(2n + 1)\times(2n - 1)\times\cdots\times 3 \times 1} \to 0$$ as $n \to ...


2

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 ...



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