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## Hot answers tagged factorial

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$N=K. Then, examine \begin{align} ... 3 For trailing zeroes its easy. A number will end in 0 if it is a multiple of 2 and 5. The multiples of 5 between 0 and 30 are: 5, 10, 15, 20, 25, 30 so you should expect there to be 7 zeroes at the end of 30!. (Notice 25 = 5^2) For the interior zeroes there's not short cut. You have to multiply out to discover both of them. 3 For every number n\in\mathbb{N} that you can think of, I can give you a sequence of n-1 consecutive numbers, none of which is prime: n!+2 (divisible by 2) n!+3 (divisible by 3) \dots n!+n (divisible by n) BTW, this proves that there is no finite bound on the gap between two consecutive primes. 3 Note that with the usual definition, n!=n(n-1)(n-2)\cdots1, we have n!=n(n-1)! If we extend this using n=1, we get 1!=1\cdot0!. It is also useful to define a product of 0 numbers to be 1. This also works for 0! which would be a product of 0 integers. The Binomial Theorem also works when we define 0!=1, for example ... 3 You should watch this Numerphile video on Youtube: https://www.youtube.com/watch?v=Mfk_L4Nx2ZI Here the argument is made that0!$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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