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80

We have $$3!\cdot 5!\cdot 7!=(1\cdot 2\cdot 3)\cdot (1 \cdot 2\cdot 3\cdot 4\cdot 5)\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7,$$ and combining some of those gives $$1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot \underbrace{(2\cdot 4)}_8\cdot \underbrace{(3\cdot 3)}_9\cdot \underbrace{(2\cdot 5)}_{10}=10!$$

55

If the formula is true it can only possibly be $8!$, $9!$, or $10!$ because $11!$ and larger have a factor of $11$. $8!$ doesn't have a large enough power of $3$, and $9!$ doesn't have a large enough power of $5$, so if the formula holds it must be $10!$.

18

You already have some clever answers; here's less clever approach. You want $$3!\cdot5!\cdot7!=n!=7!\cdot8\cdot9\cdots\cdot n,$$ so, cancelling 7! and computing that $3!\cdot 5!=6\cdot120=720$, you want $$720=8\cdot9\cdots\cdot n.$$ Now start dividing both sides by 8, then 9, etc. until you get the answer. Dividing by 8 gives you $90=9\cdots n$, and ...

11

\begin{align} 3! \cdot 5! \cdot 7! &= 6 \cdot 120 \cdot 7! \\ &= 6 \cdot 15 \cdot 8! \\ &= 2 \cdot 5 \cdot 9! \\ &= 10 \cdot 9! \\ &= 10! \end{align}

10

\begin{align} \frac{x^x}{(2x)!} & = \frac{\overbrace{x\cdots x}^{x\text{ factors}}}{\underbrace{1\cdot2\cdot3\cdots x}_{x\text{ factors}} \cdot \underbrace{(x+1) \cdot (x+2) \cdots (2x)}_{x\text{ factors}}} \\[10pt] & = \frac 1 {x!} \cdot \underbrace{\frac x {x+1} \cdot \frac x {x+2} \cdot \frac x {x+3} \cdots \frac x {x+x}}_\text{This is $<1$.} ...

8

Every number in the product $n!$ is less than or equal to $n$. In particular, one of them is $1$. Can you use this to show $n! / n^n \leq 1/n$?

8

It should work if we start factoring consecutive integers out of the expression. You have it boiled down to this $$6 \cdot 6^3\cdot5^2 \cdot 4^2 \cdot 7 =$$ $$2\cdot 3 \cdot (6^3 \cdot 5^2 \cdot 4^2 \cdot 7)=2 \cdot 3 \cdot 4\cdot 5\cdot 6\cdot 7\cdot (6^2 \cdot 5\cdot 4)$$ This is a good start, but now we need an 8. we get an 8 by using our 4 and a ...

7

Note that $$\frac{n^n}{(2n)!}=\frac{\overbrace{n\cdot n\cdot\ldots \cdot n}^{n \text{ factors}}}{n!\cdot\underbrace{(n+1)(n+2)\ldots(2n)}_{n\text{ factors}}}\le \frac1{n!}$$

7

Let $$a_n = \frac{n!(2n)!}{(3n)!} = \frac{1}{\binom{3n}{n}}.$$ We have, for every $n\geq 1$: $$0\leq \frac{a_{n+1}}{a_n}=\frac{1}{3}\cdot\frac{(2n+2)(2n+1)}{(3n+2)(3n+1)}\leq\frac{1}{3}\cdot\frac{3}{5}=\frac{1}{5}$$ hence our limit is just zero.

6

Let $p=73$. $p$ is a prime number. Thus by Wilson's theorem $$72! = (p-1)!\equiv -1 \mod p$$ Notice that $\forall 1\le n\le 36$, $73-n \equiv -n \mod p$, thus $$\frac{72!}{36!}\equiv (37\cdot 38\cdots 72)\equiv (-1\cdot -2\cdots -36) \equiv (1\cdot 2\cdots 36)(-1)^{36} \equiv (1\cdot 2\cdots 36) \equiv 36! \mod p$$ Suppose for the sake of contradiction ...

5

Note that every prime divisor of $1!3!5!\cdots(2n-1)!$ occurs at least twice, except for perhaps $2n-1$ (if it's prime). By Bertrands Postulate, we already know that $m!$ has one prime divisor that occurs only once. (See Can n! be a perfect square when n is an integer greater than 1?) A slightly stronger version of the theorem will give that there are at ...

5

Notice, $\color{red}{3}$, $\color{red}{5}$, $\color{red}{7}$ are prime numbers (can't be factorized), then the factorials can be successively reduced as follows $$3!5!7!=(3\cdot 2!)\cdot (5\cdot 4!)\cdot (7\cdot 6!)$$ $$=\color{red}{3}\cdot \color{red}{5}\cdot \color{red}{7}\cdot (2!)\cdot (4!)\cdot (6!)$$ $$=\color{red}{3}\cdot \color{red}{5}\cdot ... 4 One can (and possibly should) be even less clever than Andreas proposes. Just compute 3! 5! 7! (getting 3628800) and then use brute force to find a number whose factorial that is. E.g., obviously the number is bigger than 7!, so start with 8! and work upwards. 8! = 40320, too small. 9! = 362880, too small. 10! = 3628800, just right -- and we're done. If the ... 4 Let$$ f(a,n) = (a+1)(a+2)\cdot\ldots\cdot(a+n).\tag{1}$$We have:$$ f(a+1,n)-f(a,n) = n\cdot f(a+1,n-1) \tag{2} $$hence the claim follows by applying a double induction. 4 When you factorize 625!, you are interested in the number of 10s you can divide it by until it is no longer possible to divide evenly. But 10=2\cdot 5, so you actually want to know how many 2s and 5s are in the prime factorization of 625!. You will quickly realize if you start listing the product that there are many many more factors of 2s ... 3 Start with Stirling approximation$$n!\sim \sqrt{2\pi n}\left({n\over e}\right)^n$$Replace n by 2n et 3n we get$${n!(2n)!\over (3n)!}\sim 2\sqrt{\pi}\sqrt{{n\over 3}}\left({4\over 27}\right)^n$$So the limit we are looking for is 0 3 To prove P(k+1), start with your left hand side, i.\,e.$$ \def\P#1{\binom 22 + \binom 32 + \cdots + \binom{#1}2}\P{k+1}$$Write it as$$ \P k + \binom{k+1}2 \tag+ $$Now use, that by the induction hypothesis, we have$$ P(k): \P k = \binom{k+1}3 $$Hence, we can write (+) as$$ \P k + \binom{k+1}2 = \binom{k+1}3 + \binom{k+1}2 $$But the right hand ... 3 \displaystyle \sum^{n}_{r=0}r\cdot r! = \sum^{n}_{r=0}\left[(r+1)-1\right]\cdot r! = \sum^{n}_{r=0}\left[(r+1)!-r!\right] Now Using open Summation (Telescopic Sum), We get \displaystyle \sum^{n}_{r=0}\left[(r+1)!-r!\right] = (n+1)!-1 3 Let$$N:=m! =1!\>3!\>5!\cdots(2n-1)!$$for certain numbers m, n\in{\mathbb N}_{\geq1}, and denote by p the exponent of 2 in the prime decomposition of N. Then one has on the one hand$$p=\left\lfloor{m\over2}\right\rfloor+\left\lfloor{m\over4}\right\rfloor+\ldots <m$$and on the other hand ... 3 There is a very simple approach: by the AM-GM inequality,$$ n!^2 \leq \left(\frac{n+1}{2}\right)^{2n},$$hence the wanted limit is clearly \color{red}{0}. 3 Hint In mod 10 we have n!=0 for n\ge5. 2 Hint: consider \pmod{10}. What is 5! \pmod{10}? 2 Unless otherwise specified, I'd take each girl and boy as distinct. After all, we aren't talking of apples and oranges. (a) 2 choices of ends for girl/boy. 4*3 = 12 ways to fill the ends with particular girl/boy 5! ways to permute the rest, thus 2*12*5! = 2880 (b) Your ans is correct, but a simpler way is to treat the 4 girls as an internally ... 2 We already know that 100! \equiv -1 (mod 101). Moreover, we have i \equiv -(101-i) (mod 101), for 1\leq i\leq 50. Then, one has 50! \equiv (-1)^{50} 51\times 52\times \dots \times 100, it implies that 50! \equiv 10 (mod 101) or 50! \equiv 91 (mod 101). But 50! \not\equiv 10 (mod 101) (Someone helps me?). So, we get 51! \equiv ... 2 You can also use Stirling's approximation: n!\sim \left(\dfrac{n}{e}\right)^n\sqrt{2\pi n} as n\to \infty 2 Hint: Set p_n = n!. Then what is$$\lim_{n \rightarrow \infty}\frac{p_{n+1}}{np_n}?$$Can you take it from here? 2 Idea: show that 1!3!5!\cdots(2n-1)!\geqslant(4n-1)!, so that m\geq4n-1. By Bertrands Postulate it will follow that there are no solutions because there is a prime between 2n-1 and m. It suffices that 1!3!5!\cdots(2n-3)!\geqslant2n(2n+1)\cdots(4n-1). Let n\geq12. We have ... 2 As you've found, n=1,2,3 give solutions. In fact, the only solutions. If n=4, then 3!5!7!9!=k!> 11!, so 11\mid k! but 11\nmid 3!5!7!9!. If n=5, then 3!5!7!9!11!=k!> 13!, so 13\mid k! but 13\nmid 3!5!7!9!11!. If n\ge 6, then 3!5!\cdots (2n+1)!=k!>(4n+2)!. By Bertrand's Postulate exists a prime 2n+1<p<4n+2. But then ... 2 Assuming x is an integer in your question (so that I'll use n instead of x, for the sake of my own ease of mind): a simple way, which uses a big (and quite overkill) "hammer," is to invoke Stirling's approximation:$$ (2n)! \operatorname*{\sim}_{n\to\infty} 2\sqrt{\pi n}\frac{(2n)^{2n}}{e^{2n}} $$and look at the limit (when n\to\infty) of$$ ...

2

After cancellation, we have $$\prod_{k=0}^{n-1}\frac{n-k}{3n-k} \le \prod_{k=0}^{n-1}\frac{n-k}{3n-3k} = \prod_{k=0}^{n-1}\frac{1}{3} = \frac{1}{3^n} \to 0.$$

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