Is there any way I can 'undo' the factorial operation? JUst like you can do squares and square roots, can you do factorials and factorial roots (for lack of a better term)?   Here is an example: 5! = 120.  Is there a way I can work out the number that must be factorialed (??) to give the answer 120?
95395
I don't want to do this through trial and error, and the best way I have found so far was to start dividing from 1.     $n! = \text {a really big number}$   Ex. $n! = 9999999$   Is there a way to approximate n or solve for n through a formula of some sort?  Update (Here is my attempt):  Stirling's Approximation: $$n! \approx \sqrt{2 \pi n} \left( \dfrac{n}{e} \right ) ^ n$$  So taking the log:     $(2\cdot \pi\cdot n)^{1/2} \cdot n^n \cdot e^{-n}$       $(2\cdot \pi)^{1/2} \cdot n^{1/2} \cdot n^n \cdot e^{-n}$       $.5\log (2\pi) + .5\log n + n\log n \cdot -n \log e$       $.5\log (2\pi) + \log n(.5+n) - n$   Now to solve for n:     $.5\log (2\pi) + \log n(.5+n) - n = r$       $\log n(.5+n) - n = r - .5 \log (2\pi)$   Now I am a little caught up here.
27225
Can a number with $100$ $0$'s $100$ $1$'s and $100$ $2$'s ever be a perfect square? I have no idea where to even start. Some hints would be nice. Thanks in advance :)
14652
If I want to find the continued fraction of $\sqrt{n}$ how do I know which number to use for $a_0$? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. I tried Googling but couldn't find a website that actually explained this question.  If anyone has a good site that answers these questions either, please let me know. Thanks!
72373
Without strenuous arithmetic.   Is there a program I can download to do so?  What are the quadratic residues modulo $5^4$ or $5^5$?  Thanks!
99159
I didn't understand the permutation and of course, I got this question wrong.     Compute the inverse of the following permutation:   $$ \begin{pmatrix}  1&amp;2&amp;3&amp;4&amp;5&amp;6\\ 2&amp;5&amp;4&amp;1&amp;6&amp;3 \end{pmatrix} $$   Could you please help me with this question and provide an explanation?  Furthermore, is it possible to compute the direct permutation, not the inverse one? If yes, how?  Thank you in advance!
101930
I have this algorithm and Im searching p and q:  $n=p^2 * q $  $l=(p-1)*(q-1) / \gcd (p-1,q-1)$  $d\equiv l^{-1} \pmod n$  And the values are:  For $n$:  1043247381320041410146374847683905608886067988802815759655490071506964811427733496668055364700822779908166748193534341903285387332658081562041534228961538451830025041762537639985150344917584064827615356535763872539925989984117068207555498700585376330731428359489791600319636079777025344217248475485835298689437926003958578141784379857951264973109214803916445702964047899447002191548965239884095802091798787988745415178578597206815122915438351636672681355517982410123569961242563194480399347961753678159210217993526738306796503660563163029287419732210247579793266891085178817047536758465239950861266592524206390869822972469137566045176690277813352563392838847417674529251627419479815097868890993153215515488877097760272399483111013602618554846096893805108827211344467797072352072430591318485091677011261912992157762106672216780766138057271434398274272110062768324672899209610001835423246735718435995420720133657793744810610657   for $d$:  105717877672224295784291675385546750129601472163379527375367128238003471793097341763391999503913659542087520808364287167645349555897774428176288829031477703594776341268160433668976237198330124904372546897953955241778475322074160469767162566234358907740412805211247598698481640207956827199048419949848619731050860757808082106844953602176503247786535423423104239566910713528691298692332149783689032722772009974404716783312200885555343877611094998117307970399241640213553820043460140493803404190774171912962651075100983834071259021038322886114660028788229094872545516054757700068264628024524394995795135464746570071201   How can I get $p$ and $q$?   Thanks!
11059
I successfully thought of bounding our expression examining consecutive squares that attain values close to it, and this led to the solution I'll post as an answer, which was the one reported. However, before that, I had briefly tried manipulating $$n^4+n^3+n^2+n+1=m^2.$$Since $n=1$ is not a solution, I rewrote this as $$\frac{n^5-1}{n-1}=m^2 \\ n^5-1=nm^2-m^2 \\ m^2-1=n(m^2-n^4) \\ (m-1)(m+1)=n(m-n^2)(m+n^2),$$ but in vain. Does manipulation lead somewhere? Is there a different approach from both of mine?
69468
I solved this problem by manually adding parentheses and counting them, and got correct answer of 32. Is there a simple to find the answer? Thanks.     The value of the expression $1÷2÷3÷5÷7÷11÷13$ can be altered by including parentheses. If we are allowed to place as many parentheses as we want, how many distinct values can be obtained for this expression?
87505
I've been searching the internet for quite a while now to find anything useful that could help me to figure out how to calculate factorial of a certain number without using calculator but no luck whatsoever.  I'm well aware of the fact that there is a way to calculate any number of sigma (summation notation sigma) but haven't figured out anything for factorials yet. Could you please show me any method that should do the trick.  E.g. factorial of 10! is 3628800 but how do I calculate it without using any sorts of calculator or calculate the numbers from 10 to 1?  Thanks in advance!
51954
I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:  $$\sqrt{x+5} = x - 1$$  So I used this logic:  $$ \begin{align} \sqrt{x+5} &amp;= x - 1 \\ x + 5 &amp;= (x-1)^2 \\ x + 5 &amp;= (x-1)(x-1) \\ x + 5 &amp;= x^2 - 2x + 1 \\ 0 &amp;= x^2 - 3x - 4 \\ 0 &amp;= (x-4)(x+1) \\ \end{align} $$  Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.  But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:  $$ \begin{align} \sqrt{4 + 5} &amp;= 4 - 1 \\ \sqrt{9} &amp;= 3 \\ 3 &amp;= 3 \end{align} $$  So using 4 works as expected, but when using $-1$:  $$ \begin{align} \sqrt{-1 + 5} &amp;= -1 - 1 \\ \sqrt{4} &amp;= -2 \\ 2 &amp;\ne -2 \end{align} $$  At what stage am I going wrong?  And according the , the solution is $x = 4$.
76825
Possible Duplicate:             If I'm right, factorial $!$ means:  $$n!=1 \cdot 2 \cdot 3 \cdot 4 \cdots n $$  so:  $$ \begin{align} 5!&amp;=1\cdot2\cdot3\cdot4\cdot5=120\\ 4!&amp;=1\cdot2\cdot3\cdot4=24\\ 3!&amp;=1\cdot2\cdot3=6\\ 2!&amp;=1\cdot2=2\\ 1!&amp;=1 \end{align} $$  But what is $n!$ when $n=0$?  It can't be undefined and it can't be $n!=0$, since those are illegal in known equations like:  $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$  So what is it?
41621
Is it possible for some integer $n&gt;1$ that $2^n-1\mid 3^n-1$ ?   I have tried many things, but nothing worked.
21280
I just want to know if my proof to this question is correct. First, I proved it was true for $n = 4$. $$4!&gt;2^4$$ $$24&gt;16$$ Then, I assumed that it was true for $n=k$. $$k!&gt;2^k$$ Afterwards, I tried to prove it for $n=k+1$. $$(k+1)!&gt;2^{k+1}$$ $$(k+1)k!&gt;(2)(2^k)$$ Since $k!&gt;2^k$, am I allowed to simply do $$(k+1)(2^k)&gt;(2)(2^k)$$ $$(k+1)&gt;2$$ and this is true for all $k\geq3$. Would the last two steps be illegal because I only assumed that $k!&gt;2^k$ is true?
29565
Find the values of $a,b,c,d\in\mathbb{N}$ such that      $$ 34!=295232799cd9604140847618609643ab0000000 $$   My Attempt:  The factorial of $34$ contains a $3$, so the RHS must be divisible by  $3$. Similarly, it must be divisible by $7$, $11$, $13$, $19$ etc.  But I do not understand how can I calculate $a,b,c,d$ in this equation.
47491
What are the integer numbers $x$ such as $x^4+x^3+x^2+x+1$ is the square of an integer?  $0$ and $-1$ are obvious answers, but how many are there?
33187
Let $\theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+\theta$ in $\mathbb{Q(\theta)}$.  So problems like this really annoy me but I did crappy on the last homework after making a lot of arithmetic mistakes so I want to run everything by you guys. This one isn't actually for homework it's just a suggested practice problem but okay lets go!!!  So I'm not sure if the method I used to do this was standard or not, I don't remember my professor showing me, but what I did first was used the canonical euclidean algorithm in $\mathbb{Q[x]}$ and wrote:  $x^3+9x+6$  $=(1+x)(x^2-x+10-\frac{4}{x+1})$  $=(1+x)(x^2-x+10)-4$  the reduced modulo  the minimal polynomial of $\theta$ i.e. $x^3+9x+6$ and so  $4=(1+x)(x^2-x+10)$  $\rightarrow$  $1= \frac{1}{4}(1+x)(x^2-x+10)$ and sooo ah $(\theta^2-\theta+10)\frac{1}{4}$ is what I believe is the inverse of $(1+\theta)$ in the quotient field... Like i said I hate these problems but ehh is this correct? On a lighter note it's finally getting cool enough for me to go on proper runs here and so that's pretty bomb!
47399
I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: $$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $$