I did:  $$105 = 4*26+1\\ 1 = 105*1-4*26$$  So the inverse should be 5 as it is $\equiv 1 \pmod 4$ but it's 1. Why?
121669
I did:  $$gcd(84,5) = \\ 84 = 5*16+4 \\ 5=4*1+1$$  $$4 = 84-5*16 \\ 1 = 5-4*1 \\ 1 = 5-(84-5*16)\\ 1 = 17*5-1*84$$  So the answer should be $1$ but it's 4, what went wrong?
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!
49206
Most efficient way to do this question?  $25^{37}\cong (mod 55)$  Attempt:  $25^2 \cong 20 (mod55)$  Then I don't know the best way to continue... I know how to brute force it but I'm just curious what the best way is?
64155
I was working on a problem that resulted in the calculation: $20 \equiv 10x \pmod{11}$.  I got the answer $x \equiv 2 \pmod{11}$ with the thought process: Since $10$ is a factor of $20$ I can rewrite $20 \pmod{11}$ as $10x \pmod{11}$ with $x=2$. But isn't this basically the same as using division, which we aren't supposed to do in modular spaces?  If I solve by multiplying $20$ by the multiplicative inverse of $10 \pmod{11}$ which is $10$, I get $200 \pmod{11}$ which simplifies to $2 \pmod{11}$ anyway. Did these numbers just happen to be the same or is my original method valid?
3509
What is the best way to solve equations like the following:  $9x \equiv 33 \pmod{43}$  The only way I know would be to try all multiples of $43$ and $9$ and compare until I get $33$ for the remainder.  Is there a more efficient way ?   Help would be greatly appreciated!
59458
Not sure what to do / how to start this... I have equcation of 504 is: $2 \cdot2 \cdot 2 \cdot 3 \cdot 3 \cdot 7$
15525
Let $a,b,c,x$ be elements of a unital non-commutative ring. Assume $c$ is an inverse of $1-ab$: $$ c(1-ab) = 1$$  How can I find an inverse for $1-ba$? What I tried:   Denote the unknown by $x$. Then $x (1-ba) = 1 = x - xba$. I tried to replace $1$ with $c(1-ab)$ but it didn't help because I cannot solve for $x$. I also tried to subtract $x (1-ba) $ from $ c(1-ab)$ but couldn't solve for $x$ either. Any suggestions?
112232
If percentage points can equal decimal points, then $1.25$ should equal $125$% when multiplying $1.25$ by any number.  Therefore $125$% of $0.75$ should equal $1$ if $0.75$ is also $75$% of $1$.   Am I missing something super obvious here or what?!  Please help. Thank you!
15178
When I am faced with a simple linear congruence such as  $$9x \equiv 7 \pmod{13}$$ and I am working without any calculating aid handy, I tend to do something like the following:  "Notice" that adding $13$ on the right and subtracting $13x$ on the left gives: $$-4x \equiv 20 \pmod{13}$$  so that $$x \equiv -5 \equiv 8 \pmod{13}.$$  Clearly this process works and is easy to justify (apart from not having an algorithm for "noticing"), but my question is this: I have a vague recollection of reading somewhere this sort of process was the preferred method of C. F. Gauss, but I cannot find any evidence for this now, so does anyone know anything about this, or could provide a reference? (Or have I just imagined it all?)  I would also be interested to hear if anyone else does anything similar.
89983
I see that it has something to do with the variance and one has 12 on the numerator and the other denominator, but why is this the same 12 and how do they become inverse?  Equations:     Kruskal Wallis: The first equation under 3  $$ H = \frac{12}{N(N+1)} \sum_{i=1}^g n_i\bigg(\bar r_{i\cdot}-\frac{N-1}{2} \bigg)^2 $$ Wilcoxon: The one for the variance  \begin{align} z &amp;= \frac{R - \mu_R}{\sigma_R}  \\[5pt] &amp;\text{where}  \\[5pt] \mu_R &amp;= \frac{n_1(n_1 + n_2 + 1)}{2}  \\[8pt] \sigma_R &amp;= \sqrt{\frac{n_1n_2(n_1 + n_2 + 1)}{12}} \end{align}
55740
So, clearly, 101 is a solution.  But $n=125$ is also a solution, so is 250.  How would one come to that solution without knowing these values beforehand?
120507
What is the correct answer to this expression:    $26^{32}  \pmod  {12}$    When I tried in Wolfram Alpha the answer is $4$, this is also my answer using Fermat's little theorem, but in a calculator the answer is different, $0.$
36466
Prove that $n=\dfrac{5^{125}-1}{5^{25}-1}$ is a composite number   My attempt,  Let $x=5^{25}$, so that $5^{125}-1=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$  $=(x^4+9x^2+1+6x^3+6x+2x^2-5x^3-10x^2-5x)(x-1)$  $=((x^2+3x+1)^2-5x(x+1)^2)(x-1)$  I'm stuck at this point and don't know how to continue anymore. Hope someone can provide a detailed solution. Thanks a lot.
60872
I've tried solving it, but I'm not familiar with these kind of progressions and I couldn't get to a result. Also, I've found that the result is just $N/2$, but I don't get why would that be the answer, as it doesn't work for $N=1$ or $N=2$. Appreciate your help :)
78226
Use Euclid's algorithm to find a multiplicative inverse of 15 mod 88, hence solve the linear congruence $15x\equiv 20 \pmod{88}$  So far I have:   $88=5\cdot15+13$  $15=1\cdot13+2$  Backwards substitution gives  $15v+18w=1$   $1=15-1\cdot13$   $=15-1(88-5\cdot15)$   $=15\cdot15-1\cdot88$   now very stuck what to do from here:(
35686
The question is as follows:      Find all integers $k \geq 2$ such that $k^2 \equiv 5k \pmod{15}$.   I have an issue related to this question (its not about the solution to the question):  I know that $\overline{k} \in \mathbb{Z}_{15}$ is invertible if and only if $k$ and $15$ are relatively prime. So, assume $\overline{k}$ is invertible. Then, $\overline{k}^2 = \overline{5}\overline{k}$ implies $\overline{k} = \overline{5}.$ But isn't $\overline{5}$ not invertible, since $5$ is not relatively prime with 15? What am I missing?
1672
I have had a go at this for a while now and can't seem to get anywhere.  $$2^{(2^{517})} \pmod {23}$$  I see that Fermat's Little Theorem must come into play but I can't see where to pull $22$ from. Any help would be appreciated.  Here is my attempt:  $$2^{(2^{517} \!\!\mod {22})} \equiv  2^{(2^{517})} \pmod {23}$$ Then proceed to do Chinese remainder theorem to find the exponent?