Rom

\(8x + 1 \equiv 5 \pmod{12}\\ 8x \equiv 4 \pmod{12}\\ 8x = 12k + 4,~k \in \mathbb{Z}\\ 2x = 3k + 1,~k \in \mathbb{Z}\\ 2x \equiv 1 \pmod{3} \\ \text{now the tricky part}\\ 2^{-1} \pmod{3} = 2\\ x \equiv 2 \pmod{3}\\ \text{the problem states }a<2, \text{ this is an error, it should be }a\leq 2\)

\(\text{looking at the 2nd half of the show first}\\ \text{the songs will be AAAA!A}\\ \text{there are }4!=24 \text{ ways to arrange the 4 acoustic songs, and}\\ \text{10 ways to select the !A song at the end}\\ \text{this is then 240 ways to arrange the 2nd half of the show}\\ \text{the first half of the show is then 5 non-acoustic songs chosen from 9}\\ \text{since we care about order there are }\dbinom{9}{5}5! = 15120\\ \text{ arrangements for the first half} \\ \text{so in total we have}\\ 240 \times 15120 = 3628800 \text{ possible concert arrangements}\)

\(\text{well the easiest way to do this is just list out the 8 digit powers of 2}\\ \left( \begin{array}{cc} 24 & 16777216 \\ 25 & 33554432 \\ 26 & 67108864 \\ \end{array} \right)\)

 \(\text{looks like }33554432=2^{25} \text{ is the winner}\)

\(\text{Split the team up into 2 subteams by height}\\ \text{The tallest 5 is one subteam, the shortest 5 the other}\\ \text{Any member of the tall team can stand in the back}\\ \text{Any member of the short team can stand in front}\\ \text{There are }5! \text{ ways for the 5 tall team members to arrange themselves}\\ \text{There are also }5! \text{ arrangements for the short team members}\\ \text{Thus there are }5! \times 5! = 120^2 = 14400 \text{ different arrangements}\)

Let me echo Melody's sentiments.

If you don't know how to solve a problem don't guess at it.  It just confuses and frustrates the posters.

\(\text{This is a subtle problem with an obvious solution of }n=1 \\ \text{and not so obvious solutions of } n=6k,~k \in \mathbb{N}\)

\(\text{A simple bit of algebra shows that }\\ \left(z + \dfrac 1 z\right)^n = 1 \Rightarrow \Large z = e^{\pm i \frac \pi 3}\)

\(\text{On the other hand the roots of }z^n = 1 \text{ are }\\ \Large z = e^{i \frac{2\pi}{n}k},~\normalsize k \in \mathbb{Z}\)

\(\text{So we are looking for }n \text{ such that }\\ 2\pi \dfrac k n = \dfrac \pi 3\\ \dfrac{6k}{n}=1\\ n=6k\)

\(|M|=2|T|\\ |M\cup T| = |M|+|T|-|M\cap T|\\ |M\cup T|=2x+x - y = \\ 3x-y\)

(61,61,61,178,179)

is the only combo that does the job.

\(\text{well the largest 4 base 3 digit palindrome is }\\ p = 2222 = 80d\\ \text{I don't see anything more clever than listing this in a few bases until we find a palindrome}\\ \text{In order to have 3 digits it must be that }b^2 < 80d < b^3 \text{ so }\\ 5 \leq b \leq 8\\ \left( \begin{array}{cc} 5 & \{3,1,0\} \\ 6 & \{2,1,2\} \\ 7 & \{1,4,3\} \\ 8 & \{1,2,0\} \\ \end{array} \right)\\ \text{and we see that base 6 is the only base that produces a 3 digit palindrome of }80d \)

\(\text{let }N \text{ be the number of planets in the solar system}\\ N \pmod{13} = 6\\ N \pmod{14}=5 \\ \text{13 and 14 have no common factors so any solutions will be of the form }\\ x= r + 13\cdot 14 k = r+182k,~r,k \in \mathbb{N}\\ \text{we can eyeball that 19 is a solution and so all solutions will be }\\ N = 19 + 182k \text{, and the first of these such that N > 100 is }\\ N = 19+182 = 201\)

\(P[A\cup B] = P[A]+P[B] - P[A \cap B]\\ A= \{\text{those who like to read historical fiction}\}\\ B = \{\text{those who like to read autobiographies}\}\\ P[A\cup B] = 0.47 + 0.42 - 0.14 = 0.75\)

Put the 0.75 in the box... (or it gets the hose)