MaxWong

Oops it was googol. I was wrong. I didn't realize that. And 10 decillion was definitely 2.95 times less 0s that googol xD

It's 10 decillion.

$a^2\equiv 0 \pmod{12}\\ \text{That means }\dfrac{a^2}{12}\mbox{ is a positive integer or 0}\\ \mbox{However }a\neq 0 \mbox{ so }\dfrac{a^2}{12} \mbox{ can't be 0.}\\ \begin{array}{rcl}\\&&a^2=12x\end{array}\\ \text{Check the multiples of 12 from 12 to 144}\\ \text{12 is not a square number}\\\text{24 is not a square number}\\\color{red}\text{36 is a square number} \\\text{48 is not a square number}\\ $

$\text{60 is not a square number}\\\text{72 is not a square number}\\\text{84 is not a square number}\\\text{96 is not a square number} \\\text{108 is not a square number}\\\text{120 is not a square number}\\\text{132 is not a square number}\\ \text{144 is not a square number}$

$\begin{array}{rllll}&a^2&=&36&\\&a&=&\pm 6&\\&\text{a must not be -6}\\\therefore&a=6.\end{array}$

$\huge{2)}$

$a^2\equiv 16 \pmod {10}\\a^2\equiv 6 \pmod {10}\\ a^2=\overline{X6}\;\;\mbox{<- This notation means 10X + 6 where }1\leq X\leq 9\\ \mbox{In 1 - 10 there are only 2 numbers whose squares end with 6, namely 4 and 6}\\ \therefore a=6$

I guess you have spelt Christmas wrongly. And I think it's not Christmas today?

Other than the method Guest#1 suggested, you can just type atan(number) in the designated bar.

$\text{Example:}$

$\begin{array}{rl}&\dfrac{\ln b^2}{2}\\=&\dfrac{2 \ln b}{2}\\=&\ln b\end{array}$

I assume the case is , is the decimal point.

$\begin{array}{rllll}\\4(38+x)&-&26,729&=&209,423\\&&4(38+x)&=&236,152 \\&&38+x&=&59,038\\&&x&=&21,038\end{array}$

 cos^-1(1.55720185) is impossible as cosine function have a range of -1<x<1

$\begin{array}{rll}&&2\\+&2&3\\&2&5&\end{array}$

Yay I finally learnt how to use \begin {array} and \end {array}!! :D I learnt it all by myself :D