what is wrong with this proof?

Question

usukidoll · Answer

\documentclass[12pt]{article}
\begin{document}
Use the techniques outlined in the solution of the pirate problem to solve the following.\
a. Find the least positive integer with remainders 1,2, and 3 when divided by 7,8, and 9, respectively.\

The three congruences are

$\indent x \equiv 1 $ (mod 7)
$\indent x \equiv 2 $ (mod 8)
$\indent x \equiv 3 $ (mod 9)

We solve the system by letting 

$\indent x =f_1+2f_2+3f_3$

where $f_1.f_2,$ and $f_3$ satisfy


$\indent f_1 \equiv 1 $ (mod 7)
$\indent f_2 \equiv 0 $ (mod 7)
$\indent f_3 \equiv 0 $ (mod 7)

$\indent f_1 \equiv 0 $ (mod 8)
$\indent f_2 \equiv 1 $ (mod 8)
$\indent f_3 \equiv 0 $ (mod 8)

$\indent f_1 \equiv 0 $ (mod 9)
$\indent f_2 \equiv 0 $ (mod 9)
$\indent f_3 \equiv 1 $ (mod 9)

Notice that, under these conditions, by the congruences

$\indent x =f_1+2f_2+3f_3 \equiv 1$ (mod 7)
$\indent x =f_1+2f_2+3f_3 \equiv 2$ (mod 8)
$\indent x =f_1+2f_2+3f_3 \equiv 3$ (mod 9)

To compute $f_1$ we set $f_1 = 8 \times 9 \times b_1$ where $b_1$ satisfies the single congruence.

$72b_1 \equiv 1 $ (mod 7)

$2b_1 \equiv 1 $ (mod 7)

$2b_1-1 = 7k$

$2b_1=7k+1$

$b_1 = \frac{7k+1}{2}$

If we let $k = 1$, then

$b_1 = \frac{7+1}{2}$

$b_1 = \frac{8}{2}$

$b_1 = 4$

Thus $f_1 = 8 \times 9 \times 4 = 288$ 

Similarly, set $f_2 = 7 \times 9 \times b_2$

$63b_2 \equiv 2$ (mod 8)

$7b_2 \equiv 2$ (mod 8)

$7b_2 - 2 =8k$

$7b_2 =8k+2$

$b_2 = \frac{8k+2}{7}$

If we let $k=5$, then

$b_2 = \frac{40+2}{7}$

$b_2 = \frac{42}{7}$

$b_2 =6$

Thus $f_2 = 7 \times 9 \times 6 =378$


Also, set $f_3 = 7 \times 8 \times b_3$

$56b_3 \equiv 3 $ (mod 9)

$2b_3 - 3 =9k$

$2b_3  =9k +3$

$b_3  =\frac{9k+3}{2}$

If we let $k = 1$, we have $b_3	=\frac{9+3}{2}$

$b_3 =\frac{9+3}{2}$

$b_3 =\frac{12}{2}$

$b_3 =6$

Thus $f_3 = 7 \times 8 \times 6 = 336$


This means $x = 288+2(378)+3(336)$

$x = 288+756 + 1008$

$x=2052$

Since $ 7 \times 8 \times 9 = 504$, we need to reduce $2052$ modulo $504$

usukidoll · Answer

oy latex come on!

jack1 · Answer

...therrrrrre's a lot of brackets there...

usukidoll · Answer

why is the latex broken ? ugh view it here https://www.writelatex.com/823880mgqjsf#/1828923/

usukidoll · Answer

@hartnn

anonymous · Answer

rip lol

usukidoll · Answer

I have to write a proof that's similar to the pirate problem but th eproblem is that it's a lot harder than the pdf I found

usukidoll · Answer

got it... I was multiplying 2 and 3 twice... the x was supposed to be 1002 
then 1002 modulo 504 = 498