Question

x=2(mod 3)
x=3(mod 7)
x=7(mod 5)

find a unique incongruent solution to modulo n.

(number theory problem using chinese remainder theorem)

please help!!

Answer 1

I did not know about the chinese remainder theorem but searched for it online and learnt it as I found it interesting - so please treat my solution with a "pinch of salt".

Note that your third equation can be simplified to x=2(mod 5) as you can remove a 5 from 7 to get 2. But I'll work with what you have given.
\[\begin{align}
b_1&=3,b_2=7,b_3=5\\
C_1&=2,C_2=3,C_3=7\\
\\
B&=b_1*b_2*b_3=3*7*5=105\\
B_1&=B/b_1=B/3=105/3=35\\
B_2&=B/b_2=B/7=105/7=15\\
B_3&=B/b_3=B/5=105/5=21\\
\\
B_iX_i&=1(\text{mod }b_i)\\
B_1X_1&=1(\text{mod }b_1)\\
\therefore 35X_1&=1(\text{mod }3)\\
\therefore 2X_1&=1(\text{mod }3)\quad\text{(as 33 is largest multiple of 3<35, 35-33=2)}\\
\therefore 2X_1&=4(\text{mod }3)\quad\text{(1+3=4)}\\
\therefore X_1&=2\\
\\
B_2X_2&=1(\text{mod }b_2)\\
\therefore 15X_2&=1(\text{mod }7)\\
\therefore X_2&=1(\text{mod }7)\quad\text{(as 14 is largest multiple of 7<15, 15-14=1)}\\
\therefore X_2&=1\\
\\
B_3X_3&=1(\text{mod }b_3)\\
\therefore 21X_3&=1(\text{mod }5)\\
\therefore X_3&=1(\text{mod }5)\quad\text{(as 20 is largest multiple of 5<21, 21-20=1)}\\
\therefore X_3&=1\\
\\
x&=B_1X_1C_1+B_2X_2C_2+B_3X_3C_3\\
&=35*2*2+15*1*3+21*1*7\\
&=140+45+147\\
&=332
\end{align}\]this is just one possible solution. in general you can add or subtract multiples of \(B=105\) from this value to get other solutions. so we could write the general solution as:\[x=332+105n\qquad n=-\infty,...,\infty \]