# Question 2 On Congruence

Solve the following system of congruences: $$x \equiv 2 \pmod 5$$ $$x \equiv 1 \pmod8$$ $$x \equiv 7 \pmod 9$$ $$\quad x \equiv -3 \pmod {11}$$

i have no idea how to go about starting this, any help would be much appreciated.

• Can you give at least some ideas? – DonAntonio Nov 28 '12 at 13:44

For example

$$x=2\pmod 5\Longleftrightarrow \,\exists\,\, k\in\Bbb Z\,\,\,s.t.\,\,x=2+5k$$

$$x=1\pmod 8\Longleftrightarrow\,\,\exists\,\,m\in\Bbb Z\,\,\,s.t.\,\,\,x=1+8m$$

and etc.

You can continue with this, form some equations and find some solution...or else you can use the powerful and nice CRT = Chinese Remainder Theorem, and in particular its proof , and do it faster, cleaner and nicer.

• how would i go about using the CRT method?? i have the M value = 2 x 8 x 9 x 11 = 3960 ... i divide the M value by each mod value and find its congruents ... i am left with 792=2 mod 5, 495=7 mod 8, 440=8 mod 9, 360=8 mod 11... by dont know where to go from here?? – jill Nov 28 '12 at 15:22
• In the site I linked to you look under"A constructive algorithm to find the solution". It's pretty simple, but of course: there's some work to do. I got the solution $\,x= 1,537\,$ – DonAntonio Nov 28 '12 at 17:23

Another option is using the linear diophantine equation $5a-8b=-1$. Because $x=5a+2=8b+1$.

This will give a certain condition for $a$ of the shape $a=8k+k_a$, where $k_a$ is a constant. So $x=40k+(5k_a+2)$, or $x\equiv 5k_a+2\pmod{40}$.

This reduces the first two equations to only one. If you repeat this twice you will finally obtain something of the shape $x\equiv p\pmod{q}$

Solving the first and second equations simultaneously, you get x= 17 mod 40. Solving the third and fourth equations simultaneously, you get x = 52 mod 99. Solving these two results simultaneously, you get x = 1537 mod 3960 which gives you all possible solutions.

Hint

Consider the solution to \begin{align} x&\equiv1\pmod{5}\\ x&\equiv0\pmod{8}\\ x&\equiv0\pmod{9}\\ x&\equiv0\pmod{11}\\ \end{align}\tag{1} Suppose we can solve $$5a+792b=1\tag{2}$$ then $x\equiv792b\pmod{3960}$ would satisfy $(1)$. We can use the Euclid-Wallis Algorithm to solve $(2)$: $$\begin{array}{r} &&158&2&2\\\hline 1&0&1&\color{#C00000}{-2}&\color{#C00000}{5}\\ 0&1&-158&317&-792\\ 792&5&2&1&0\\ &&&{\uparrow}&{\star} \end{array}\tag{3}$$ $(3)$ tells us that $b\equiv-2\equiv3\pmod{5}$ satisfies $(2)$. Thus, $x\equiv2376\pmod{3960}$ satisfies $(1)$.

There are three other sets of equations corresponding to $(1)$ that can be solved and then combined to get the solution to your problem.