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I have found the following congruences:

  1. $x \equiv 2\mod 5$
  2. $x \equiv 12\mod27$
  3. $x \equiv 2\mod4$

How can I solve for x using the Chinese Remainder Theorem? Please include justifications for the steps you take.

How can I find $x$?

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Hint $\ $ Note that by the constant-case optimization of CRT we have

$$\rm\begin{eqnarray} x\equiv 2\pmod 4\\\rm x\equiv 2\pmod 5\end{eqnarray}\ \iff\ x\equiv 2\pmod{4\cdot 5}$$

Now applying Easy CRT to the RHS congruence and the remaining congruence we obtain

$$\rm\begin{eqnarray} x\equiv \color{#C00}2\pmod{\color{blue}{20}}\\\rm x\equiv \color{#0A0}{12}\pmod{27}\end{eqnarray}\ \iff\ x\equiv \color{#0A0}{12}+27\,\left[\dfrac{\color{#C00}2\!-\!\color{#0A0}{12} }{27}\ mod\ \color{blue}{20}\right]\equiv 282\pmod{\color{blue}{20}\cdot 27}$$

where the above fraction is computed as $\rm\, mod\ 20\!:\ \dfrac{10}{27}\equiv \dfrac{10}{7}\equiv\dfrac{30}{21}\equiv \dfrac{10}1,\ $ and $\rm\:-10\equiv 10$

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I like the look of this "Easy CRT" method. However, could you apply some intuition on why the formula works? I read the hyperlinked proof you gave in another question and it is not clear for a novice like myself. – CodeKingPlusPlus Sep 23 '12 at 22:39
@Code I added a note to the Easy CRT link showing how the solution is derived and easily remembered. – Bill Dubuque Sep 23 '12 at 23:11

The first thing to look for is that your moduli are pairwise coprime - they are. Good. We consider that our system is of the form $x \equiv a_i \pmod{m_i}$. So, let $a_1 = 2$, $a_2 = 12$, and $a_3 = 2$. Further, let $m_1 = 5$, $m_2 = 27$, and $m_3 = 4$, then

$M_1 = m_2 m_3$,

$M_2 = m_1 m_3$, and

$M_3 = m_1 m_2$.

Now we solve (usually by the Euclidean algorithm, or by inspection for smaller numbers that are somewhat obvious) for each $y_i$ where

$y_1 M_1 \equiv 1 \pmod{m_1}$,

$y_2 M_2 \equiv 1 \pmod{m_2}$, and

$y_3 M_3 \equiv 1 \pmod{m_3}$.

Then our solution that satisfies all three equivalences is

$x \equiv a_1 y_1 M_1 + a_2 y_2 M_2 + a_3 y_3 M_3 \pmod{m_1 m_2 m_3}$

I trust you can take the arithmetic from here.

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This particular one can go manually:

From 1. and 3. we get $x\equiv 2 \mod 20$, so we need an $x$ among $2,22,42,62,\ldots$ which is $\equiv 12 \mod 27$. Subtracting $12$, it is $-10,10,30,50,\ldots,270,\ldots$.

So, $x\equiv 282 \mod (20\cdot 27)$.

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I will assume that you are partly familiar with the machinery of CRT.

We want to first find $a, b,c$ such that $a(27)(4)\equiv 1\pmod{5}$, and $b(5)(4)\equiv 1\pmod{27}$, and $c(5)(27)\equiv 1\pmod{7}$. Then a solution of our system of congruences is $$x=2(a)(27)(4)+12(b)(5)(4)+2(c)(5)(27).$$ It may be that the $x$ we obtain is kind of large, and we may want to replace it by something congruent to it modulo $(5)(27)(4)=540$ that is, say, in the interval $[0,539]$.

Let's find suitable $a$, $b$, and $c$. In general for this work, and with large moduli, we would use the Extended Euclidean Algorithm. Let's not bother, since we are concentrating on CRT.

For $a$, work modulo $5$. Since $27$ is congruent to $2$, we want $8a\equiv 1\pmod{5}$, or equivalently $3a\equiv 1\pmod{5}$. So we can take $a=2$. Or maybe $a=-3$. Doesn't matter.

For $b$, we want $20b\equiv 1\pmod{27}$. Slightly unpleasant, but note that $(20)(4)\equiv -1\pmod{27}$, so we can take $b=-4$.

For $c$, the modulus is tiny. We want $c(5)(27)\equiv 1\pmod{4}$. But $5$ is congruent to $1$, and $27$ is congruent to $3$, so we can take $c=3$.

Remark: But we can take a bit of a shortcut. Note that we wanted $x\equiv 2\pmod{5}$ and $x\equiv 2\pmod{4}$. This is the case iff $x\equiv 2\pmod{20}$. So we want to solve the system of congruences $x\equiv 2\pmod{20}$, $x\equiv 12\pmod{27}$. This is a two congruence CRT problem. We find $a$ and $b$ such that $a(27)\equiv 1\pmod{20}$ and $b(20)\equiv 1\pmod{27}$. Then we can take $$x=2(a)(27)+12(b)(20).$$ For $a$, it is clear that $3$ will do. For $b$, $-4$ works. This gives answer $2(3)(27)+12(-4)(20)$. For something in the range $[0.539]$, add $(2)(540)$. We get $x=282$. So the general solution is $x\equiv 282\pmod{540}$.

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