Chinese Remainder Theorem for $x\equiv 0 \pmod{y}$

Question

Can anyone solve the following system of congruences using CRT step-wise, without skipping any part?

$$\begin{cases} x\equiv 3 \pmod{7}\\ x\equiv 3 \pmod{13}\\ x\equiv 0 \pmod{12}\end{cases}$$

The answer to this problem $276$ but, I keep getting the answer $264$. What is it that I am doing wrong?

Answer 1

$x$ is a multiple of $12$ so $x=12k$. We also have $12k\equiv 3\bmod 7$, $12$ is congruent to $5\bmod 7$ and the inverse of $5\bmod 7$ is $3$ by inspection. So $k\equiv 3\cdot5\cdot k\equiv 3\cdot3\equiv 2\bmod 7$. Therefore $k$ is $7l+2$.

So $x=12(7l+2)=84l+24$.

We now have $84l+24\equiv 3 \bmod 13$. This tells us $84l\equiv 5 \bmod 13\implies 6l\equiv 5 \bmod 13$. The inverse of $6\bmod 13$ is $11$ by inspection, therefore $l\equiv 11\cdot 6 \cdot l\equiv 11\cdot 5\equiv 3 \bmod 13$. Therefore $l=13j+3$

And so $x=84l+24=84(13j+3)+24=1092j+276$.

So indeed the solution is to take $n\equiv 276\bmod 1092$. And by the chinese remainder theorem the solution is unique.

Answer 2

$12\mid x\,\Rightarrow\, x=12y.\ $ And $\ 7,13\mid x\!-\!3\!\iff\! \overbrace{{\rm lcm}(7,13)}^{\large 91}\mid x\!-\!3\!\iff\! x\equiv\color{#c00} 3\pmod{91}$

$\ {\rm mod}\ 91\!:\ x = 12y\equiv\color{#c00} 3\iff y\equiv \dfrac{3}{12}\equiv \dfrac{1}4\equiv \dfrac{92}4\equiv \color{}{23}\!\iff\! \color{#0a0}{y = 23+91n} $

Therefore $\, x = 12\color{#0a0}y = 12(\color{#0a0}{23 + 91n}) = 276 + 1092k$

Answer 3

The first pair of congruences would give the result of:

$x \equiv3 \mod 91$

Thus, the next part is solving 3+91a being a multiple of 12 which could be reduced to 3+7a where one could do trial and error to find a=3 would be a solution as 3+7*3=3*8=24.

3+91*3=3*92=276.

Answer 4

I directly use the result (formula) for Chinese Remainder Theorem to solve this problem. By the way, this method needs a little knowledge of group theory.

For the system $$x\equiv a_1\pmod{m_1}\\ x\equiv a_2\pmod{m_2}\\ \vdots\\ x\equiv a_n\pmod{m_n}$$ The general solution is given by $$x=\sum_{i=1}^na_iM_iy_i+t\cdot \prod_{i=1}^nm_i,$$ where $$t\in\mathbb{Z},\qquad M_i=\prod_{j=1,\ j\neq i}^nm_j\qquad \text{and}\qquad [M_iy_i]_{m_i}=[1]_{m_i}.$$

Therefore, in this problem, $a_1=3,\ m_1=7, a_2=3,\ m_2=13,\ a_3=0,\ m_3=12$,
we have $$M_1=13*12=156,\quad y_1=4$$ $$M_2=7*12=84,\quad y_2=-2$$ $$M_3=7*13=91,\quad y_3=-5$$ So the solution is $x=3*4*156-3*2*84-0*5*91+7*13*12*t=1368-1092t$. For your answer, it is just the smallest positive solution, which can be obtained from letting $t=1$, $1368-1092=276$.