Use the Chinese Remainder Theorem to determine the value of $x$.

Question

I'm trying to solve the following modular arithmetic question using the Chinese Remainder Theorem, using this link. (We learned a different method in our class, but I found this easier to grasp). $$x \equiv 1 (\text{mod} \ 5)$$ $$x \equiv 2 (\text{mod} \ 7)$$ $$x \equiv 3 (\text{mod} \ 9)$$ $$x \equiv 4 (\text{mod} \ 11)$$

I then represented $x$ as a sum of $4$ boxes, such that the first term is "related" to $\text{mod} \ 5$ (i.e. the $1^{st}$ term will not be made $0$ due to the $\text{mod} \ 5$), the second term is related to $\text{mod} \ 7$ and so on. Here's what I mean by "related":

If we only consider $\text{mod} \ 5$, the value of box $1$ is $693$, the value of box $2$ is $495$, then $693 \ \text{mod} \ 5 = 3$ but $495 \ \text{mod} \ 5 = 0$. Likewise, if we only consider $\text{mod} \ 7$, then the value of box $1$ is $693 \ \text{mod} \ 7 = 0$ but $495 \ \text{mod} \ 7=5$. And so on...

After doing all that, I have $$x = (7 \times 9 \times 11) + (5 \times 9 \times 11) + (5 \times 7 \times 11) + (5 \times 7 \times 9)$$

The next step is applying the $\text{mod} \ 5$ to $x$: $$\begin{align} x \ \text{mod} \ 3 &\equiv 691 \ \text{mod} \ 5 + 495 \ \text{mod} \ 5 + 385 \ \text{mod} \ 5 + 315 \ \text{mod} \ 5 \\ &\equiv 693 \ \text{mod} \ 5 + 0 + 0 + 0 \\ &\equiv 693 \ \text{mod} \ 5 \\ &\equiv 3 \ (\text{mod} \ 5) \end{align}$$

This is where I get stuck. In the video, and the video doesn't explain how to deal with such a scenario.

PS - If there is a more "intuitive" or more efficient version of the Chinese Remainder Theorem, I'd be grateful if you could share it.

PPS - Sorry if the question is a bit awkwardly formulated. As you can guess this is my first doing this.

Answer 1

There should be $x = (7 \times 9 \times 11)\cdot(7 \times 9 \times 11)^{-1}_5\cdot 1 $ ${}+ (5 \times 9 \times 11)\cdot(5 \times 9 \times 11)^{-1}_7\cdot 2 $ ${}+ (5 \times 7 \times 11)\cdot(5 \times 7 \times 11)^{-1}_9\cdot 3 $ ${}+ (5 \times 7 \times 9)\cdot (5 \times 7 \times 9)^{-1}_{11}\cdot 4$ for this approach.

Answer 2

That is a TERRIBLE video. But the technique is interesting.

SO we have

$x = a*693 + b*495 + c*385+d*315$.

First we do $\mod 5$.

$x \equiv 3*a + 0 +0 +0\equiv 3a \pmod 5$ and we need $3a \equiv 1 \pmod 5$. Now trial and error shows us that $3*2 = 6 \equiv 1 \pmod 5$ so $a=2$ will do.

Now $\mod 7$

$x\equiv 0 + b*5 + 0 + 0\equiv 5b \pmod 7$. So we need $5b\equiv 2\pmod 7$.

He doesn't explain how do do this. Trial and error shows us that $5*6 =30\equiv 2 \pmod 2$ so $b=6$ will do.

Then we $\mod 9$ (not $3$)

$x \equiv 7c \pmod 9$ and we need $7c \equiv 3\pmod 9$.

Okay. No trial and error any more.... $7c = 3 + 9k$ so $7\frac c3= 1+ 3k$ so $3|c$. Le $c = 3e$. $7e = 1+3k$ so $(2*3+1)e= 1+3k$ so $e = 1 + 3(k-2)$ so we can have $e=1$ and $c = 3$. $7*c = 21 =3+18 \equiv 3 \pmod 9$.

So $c= 3$ will do.

And finally $\mod 11$ we have $x \equiv 315d\equiv 7d\pmod {11}$ so we need $7d\equiv 4\pmod 11$.

$7d = 4 + 11k$

$(11-4)d= 4 + 11k$

$-4d = 4 + 11(k+d)$ so $d=-1$ will do.

So we can have $x = 2*693 + 6*495+ 3*385 - 315=5196$

Of course that not the smallest positive answer.

To get a reasonable answer I'd alternate a few negative and positive values.

Instead of $b=6$ we can have $b\equiv 6 \equiv -1 \pmod 7$ and use $b=-1$ to get

$x = 2*693 -495 + 3*385 -315=1731$ will do. (And if my instincts are right that is smallest value between $0$ and $5\times 7\times 9\times 11 = 3465$

$2*693 -495 + 3*385 -315\equiv 2*3 + 0 + 0 + 0 \equiv 1 \pmod 5$.

And $2*693 -495 + 3*385 -315\equiv 0-5 + 0 + 0 \equiv 2\pmod 7$.

And $2*693 -495 + 3*385 -315\equiv 0+0+3*7 +0+0\equiv 21 \equiv 3 \pmod 9$

And $2*693 -495 + 3*385 -315 \equiv 0+0+0-7\equiv 4 \pmod {11}$.

.....

I've never seen this method before.... but I ... sort of liked it. But the presentation in that video was terrible.

Answer 3

I think the best way for me to solve a CRT problem is like this: $$x \equiv1 \pmod{5} \implies x \in \{1,6,11,16,21,26\dots\}$$ $$x \equiv4 \pmod{11} \implies x \in \{4,15,26,\dots\}$$

Now one can immediately see the intersection at $x=26$, and indeed $x \equiv 26 \pmod{55}$ satisfies both $x \equiv1 \pmod{5}$ and $4 \pmod{11}$.

Similarly, $$x \equiv 2 \pmod{7} \implies x \in \{2,9,16,23,30,\dots\}$$ $$x \equiv 3 \pmod{9} \implies x \in \{3,12,21,30,\dots\}$$ So $x \equiv 30 \pmod{63}$

Now, from there I can solve it with the casual method: $$x \equiv26 \pmod{55} \implies x=55k+26$$ $$\implies55k+26 \equiv30 \pmod{63} \implies 55k \equiv4 \equiv 130 \pmod{63}$$ $$\implies 11k \equiv 26 \equiv 341 \pmod{63} \implies k \equiv 31 \pmod{63} \implies k=63j+31$$ $$\implies x=55(63j+31)+26=3465j+1731 \implies x \equiv 1731 \pmod{3465}$$ Noting, of course, that $3465=5\cdot7\cdot9\cdot11$

Answer 4

The best way to do the Chinese Remainder Theorem is to do it one at a time, and merge two conditions repeatedly.

For two values, the best way to compute is given in the Wikipedia page, under the "Case of two moduli" section.

From here, you want to contract conditions: you can convert $x \equiv 1 \pmod 5, \; x \equiv 2 \pmod 7$ into $x \equiv 16 \pmod {35}$ using this technique, and then repeat on $35$ and $9$ to find a condition for $x$ modulo $315$, and finally finish using the modulo $315$ condition and the modulo $11$ condition.