Solving a system of congruences using the Chinese Remainder Theorem

Question

Suppose I have the congruences:

$x \equiv 3 ($mod $7)$

$x \equiv 8 ($mod $9)$

$x \equiv 1 ($mod $5)$

$x \equiv 1($mod $16)$

The Chinese Remainder Theorem says I will have a solution $($mod $5040)$.

Basically what I did was create $4$ combinations.

$720 ($mod $7) = 6$.

$560 ($mod $9) = 2$.

$1008 ($mod $5) = 3$.

$315 ($mod 16$) = 11$.

I then proceeded to multiply out $3 \cdot 720 \cdot 6 + 8 \cdot 560 \cdot 2 + ...$, and then take that answer mod $5040$, but got an answer that does not satisfy the congruences. I recall doing this same exact procedure when doing the egg problem figuring out how many eggs the lady started with, and it worked out fine. What am I doing wrong? The help would be appreciated.

Answer 1

I don't understand your procedure, but it seems to be something like this method: write $$x=720a+560b+1008c+315d\ ,$$ substitute into the four congruences and simplify. For example, the first one gives $$6a\equiv3\pmod7\ ,$$ which is easily solved by trial and error to get $a=-3$, among other possibilities. Similarly you get $b=4$, $c=2$, $d=3$ and hence $$x\equiv -3\times720+4\times560+2\times1008+3\times315\equiv3041\pmod{5040}\ ,$$ assuming that my mental arithmetic is correct ;-)

Answer 2

The trick to these is to take them two at a time. We know x=17 is a solution to the first two by checking, so $x\equiv 17 \pmod{63}$ by CRT. Also, looking at equations 3 and 4, we realize that if x is one more than a multiple of 5 and 1 more than a multiple of 16, x must be 1 more than a multiple of 80, or $x\equiv 1 \pmod{80}$. The last step, combining $x\equiv 1 \pmod{80}$ and $x\equiv 17 \pmod{63}$ is a little harder, but you should be able to get there.

Answer 3

It is much easier to successively solve the congruences (as here).

${\rm mod}\ 80\!:\ x\equiv 1\iff x = 1+80i$

${\rm mod}\ 9\!:\ {-}1\equiv x\equiv 1+80i\equiv 1-i\iff i\equiv 2\iff i=2+9j$

${\rm mod}\ 7\!:\ 3\equiv x\equiv 1+80(2+9j)\equiv -j\iff j\equiv -3\equiv 4\iff j = 4+7k$

Therefore, $\ \ x = 1+80(2+9(4+7k)) = 3041+5040k$

Answer 4

$720 \equiv 6 \pmod 7$ is a good start, but you want $\equiv 3 \pmod 7$

Since $\dfrac 12 \equiv 4 \pmod 7$, you could multiply both sides by 4 and get $2820 \equiv 3 \pmod 7$.

Another way would be to get $\equiv 1 \pmod 7$ Since $\dfrac 16 \equiv -1 \pmod 7$, you would get $-720 \equiv 1 \pmod 7$, so $-2160 \equiv 3 \pmod 7$

You would have to do the same sort of thing to the other numbers.