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I see $\gcd(2,36)=2$ and divides $6$ so I divide and receive $z\equiv3 \pmod {18}$.

For $3z\equiv7\pmod 8$ I see $\gcd(3,8)=1$ and 1 divides 7. I also see that to reduce $3z\equiv7\pmod8$, I multiply by 3 which gives me $21\pmod8$. I am then left with $z\equiv3\pmod{18}$ and $z\equiv21\pmod 8$.

I then proceeded to use the Chinese Remainder Theorem on these two congruences and receive my answer.

My CRT work: 18 x 8=144 a=3 b=21 8(mod 18) = 8 18(mod 8) = 2, 3 x 8 x 8 = 192, 21 x 18 x 2 = 756, 192+756 = 948, 948(mod 144) = 84(mod 144)

However, I was told this was wrong but I am unsure why, if anyone can help point me in the right direction, that would be great.

Pauliene
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  • Who asked you to solve the system of congruences? Name of class? Name of text book? Very passive voice: "I am asked", "I was told": Who asked you, and who told you..? – amWhy Oct 25 '21 at 19:32
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    always split into prime powers for problems done by hand. You have $2z \equiv 6 \pmod 9, $ then $2z \equiv 6 \pmod 4, $ as well as $3z \equiv 7 \pmod 8. $ Then you need to check that the examples with powers of $2$ are consistent. In the end you have someting mod 9 and something mod 8, together something mod 72 – Will Jagy Oct 25 '21 at 19:34
  • Dana Please show your work following" "I then proceeded to use the Chinese Remainder Theorem, ... and receive my answer." Please show us your work and your answer. – amWhy Oct 25 '21 at 19:34
  • Hello, I added my CRT work but do not know how to format it properly – Pauliene Oct 25 '21 at 19:47
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    Dana, I recommend against attempting to use CRT with moduli that share a gcd. Try the bit with moduli 4,8 and 9, combine the 4,8 into a single mod 8, then combine moduli 8,9 into mod 72 – Will Jagy Oct 25 '21 at 19:49
  • Hi, I was unaware that you can combine 2 moduli into a single mod as you suggest combining the 4,8 into a single 8. How would I go about doing that? – Pauliene Oct 25 '21 at 19:56
  • Dana, I wrote them above; recommend you solve $2z \equiv 6 \pmod 4$ and $3z \equiv 7 \pmod 8$ by simply trying $z$ values from $0$ to $7;$ or $1$ to $8$ if you prefer. Note detail: $2z \equiv 6 \pmod 4$ means the same thing as $2z \equiv 2 \pmod 4$ – Will Jagy Oct 25 '21 at 20:00
  • Hello thank you so much for your help! With your advice, I received z=5. I then have the congruences: z≡5(mod 8) and z≡3(mod 9). Using CRT I eventually got the answer 21(mod 72). Is this correct? – Pauliene Oct 25 '21 at 20:14
  • Dana yes, that is correct – Will Jagy Oct 25 '21 at 21:51
  • And this follows from the LCM generalization of CRT. The coprime case has LCM equal to product. – Roddy MacPhee Oct 25 '21 at 23:03
  • By General Easy CRT in the linked dupe: $$\begin{align} \begin{array}{}&x\equiv 3\pmod{!18}\&x\equiv 5\pmod{!8}\end{array}!!&\iff x\equiv 3+18\left[\dfrac{(5!-!3)/2}{\ \ \ \ \ \ 18/2}\bmod 8/2 \right]!!!\pmod{!18(8)/2}\[.4em] &\iff x\equiv 3+18,[,2/1,\bmod, 4,],\equiv, 21\pmod{72}\end{align}$$ – Bill Dubuque Jun 02 '22 at 08:14

2 Answers2

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$2z\equiv 6\bmod 36\Rightarrow 2z=36t+6\Rightarrow z=18t+3\rightarrow 3z=54 t+9$

$3z\equiv 7\bmod 8 \Rightarrow 3z=8t_1+7$

So we have to solve following equation:

$54t+9=8t_1+7, Or:, -27t+4t_1=1$

One solution to this equation is $(t, t_1)=(1, 7)$ so we have:

$t=4s+1$ and $t_1=27 s+7$ such that:

$3z=54(4s+1)+9=8(27 s+7)+7=216 s+63$

Or:

$z=72 s+21\rightarrow z\equiv 21\bmod 72$.

sirous
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$$\begin{cases}2z\equiv6\pmod{36}\\3z\equiv7\pmod8\end{cases}\iff\begin{cases}2z=6+36X\\3z=7+8Y\end{cases}\Rightarrow8Y-54X=2\iff4Y-27X=1$$ The general solution of this last equation is get standard way by the help of a single solution, in this case$(X,Y)=(7,1)$ so $X=4n+1$ (not necessary to get $Y$)so $$z=3+18X=72n+21\text{ where } n\in\mathbb Z$$

Piquito
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