# Obtain the two incongruent solutions modulo 210 of the system

• Math100
In summary, the conversation discusses a system of congruences and applies the Chinese Remainder Theorem to find the solutions. There is a minor typo at the end, but the main point is that the solutions are ##x \equiv 59, 64 \pmod {210}##. The conversation also mentions the possibility of making mistakes and the difficulty of imagining such errors.
Math100
Homework Statement
Obtain the two incongruent solutions modulo ## 210 ## of the system
## 2x\equiv 3\pmod {5} ##
## 4x\equiv 2\pmod {6} ##
## 3x\equiv 2\pmod {7} ##.
Relevant Equations
None.
Consider the following system of congruences:
## 2x\equiv 3\pmod {5} ##
## 4x\equiv 2\pmod {6} ##
## 3x\equiv 2\pmod {7} ##.
Then
\begin{align*}
&2x\equiv 3\pmod {5}\implies 6x\equiv 9\pmod {5}\implies x\equiv 4\pmod {5}\\
&4x\equiv 2\pmod {6}\implies 2x\equiv 1\pmod {3}\\
&3x\equiv 2\pmod {7}\implies 6x\equiv 4\pmod {7}\implies -x\equiv 4\pmod {7}\implies x\equiv 3\pmod {7}.\\
\end{align*}
Note that ## 2x\equiv 1\pmod {3}\implies x\equiv 2\pmod {6} ## or ## x\equiv 5\pmod {6} ## because
## gcd(4, 6)=2 ##, which implies that there are two incongruent solutions ## x_{0}, x_{0}+\frac{6}{2} ## where ## x_{0}=2 ##.
Applying the Chinese Remainder Theorem produces:
## n=5\cdot 6\cdot 7=210 ##.
This means ## N_{1}=\frac{210}{5}=42, N_{2}=\frac{210}{6}=35 ## and ## N_{3}=\frac{210}{7}=30 ##.
Observe that
\begin{align*}
&42x_{1}\equiv 1\pmod {5}\implies 2x_{1}\equiv 1\pmod {5}\\
&\implies 6x_{1}\equiv 3\pmod {5}\implies x_{1}\equiv 3\pmod {5}\\
&35x_{2}\equiv 1\pmod {6}\implies -x_{2}\equiv 1\pmod {6}\\
&\implies x_{2}\equiv -1\pmod {6}\implies x_{2}\equiv 5\pmod {6}\\
&30x_{3}\equiv 1\pmod {7}\implies 2x_{3}\equiv 1\pmod {7}\\
&\implies 8x_{3}\equiv 4\pmod {7}\implies x_{3}\equiv 4\pmod {7}.\\
\end{align*}
Now we have ## x_{1}=3, x_{2}=5 ## and ## x_{3}=4 ##.
Thus
\begin{align*}
&x\equiv (4\cdot 42\cdot 3+2\cdot 35\cdot 5+3\cdot 30\cdot 4)\pmod {210}\equiv 1214\pmod {210}\equiv 164\pmod {210}\\
&x\equiv (4\cdot 42\cdot 3+5\cdot 35\cdot 5+3\cdot 30\cdot 4)\pmod {210}\equiv 1739\pmod {210}\equiv 59\pmod {210}.\\
\end{align*}
Therefore, the two incongruent solutions modulo ## 210 ## are ## x\equiv 59, 64\pmod {210} ##.

Looks good, except for the typo at the end.

fresh_42 said:
Looks good, except for the typo at the end.
What's the typo at the end?

Math100 said:
What's the typo at the end?
##64## is no solution.

Math100
fresh_42 said:
##64## is no solution.
I was careless.

Math100 said:
I was careless.
I have made mistakes worse than that!

fresh_42 said:
I have made mistakes worse than that!
Really? That's difficult to imagine.

Math100 said:
Really? That's difficult to imagine.
In chess speak, I'd say: you look at the position, into the position, and do not see the obvious. Even grandmasters have run into knight forks. Not that I am one, but **** happens.

SammyS and Math100

## 1. What does it mean to obtain two incongruent solutions modulo 210?

To obtain two incongruent solutions modulo 210 means to find two different solutions to a system of equations, where the solutions are not equivalent when taken modulo 210. This means that the solutions have different remainders when divided by 210.

## 2. Why is it important to obtain two incongruent solutions modulo 210?

Obtaining two incongruent solutions modulo 210 allows for a more accurate and complete solution to the system of equations. It also helps to avoid errors or inaccuracies that may arise from using only one solution.

## 3. How is obtaining two incongruent solutions modulo 210 different from obtaining two solutions without considering modulo 210?

Obtaining two incongruent solutions modulo 210 takes into account the concept of remainders and congruence, while obtaining two solutions without considering modulo 210 may only focus on finding any two solutions that satisfy the system of equations.

## 4. Can there be more than two incongruent solutions modulo 210?

Yes, there can be more than two incongruent solutions modulo 210. The number of incongruent solutions will depend on the specific system of equations and the values of the variables involved.

## 5. How can one obtain the two incongruent solutions modulo 210 of a system of equations?

The process of obtaining two incongruent solutions modulo 210 involves solving the system of equations using methods such as substitution, elimination, or graphing. It is important to keep in mind the concept of congruence and use modular arithmetic to find two solutions that are not equivalent when taken modulo 210.

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