Congruence Solution: 42x-90x=-48x Simplified

[Math100]

Homework Statement

Solve the congruence 42x \equiv 12 (mod 90).

Relevant Equations

None.

42x-90x=-48x
x \equiv 12/-48 (mod 90)
Simplify
x \equiv -1/4 (mod 90)
Is this the correct answer for the given abstract algebra problem?

 

[fresh_42]

Homework Statement:: Solve the congruence 42x \equiv 12 (mod 90).
Relevant Equations:: None.

42x-90x=-48x
x \equiv 12/-48 (mod 90)
Simplify
x \equiv -1/4 (mod 90)
Is this the correct answer for the given abstract algebra problem?

I guess the result is expected to be an integer between $0$ and $89$. What does $42x \equiv 12 \mod 90$ mean as an equation?

 

[Math100]

So a fraction in answer isn't allowed in this type of problem?

 

[fresh_42]

So a fraction in answer isn't allowed in this type of problem?

No. Your fraction is $-(1/4)=(-1)\cdot 4^{-1}=89 \cdot 4^{-1}$ and you have to calculate the inverse of $4$ modulo $90$. However, $4$ has no inverse modulo $90$, i.e. your result doesn't even exist.

Again: what do we have when we write it as an equation?

 

[Math100]

42x \equiv 12 (mod 90)

 

[Math100]

I really don't know the next step.

 

[fresh_42]

We have $42x=90n+12$ with some integer $n$. Now whatever you made to get $1/4$, we can certainly divide the equation by $6$ and solve the smaller problem.

 

[Math100]

I got 7x=15n+2. And how do I solve it from here?

 

[fresh_42]

I got 7x=15n+2. And how do I solve it from here?

Now we need $7^{-1} \mod 15$, i.e. which number multiplied by $7$ leaves the remainder $1$ modulo $15\,?$

 

[Math100]

2, because 2 multiplied by 7 is 14.

 

[fresh_42]

2, because 2 multiplied by 7 is 14.

That would be $-1$. But it will do. What do we get from multiplication by $2$ on both sides?

 

[Math100]

14x=30 n+4

 

[fresh_42]

14x=30 n+4

And this means?

$14x \equiv -x \equiv 4 \mod 15$. Now what is $x$?

 

[Math100]

x \equiv 11 (mod 15)

 

[Math100]

Is that, above, the correct answer?

 

[fresh_42]

Is that, above, the correct answer?

Check it out: What is $42\cdot 11$ divided by $90$? Which remainder do you get?

 

[Math100]

462/90=5.13333333...
5×90=450
462-450=12
So the remainder is 12.

 

[fresh_42]

462/90=5.13333333...
5×90=450
462-450=12
So the remainder is 12.

... which is what we were looking for.

 

[Math100]

Okay. Now we just solved the problem. But is there an actual, formal formula for this type of abstract algebra problem?

 

[fresh_42]

Okay. Now we just solved the problem. But is there an actual, formal formula for this type of abstract algebra problem?

If you go through this thread, then you'll see that we only used basic arithmetic. Originally we had $42x\equiv 12\;(90)$ and the first impulse would be to ask for the inverse of $42$ in order to divide the equation. But there is no such inverse as $42$ and $90$ aren't coprime. But division by $6$ yields the same problem: $7x\equiv 2 \;(15)$ and now $7$ and $15$ are coprime. This means there is an inverse of $7$. You can use Euclid's algorithm to find this inverse, which you can find e.g. on Wikipedia (Bezout's Lemma). But for small numbers like $15$ it is easier to just check them. I mean it is not too difficult to see that $7\cdot 13 =91 =6\cdot 15 +1$ so that $7^{-1} \equiv 13\;(15)$. Now we can divide to separate $x$:
$$
7^{-1}\cdot 7 \cdot x = 1\cdot x = x = 7^{-1}\cdot 2 = 13\cdot 2 = 26 = 11\mod 15
$$

 

[Math100]

Okay. One more question, is this an example of abstract algebra problem or discrete mathematics problem? I'm considering to take an upper division mathematics course at a four year institution this Fall but don't know if Abstract Algebra will be the next course after Ordinary Differential Equations.

 

[fresh_42]

Okay. One more question, is this an example of abstract algebra problem or discrete mathematics problem?

Yes. I mean: both. Maybe a bit more discrete mathematics as it tends toward applied mathematics. Such calculations are used e.g. in the theory of error correcting codes. Abstract algebra emphasizes more the theoretical part: Why is modulo $90$ only a ring, whereas modulo $15$ is a field? The difference is: we can divide all numbers (except zero) in a field, but not necessarily in a ring.

I'm considering to take an upper division mathematics course at a four year institution this Fall but don't know if Abstract Algebra will be the next course after Ordinary Differential Equations.

This depends on your goals. I hope my answer got you an impression of the difference between the two.

 

[Math100]

Thank you so much for the help.