2
$\begingroup$

My problem says:

Give solution to this problem of congruence, with all incongruent solutions according to the requested module and all integer solutions.

$10x \equiv 15 \mod 35$

But I can not understand the steps to solve this exercise.

I would also like to know the meaning of this equation, ie:

$(n=m) \rightarrow$ $n$ is equal to $m$

$(n\gt m \lor n\lt m) \rightarrow$ $n$ is greater or lower than $m$

But, whats means $a \equiv b\mod c$?

I hope someone can recommend any book or information to solve such equations.

$\endgroup$
1
$\begingroup$

$$10x \equiv 15 \pmod{35}$$

This is equivalent to $$35 \mid 10x-15$$

There exists an integer $k \in \mathbb{Z}$ such that

$$10x-15 = 35k$$

$$2x-3 = 7k$$

Going back to the notation of modular arithmetic, this is the same as:

$$2x \equiv 3 \pmod{7}$$

(We could have skipped the intermediate steps of converting to a normal equation, but I wanted to show why it was possible to cancel the fives in the equivalence)

Now we can test values of $x$ so that the (easier) congruence modulo $7$ is satisfied. It suffices to test $x=0,\,1,\,2,\,3,\,4,\,5,\,6$, because those cover all of the possible remainders, modulo seven.


$$2(0) \equiv 0 \not\equiv 3 \pmod{7}$$

$$2(1) \equiv 2 \not\equiv 3 \pmod{7}$$

$$2(2) \equiv 4 \not\equiv 3 \pmod{7}$$

$$2(3) \equiv 6 \not\equiv 3 \pmod{7}$$

$$2(4) \equiv 8 \equiv 1 \not\equiv 3 \pmod{7}$$

$$2(5) \equiv 10 \equiv 3 \pmod{7}$$

$$2(6) \equiv 12 \equiv 5 \not\equiv 3 \pmod{7}$$


Thus, the solutions are exactly the integers $x$ which leave a remainder of five when divided by seven.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

What $a \equiv b \mod c$ means is that $c | (a-b).$ For the actual exercise, just write down the $35$ numbers $10 x$ for $x=0, \dots, 34$ and see which one of them gives remainder 15 when divide by $35.$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

We can define the congruence relation $a\equiv b\pmod c$ as follows: $$a\equiv b\pmod c \iff c\mid (a-b).$$

That is, $c$ divides the differences $a - b, b-a$. Put differently, $$a\equiv b\pmod c \iff a - b = kc, \; \text{for some }\,k\in \mathbb Z.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.