# Congruence equation problem

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.

-

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

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.

-

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 $35.$

-

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, $$a\equiv b\pmod c \iff a - b = kc, \; \text{for some }\,k\in \mathbb Z.$$

-