I'm trying to solve the following problem:
What remainder does integer $n$ have when divided by $142$ if we know that $20n + 4$ and $72n – 12$ have the same remainder when divided by $142$?
I 'translated' the problem to maths:
$$
20n + 4 \equiv 72n – 12 \pmod {142} \\
n \equiv x \pmod {142}
$$
And we are essentially looking for $x$. Solving the linear congruence on top:
$$
20 + 4 \equiv 72n – 12 \pmod {142} \\
52n \equiv 16 \pmod {142} \\
13n \equiv 4 \pmod {71} \\
13n \equiv 4 + 8 \times 71 \pmod {71} \\
13n \equiv 572 \pmod {71} \\
n \equiv 44 \pmod {71} \\
$$
So $n = 71k + 44$ for any $k \in \mathbb{Z}$.
The problem is that I can't answer the original question as there, the modulo is $142$ and now it is $71$. How could I somehow 'change' the modulo and get the right answer? Or what would be a way to think about this problem?
PS: I realise that I'll get more solutions. For example, if I had $2x \equiv 4 \pmod 8$, I would divide by $2$ and get $x \equiv 2 \pmod 4$. Then $x = 4k + 2$. For these numbers are for example $6, 9, 11, \ldots$. I can see that if we were dividing these by $8$, not $4$, the remainders would be $2$s and $6$s.
Best Answer
So $n=71k + 44$ for $k \in \mathbb{Z}$. If $k=2a$, where $a \in \mathbb{Z}$, then $n = 142a+44$. On the other hand, if $k =2a +1$ then $n=142a+71+44$. Therefore, $$n \equiv 44,115 \mod{142}.$$
In general, if we know that $$n \equiv \alpha \mod{\beta}$$ and want to compute $n \mod{(k \beta)}$ then we use the fact that $n = \alpha + r\beta$ for $r \in \mathbb{Z}$.
By noting that $r \beta \equiv 0, \beta, \cdots, (k-1)\beta \mod{(k \beta)}$, we get $$n \equiv \alpha, \alpha+\beta, \cdots, \alpha+(k-1)\beta \mod{(k\beta)}.$$