Find all numbers between 1900 < b < 2000 that are congruent to a mod m...

Question

So when $a = 1$ and $m = 13$ I know that by the definition of congruence, $$13 \vert (1 - b)$$ which means $$b = 13k + 1, k \in \mathbb{Z}$$

And then $$1899 < 13k < 1999$$

I'm not sure how to proceed from here.

Answer 1

For this example, simply continue solving the inequality. Diving by 13 gives $$ 146.07... < k < 153.77... $$ The valid $k$ must be integers, so we get that all possible $k$ are $147, 148,149,150,151,152,$ and $153$. To calculate the $b$ values, simply take the set of $k$, that is $\{147,\ldots,153\}$ and multiply by 13 and add 1 to each member (because $b = 13k + 1$). Then you are done.

Then simply repeat this process for b) and c).

Answer 2

$$b = 13k + 1, k \in \mathbb{Z}$$ $$1899 < 13k < 1999$$

$$146.07<k<153.76$$

But $k$ is an integer

Thus $$147\le k\le 153$$

$$b = 13k + 1,k\in[147,153]$$