To see why the second fact is true, say $a$ has a remainder of $r$ when divided by $b$ and $c$ has a remainder of $s$. Then $a=bq+r$ and $c=bt+s$ for some integers $q$ and $t$, and so $$a+c=bq+bt+r+s,$$ which has the same remainder as $r+s$ since we can subtract multiples of $b$ without changing the remainder.

The third rule is similar: writing $a$ and $c$ as above, we now have $$a\cdot c=(bq+r)\cdot(bt+s)=b^2qt+b(qs+rt)+rs,$$ and again taking away the multiples of $b$ we are left with $rs$.

Pages: 1 2

There is a Friday the 13th in October of year 1000000th?