In these two problems, (mod 9) and (mod 3) are natural to use because any number is equivalent to it's sum of digits. So, taking 5.8 as an example, even though there are 5! = 120 ways to rearrange the digits 0, 2, 4, 6, 8, the sum of the digits is ALWAYS 20. So we know that the number is 2 (mod 3) and (2 mod 9).

In general, however, there might be some trial and error in which mods to use. Further, remember that mods can give us restrictions (so they can be used to show something does NOT exist), but cannot be used to prove something does exist.

Let's take a basic example of 135 and ask is 135 a perfect square, meaning is there an integer k such that $k^2 = 135$. (Obviously the answer is no, but let's explore a bit).

Note if $k^2 = 135$ then

• Mod 2: $k^2 \equiv 1 \pmod{2}$. Note this does imply that $k\equiv 1 \pmod{2}$, but is otherwise INCONCLUSIVE in determining if $135$ is a perfect square. (Because it is possible for a perfect square to be $\equiv 1 \pmod{2}$.)
• Mod 3: $k^2 \equiv 1+3+5 \equiv 0 \pmod{3}$. Again, this does imply that $k\equiv 0 \pmod{3}$, but is otherwise INCONCLUSIVE in determining if $135$ is a perfect square.  (Because it is possible for a perfect square to be $\equiv 0 \pmod{3}$.)
• Mod 9: $k^2 \equiv 1+3+5\equiv 0 \pmod{0}$, INCONCLUSIVE.
• Mod 10: $k^2\equiv 5 \pmod{10}$, INCONCLUSIVE.
• Mod 4: $k^2\equiv 3 \pmod{4}$. Here, note none of $0^2$, $1^2$, $2^2$, or $3^2$ are $\equiv 3 \pmod{4}$, so we can CONCLUSIVELY say that $k^2$ is NOT a perfect square.

If it helps, think of this as a proof by contradiction. If we assume $k^2 = 135$ then $k^2 \equiv 3 \pmod{4}$ but this is a contradiction.