arithmetic - Number game and divisibility: what is the proof that this trick (divisible by 8) works?

The number game is part of the longlasting French game "des chiffres et des lettres", which in UK is also known as "Countdown".

Short description: you have 6 tiles corresponding to numbers, and you must use all 4 operations (add, multiply, substract, divide) to obtain a randomly chosen three digit number.

I play this game quite often and stumbled upon a page which gave this trick to tell whether a (three digit) number is divisible by 8. Say the three digits are named H, T and U (for Hundreds, Tens and Units), then to find if HTU is divisible by 8, you can do this:

• if H is even, then TU must be divisible by 8;



• if H is odd, then TU must be divisible by 4, but not 8.

And indeed this works (insofar as I didn't find any counterexample); for instance 544 is divisible by 8 (8 * 68) since 5 is odd, and 44 is divisible by 4 but not 8.

But what is the proof?

Answer

Note that $1000$ is divisible by $8$, so we can just consider $H$, $T$, and $U$.

In fact, let's group $T$ and $U$ together to R, i.e. $R=10T+U$.

So, we now need to consider $100H+R$.

If $R$ is divisible by $8$ and $H$ is odd, then the remainder would be $100H = 100(2n+1) = 200n+100 \to 100 \to 4$.

If $R$ is divisible by $4$ but not by $8$, and $H$ is odd, then we can write $R=4(2p+1)$ and $H=2h+1$, so the remainder would be $200h+100+8p+4=8(25h+p+13) \to 0$.

One can consider the other two cases to prove the algorithm.