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→100→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)→0.
One can consider the other two cases to prove the algorithm.
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