Ann and Bob are going to play a game. Ann chooses a two digit numbers from 01-99. Bob then mirrows the number and adds the checksum to this number and announces the result to Ann. The players then alternately continue with this process, while reducing any result mod 100 (hence only two digit numbers will appear). Is it possible that Bob ends up with number 00? If yes, what are the possible numbers for Ann to start with?

Example of a possible sequence: : Ann: 59 -> Bob 95+14=109=09 mod 100 -> Ann 90+9=99

  • $\begingroup$ Can you clarify what you mean by to "mirrow" and what formula they use for the checksum? $\endgroup$ – msh210 Jul 7 at 20:59
  • $\begingroup$ Both digit gets switched and the i think the checksum is the cross sum (95:5+9=14 $\endgroup$ – Matti Jul 7 at 21:14

Just a quick thing to cut our work in half: given $10x+y$ it gets mapped to $10y+x+x+y=2x+11y$, possibly minus 100. Regardless, $2x+11y \equiv 10x+y \pmod{2}$, so we only have to look at the even numbers.
enter image description here
The possible numbers that eventually reach $0$ are $\boxed{16, 56, 64, 68, 76, 80}$ (rereading the question it appears you can't start on 0, so that's ruled out).


Unless I'm mis-understanding

if Ann begins with 68, Bob would have 86+14 = 100 mod 100 = 00

looking at the full possibilities

enter image description here

And the only possibilities that lead to that are:

marked in yellow.... 16, 56, 64, 68, 76 and 80


note that some will lead to Bob giving back 00, which Ann will give 0+0=0 back.

  • 1
    $\begingroup$ correct, can you list all numbers which Ann can choose to put Bob on 00? $\endgroup$ – ThomasL Jul 7 at 21:17

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