Which number?

Dec 2019
52
1
ok
If Alpha's number is 8, Alpha knows that Beta's number couldn't be 4, as that would leave no number available for Gamma. Do you understand that?
But you said Beta's number was 4, 6, 7 or 9? I'm lost, I'm sorry. This should be easier than it looks :|
 

skipjack

Forum Staff
Dec 2006
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I was merely considering the possibility that Alpha has 8. Alpha could initially rule out two possible numbers for Beta (for example, if Alpha had 8, Beta couldn't have 4 or 8). To do that, Alpha's number must be 4, 5, 8 or 10.
 
Dec 2019
52
1
ok
If Alpha's number is 8, Alpha knows that Beta's number couldn't be 4, as that would leave no number available for Gamma. Do you understand that?
I do.. but how are 5 and 10 also possible for Alpha?
 
Jun 2019
493
260
USA
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Here's a table I made when I was first considering the problem, if it helps.
4 and 8 can't be in play at the same time. They would have to be 4+4=8 (two fours) or 4+8=12 (no 12s available).
Likewise, 5 and 10 can't be in play at the same time. 5+5=10 or 5+10=15, neither valid.

That's about as far as I got before realising there would be a lot of branches to consider and gave up. :p
 
Dec 2019
52
1
ok
View attachment 10732
Here's a table I made when I was first considering the problem, if it helps.
4 and 8 can't be in play at the same time. They would have to be 4+4=8 (two fours) or 4+8=12 (no 12s available).
Likewise, 5 and 10 can't be in play at the same time. 5+5=10 or 5+10=15, neither valid.

That's about as far as I got before realising there would be a lot of branches to consider and gave up. :p
I don't get it lol. Thanks for the effort tho. Appreciated! :)
 

skipjack

Forum Staff
Dec 2006
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. . . how are 5 and 10 also possible for Alpha?
If Alpha's number is 10, Alpha knows that Beta's number couldn't be 5, as that would leave no number available for Gamma.
 
Dec 2019
52
1
ok
If Alpha's number is 10, Alpha knows that Beta's number couldn't be 5, as that would leave no number available for Gamma.
Can I just have your brain at this point?

So how are 4, 6, 7 and 9 possible for Beta? And how did you conclude that Alpha has to have 4?
 
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skipjack

Forum Staff
Dec 2006
21,301
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Beta, knowing that Alpha's number is 4, 5, 8 or 10 , finds that there are exactly three possible candidates for Gamma's number. This is possible only if Beta's number is 4, 6, 7 or 9. At this stage, there are relatively few possibilities overall, and they can easily be listed.
 
Dec 2019
52
1
ok
Beta, knowing that Alpha's number is 4, 5, 8 or 10 , finds that there are exactly three possible candidates for Gamma's number. This is possible only if Beta's number is 4, 6, 7 or 9. At this stage, there are relatively few possibilities overall, and they can easily be listed.
But how can you conclude that?
 

skipjack

Forum Staff
Dec 2006
21,301
2,377
If Beta had 3, Gamma's number could be 1, 2, 5, 7 or 8, which is 5 possibilities, not 3.

If Beta had 7, Gamma's number could be 1, 2 or 3, which is 3 possibilities. and couldn't exceed 3, as that would require Alpha's number to be less than 4, which is already ruled out.

In a similar way, it turns out that Beta's number is 4, 6, 7 or 9.