Logical riddles • Solution for ›The election‹


For this explanation I will åshorten the names of the candidates with the starting letter. Uppercase letters symbolize that the person will be elected, lowercase letters that they will lose the election. The numbers in parantheses refer to the statements.

Again the statements in short:

 1. [f h]
 2. [A b e] 
 3. [a b D F] 
 4. [A b c d E F G H]
 5. [b C E] 
 6. [a c] 
 7. [e d H] 
 8. [A d g]
 9. [B G] 
10. [d e F]
11. [c E F h] 
12. [a f G]
13. [c D] 
14. [B C H] 
15. [g h]
16. [C D e g]

Assuming B, then after (9) g, after (15) H, (14) c, (6) A and after (13) d. But then (8) would be completely correct which isn't allowed. So b.

From (2) and (5) follows that either [a C], [A c] or [a c]. [a c] is not possible because of (6). Let's assume [a C]. Then after (5) e. If d would be correct, then after (10) f, (12) g, (15) H. But that's contradicting (7). Then D. That leads to (3) f, (1) H, (12) g. But that's also not possible because of (16). So the assumption [a C] was wrong. Leaves only A c.

Now we know b, A and c. It follows: (2) E, (13) d, (8) G.

That leaves four possibilities: [f h], [F H], [f H] and [F h]. [f h] drops out because of (1), [F H] because of (4) and [F h] because of (11). Leaves f H.

So that's the result:
Adams, Edwards, Gardiner and Harrison were elected to the city council.

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