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# Help me solve this logic puzzle?

## Question

It was the last week of the baseball season, and in the Felidae Leage the Alleycats, the Bobcats, the Cougars, and the Domestics were all tied for the first place. It was decided that there would be a series of pla-off games in which each team would play one game with each of the other three teams; the team winning the most play-off games would win the pennant.

1.) The distribution of runs scored by each team during the play-off games was as follows (listed alphabetically according to the teams' home cities):

Sexton-City Team 1 - 3 - 7

Treble-City Team 1- 4 - 6

Ulster-City Team 2 - 3 - 6

Verdue-City Team 2 - 4 - 5

2.) Each team won a different number of play-off games.

3.) The score for each play-off game was different from that of any other play-off game

4.) The greatest difference in runs scored by two teams at any one game was 3 runs; this difference occurred only once when the team that lost the greatest number of play-off games lost by 3 runs.

5.) Two teams scored the same number of runs during the first round and two teams scored the same number of runs during the second round of the play-off games.

6.) During the last round the Alleycats scored the larger off number of runs, the Bobcats scored the smaller odd number of runs, the Cougars scored the larger even number of runs, and the Domestics scored the smaller even number of runs.

Which one of the four teams won the pennant? the Alleycats, the Bobcats, the Cougars, or the Domestics

Note: Only the information presented in this puzzle and logical induction skills will help you solve this puzzle!

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cougars

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im sorry im probably wrong

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I'm confused... What does 1st, 2nd, 3rd round of play-off games mean?

Is it (assuming teams are A,B,C,D)

First round : A v/s B, C v/s D

Second round : A v/s C, B v/s D

Third round : A v/s D, B v/s C

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^ I think so.

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doesn't the clause "every team won a different number of matches" imply that one team won all 3 three, another won 2 ,the next won 1 and the last won 0?

in that case the winner has to be either ulster or verdue city.but i cant find a schedule in which either of them win all three matches and the other clauses are satisfied

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Got it.

There must be 6 games - each team is paired with every other.

There are 4 teams, which we'll abbreviate S, T, U, and V. Each won a different number of games, so one must have won zero. Furthermore, this makes ties impossible; for teams to win 3, 2, and 1 games each, we need 6 victories, which means every game has a victory.

We'll focus first on figuring out who lost all of their games. By the rules, this is the only team with a 3-point loss. We will also figure out who won all of their games.

S: 1, 3, 7

T: 1, 4, 6

U: 2, 3, 6

V: 2, 4, 5

We can eliminate scores as we assign them to teams.

S did not lose all of their games. They had to have won at least 1.

T looks like a good candidate, but if we assume that they lost all games and try to assign a different number of victories to each team, respecting the "no matching games" and "only one 3-point loss and no greater" rules, we end up with two answers that both make it impossible to set up 2 rounds where 2 teams score equal numbers of points. So T did not lose all their games.

Could U have lost all 3 games? Maaaabye, but V looks more promising.

If we assume that V lost all 3 games, they had the 3-point loss. This would have to have been to S, 7-4. The two teams that scored 6 must them have scored 6-4 and 6-5, since 6-3 would have made a second 3-point win and 6-2 is even worse. In addition, by elimination, we now know that U won all of their games. Taken together, this gives us "T-U: 6-4, U wins", "T-V: 6-5: T wins", and "U-V: 3-2, U wins". Following this logic even further, we get "S-U: 2-1, U wins" and "S-T: 3-1, S wins."

This is promising! I never proved that U wasn't the team that lost all 3, but let's keep following this hunch for now.

SV: 7-4, S wins

TU: 6-4, U wins

TV: 6-5, T wins

UV: 3-2, U wins

SU: 2-1: U wins

ST: 3-1: S wins

Notice that we can set up the first two rounds as ST;UV and SV; TU and fulfill the "two teams with equal scores" rule. This leaves the final round as SU; TV.

Final round:

S scores: 1 (Bobcats)

U scores: 2 (Domestics) - VICTORS

T scores: 6 (Cougars)

V scores: 5 (Alleycats)

To check my work for errors:

Round 1 (or Round 2):

Sexton Bobcats vs. Treble Cougars: 3-1, Bobcat victory

Ulster Domestics vs. Verdue Alleycats: 3-2, Domestics victory

Round 2 (or Round 1):

Sexton Bobcats vs. Verdue Alleycats: 7-4, Bobcat victory, Alleycats trounced 3-1

Treble Cougars vs. Ulster Domestics: 6-4, Domestics victory

Round 3:

Sexton Bobcats vs. Ulster Domestics: 2-1, Domestics victory

Treble Cougars vs. Verdue Alleycats: 6-5, Cougar victory.

Great puzzle!

Edited by WitchOfDoubt
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Got it.

There must be 6 games - each team is paired with every other.

There are 4 teams, which we'll abbreviate S, T, U, and V. Each won a different number of games, so one must have won zero. Furthermore, this makes ties impossible; for teams to win 3, 2, and 1 games each, we need 6 victories, which means every game has a victory.

We'll focus first on figuring out who lost all of their games. By the rules, this is the only team with a 3-point loss. We will also figure out who won all of their games.

S: 1, 3, 7

T: 1, 4, 6

U: 2, 3, 6

V: 2, 4, 5

We can eliminate scores as we assign them to teams.

S did not lose all of their games. They had to have won at least 1.

T looks like a good candidate, but if we assume that they lost all games and try to assign a different number of victories to each team, respecting the "no matching games" and "only one 3-point loss and no greater" rules, we end up with two answers that both make it impossible to set up 2 rounds where 2 teams score equal numbers of points. So T did not lose all their games.

Could U have lost all 3 games? Maaaabye, but V looks more promising.

If we assume that V lost all 3 games, they had the 3-point loss. This would have to have been to S, 7-4. The two teams that scored 6 must them have scored 6-4 and 6-5, since 6-3 would have made a second 3-point win and 6-2 is even worse. In addition, by elimination, we now know that U won all of their games. Taken together, this gives us "T-U: 6-4, U wins", "T-V: 6-5: T wins", and "U-V: 3-2, U wins". Following this logic even further, we get "S-U: 2-1, U wins" and "S-T: 3-1, S wins."

This is promising! I never proved that U wasn't the team that lost all 3, but let's keep following this hunch for now.

SV: 7-4, S wins

TU: 6-4, U wins

TV: 6-5, T wins

UV: 3-2, U wins

SU: 2-1: U wins

ST: 3-1: S wins

Notice that we can set up the first two rounds as ST;UV and SV; TU and fulfill the "two teams with equal scores" rule. This leaves the final round as SU; TV.

Final round:

S scores: 1 (Bobcats)

U scores: 2 (Domestics) - VICTORS

T scores: 6 (Cougars)

V scores: 5 (Alleycats)

To check my work for errors:

Round 1 (or Round 2):

Sexton Bobcats vs. Treble Cougars: 3-1, Bobcat victory

Ulster Domestics vs. Verdue Alleycats: 3-2, Domestics victory

Round 2 (or Round 1):

Sexton Bobcats vs. Verdue Alleycats: 7-4, Bobcat victory, Alleycats trounced 3-1

Treble Cougars vs. Ulster Domestics: 6-4, Domestics victory

Round 3:

Sexton Bobcats vs. Ulster Domestics: 2-1, Domestics victory

Treble Cougars vs. Verdue Alleycats: 6-5, Cougar victory.

Great puzzle!

I came to this conclusion too, but 2 of the games have the same overall score (11) so it violates rule #3. I've been waiting to see others' solutions, but I think there is a typo in the OP or something

edit: nevermind... I just realized that 'score' meant both individual numbers... jeez I spent so long trying to figure out why it wasn't working out perfectly... lol

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dear witchofdoubt u r violating the clause 6, which indirectly implies that in the third round the team with the highest score must has an odd number of runs..................................any some sort of inconsistency is there in the puzzle.......................................

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There are seven possible combinations with a difference of 3. Using the rule that the loser lost all their games.

S7-T4 - Fails since S7 is the only one that can beat T6

S7-V4 - Fails since either T6 or U6 would have to beat V5 but that equals 11 too.

T6-U3 : V5-U3 or V4-U3 & S7-U6

T4-S1 - Fails since S can't lose all the games

U6-S3 - Fails since S can't lose all the games

V5-U2 - Fails since T4-U3 equals 7

V4-T1 : S7-T6 & U6-T4

Out of six games each team won 0-3 games. U or V has to have won 3 since S & T can't.

V4-T1 : S7-T6 & U6-T4

If U=3 then U3-V2, U2-S1 and V5-S3 - Fails since V4-T1 = U3-V2

If V=3 then V2-S1, V5-U3 and S3-U2 - Fails since V4-T1 = S3-U2

T6-U3 : V5-U3 or V4-U3 & S7-U6 (9, 8 or 7, 13)

V=3 since U is the loser

V2-T1, V4-S3 or V5-S3, T4-S1 - Fails since T4-S1 is a difference of three

V5-T4, V2-S1, S3-T1 - Fails since T6-U3 = V5-T4

Conclusion: There were tie(s).

Scoring combinations:

0 - 1 - 2 - 3 - Fails due to above

0 - 0.5 - 2.5 - 3

0.5 - 1 - 1.5 - 3

0.5 - 1 - 2 - 2.5

0.5 - 1 - 1.5 - 2

And back to the beginning...

Edited by curr3nt
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I'm not seeing any of the tie scenerios working either

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Curr3nt, I think itachi-san hit the right interpretation: "same score" does not mean "same total score", but "same pair of team scores". Then

there is no conflict between a 7-4 and a 6-5 game.

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