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Mystery Master | Analysis of Dandy Salespeople | Member |
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Mystery Master | Analysis of Dandy Salespeople | Member |
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This five-star logic puzzle is a hard puzzle to solve. It has 6 noun types, 12 nouns per type, 4 links, 265 facts and 5 rules. And to make this puzzle even tougher, not all of the ages are given, so you have to calculate them! The program does this via rules. Very tough indeed. It needs 2160 marks and 15 grids.
Because all of the ages are not known, I use placeholders named A01 to A12, where A01 is with January, A02 is with February, and so on. The four ages I do know are: 20, 50, 46, 26. When the rules can determine one of these ages, the appropriate placeholder will be updated.
Here are the clues that involve age. For clues 3, 4, 5, 7, 12, and 16, I changed the relationship so the first noun is younger than the second noun.
If the puzzle was solved without rule 6, and with no assumptions, the program finds 2120/2160 marks, 165/180 pairs, and 8/12 ages calculated: 25, 50, 24, 46, 23, 26, 52, 48. So one of the placeholders must be 20 (either Feb, Apr, Jul, Oct, or Nov). The one fact still remaining is:
The man from Torrance won in August or November, while Janice won in December. Since both August and November are before December, we can say this fact has done its job.
The ages we still need to determine are from clues 5, 12, and 15. From what we know so far, the 20 year old won in either February, July, October, or November.
Clue 3 states: "Carmen Vasquez is twice as old as the woman from the Lomita lot". Since Carmen is 46, the woman from Lomita is 23. Check. Note that Carmen's age must be an even number (divisible by 2).
Clue 4 states: "The man who won the Caprice is half as old as the woman from Maywood". The Caprice winner is 25, and the woman from Maywood is 50. Check. Note that the Maywood winner's age must be an even number.
Clue 4 also states: "The woman from Maywood is two years older than the woman from La Puente." The woman from Maywood is 50, and the woman from La Puente is 48. Check.
Clue 5 states: "The February winner, who works in the Downey lot, is twice as old as Prince." Prince won in July, so the age to calculate is A07. Note that the February winner's age must be an even number, so A02 must be even.
Clue 5 also states: "The February winner is three times as old as Stuart (who won the Accord)." Stuart won the Accord in November. Both ages, A02 and A11, need to be calculated. Note that the February winner's age must be divisible by 3, so A02 must as well. This means the February winner's age is not 20.
Clue 6 states: "The man from Torrance is not 20 years old." The rules ensure this.
Clue 7 states: "The Hollywood winner is older than the Whittier winner." Both Hollywood and Whittier may be in Jan, Aug, or Nov. The rules ensure this.
Clue 8 states: "The 50-year-old won the Previa." Check.
Clue 12 states: "The winner of the Skylark is twice as old as the man who won in November." The Skylark was in October. Both ages, A10 and A11, need to be calculated. Note that the Skylark winner's age must be an even number, so A10 must be even.
Clue 15 states: "Mr. Jackson is ten years younger than Paulette." Mr Jackson won in November, while Paulette won in July. Both ages, A07 and A11, need to be calculated.
Clue 15 also states: "Paulette is half as old as Ms. Harper." Paulette won in July, while Ms. Harper won in February. Both ages, A02 and A07, need to be calculated. Note that Ms. Harper's age must be an even number.
Clue 16 states: "Sanders is twice as old as Straw." Since Straw is 26, Sanders must be 52. Check. Note that Sanders' age must be an even number.
Clue 17 states: "The Inglewood salesperson won the month before the 46-year-old." Since the Inglewood salesperson won in April, the May winner must be 46. Check.
Clue 18 states: "The woman from Bell won the month before the 26-year-old." Since the woman from Bell won in July, the August winner must be 26. Check.
Clue 20 states: "Millie is half as old as the woman who works in Gardena." Millie is 23 and the woman in Gardena is 46. Check. Note that the age of the woman from Gardena must be an even number.
Clue 22 states: "Bruce is half as old as Ms. Swan." Bruce won in April and is 24, while Ms. Swan won in December and is 48. Check. Note that Ms. Swan's age must be an even number.
Here are the equations we need to solve where the unknowns are A02, A07, A10, A11. One of these must be 20, but keep in mind the man from Torrance is not 20. (clue 6).
From the first two expressions, we see that A02 > A07 > A11. So where does A10 fit in? From the first three expression, A02 > A10 > A11. A07 is half of A02. Since A10 = 2 * A11 = 2 * A02 / 3, so A10 is two thirds of A02. Therefore A02 > A10 > A07 > A11.
Since A02 must be divisible by 6, A02 is not 20.
If A10 is 20, then A11 is 10, A02 is 30, A07 is 15, but this does not satisfy (4). So A10 is not 20.
If A07 is 20, A02 is 40, but A02 must be divisible by 6. So A07 is not 20.
If A11 is 20, A02 is 60, A07 is 30, and A10 is 40. This satisfies 60 > 40 > 30 > 20. Therefore, A11 is 20.
If I did not know that one of the ages was 20, I would need matrix math to solve 4 equations with 4 unknowns. Because I do know, I can invoke a rule when the nouns in clues 5, 12, and 15 are all paired with an age. That rule is rule 6.
Rule 6 examines the nouns in clues 5, 12, and 15 to see if they are all paired with an age. When that happens, it sets the age of the November winner to 20, and enforces this part of clue 6: "The man from Torrance is not 20 years old." By enforcing rule 6, the puzzle is solved using level 1a logic.
This is one of the hardest puzzles you will ever encounter. Don't let the numbers discourage you.