Conundrum: The English Department
The English department at the local university has nine professors. Each has been with the department a different number of years, ranging from the new-hire (zero years), all the way up to the chair who has been with the department for fifteen years. Since the university only hires at the beginning of the school year, the number of years that each person has been with the department can be expressed as a whole number.
This morning, the nine professors divided themselves into three committees and each of these committees held a meeting which lasted all morning. In the afternoon, the nine professors divided themselves into three different committees and each of these committees held a meeting which lasted all afternoon. At no point today did anybody meet with anyone outside of these six committees.
1. Irene met with Adam and Dr. Marshall in the morning, and met with Deborah and Dr. Smith in the afternoon. Both meetings were held in Conference Room A.
2. Dr. Osborne met with Charles and Dr. Kaplan in the morning, and met with Gerald and Dr. Lewis in the afternoon. Both meetings were held in Conference Room B.
3. Dr. Johnson met with Frank and Dr. Rogers in the morning, and met with Elizabeth and Dr. Nelson in the afternoon. Both meetings were held in Conference Room C.
4. Each of the six committees has the exact same combined number of years that the three committee members have been with the department, though no two of the committees are identical.
5. Harold has been with the department longer than Barbara has.
6. After the Shakespeare scholar, who has been with the department exactly four times as many years as Irene has, was hired, nobody else was hired until five years later, when the Romantic poetry expert joined the department.
7. Dr. Kaplan was hired one year before Dr. Peterson and one year after Dr. Lewis. Nobody was hired the year before Dr. Lewis. Nobody mentioned anywhere above has left the department.
The department is currently hiring for a tenure-track position for next year. They offer a competitive salary and an impressive benefits package. To apply for a position, determine the full names of all nine professors, and how many years each has been with the department.
UPDATE: Puzzle solved by ArtVark. See comments for answer.
January 15th, 2008 at 6:38 pm
Gerald Rogers — 0 years
Charles Nelson — 1 year
Irene Peterson — 3 years
Deborah Kaplan — 4 years
Adam Lewis — 5 years
Barbara Johnson — 7 years
Elizabeth Marshall — 12 years
Frank Smith — 13 years
Harold Osborne — 15 years
Looks like i dun gets a possishun on yer English staff!
January 15th, 2008 at 8:54 pm
Explanation:
Someone has been there 15 years (longest). Someone has been there 0 years (shortest). We need to figure out how long the other seven profs have been there.
Someone has been with the department four times as many years as Irene. Since the maximum stay has been 15 years, Irene can not have been there longer than 3 years. If Irene had been there 0 or 1 year, it would be impossible to have no one hired for five years after the Shakespeare Prof. Therefore Irene has to have been there 2 or 3 years.
Irene is not Marshall or Smith because she met with those professors. Irene is not Osborne Kaplan, Johnson, Rogers, Lewis, or Nelson because they met in different rooms from Irene at some point. By process of elimination Irene is Prof. Peterson.
Kaplan was there one year longer than Peterson and Lewis one year longer than that. If Irene Peterson had been there two years, then Lewis would have been there four years which would make less than a five year gap between the hiring of the person with four times the length of service as Irene (8 years) and the next hire.
Therefore Irene Peterson was there 3 years, Kaplan 4, and Lewis 5.
The Shakespeare expert has to now have 12 years service, and the Romantic Poetry person 7.
So seven of the nine possible service lengths are 0, 3, 4, 5, 7, 12, and 15. Eight through eleven can’t happen (five period of not hiring), and 6 cannot happen (No one was hired the year before Lewis). The remaining possible values are 1, 2, 13, 14.
Now since every meeting had the same sum of service values, the three meetings at one time has to consist of three different sums of service that add up to the same number. Therefore the sum of everyones’ service must add up to a number evenly divisible by 3.
0+3+4+5+7+12+15 = 46. 46 plus the other two years of service (out of 1,2,13, and 14) must add up to a number divisible by 3. This can only happen if the other two are 1 and 13 (for a total of 60). Therefore the lengths of service are 0,1,3,4,5,7,12,13,15, and the sum of service in each meeting is 20.
So room A has 3 years from Irene for both the morning and afternoon sessions. There must be two other pairs of lengths of service that add up to 17. The only two possible ones are 4 and 13 and 5 and 12. In the morning Kaplan with 4 years of service was in room B. Therefore, the years of service for the other two profs in room A in the morning must be 5 and 12. This means that Prof Lewis met in room a in the morning. By process of elimination, this must be Adam. Marshall must therefore have 12 years of service.
Now room A in the afternoon must have the profs with 4 and 13 years service. This makes Deborah Professor Kaplan (by process of elimination), and gives Prof Smith 13 years of service.
Now room B in the morning needs 16 years of service in the morning for the two profs other than Kaplan, and 15 years of service in the afternoon other than Lewis. This can only happen if Osborne had 15 years service, Charles 1 year, and Gerard 0 years. Charles had meetings with Osborne and Kaplan, and was in different rooms from Johnson, Rogers, Lewis, Marshall and Peterson. Smith has 13 year service. By process of elimination, Charles must be Nelson. Gerard had meetings with Osborne and Lewis, and was in a different room from Peterson, Kaplan, Smith, Johnson, and Nelson. Marshall has 12 years of service. By process of elimination, Gerard is Professor Rogers.
The years of service for Rooms A and B have now been completely figured, and no 7 year prof has yet to appear. Therefore the person with 7 years service must be Johnson who only met in room C. The remaining two people in each session in room C must have years of service that add up to 13. The only way that can happen with all the data provided so far is if the profs. with 13 and 0 years service were in room C in the morning and the profs with 12 and 1 year service were there in the afternoon. This means that Elizabeth must be Prof Mashall, and Frank must be Prof Smith.
The only thing that now remains unfilled are the first names of Osborne and Johnson. Since Harold and Barbara are the only two remaining first names not used, and since Harold has been around longer than Barbara, Harold must be Osborne (with 15 years service) and Barbara must be Johnson.
January 15th, 2008 at 10:00 pm
Artvark, very well done! Your answer is entirely correct, and your explanation is both elegant and solid. In fact, your method of solving is exactly the one I had intended.
I expected the most difficult part to be the “must be divisible by three” piece, which requires more out-of-the-box thinking than your typical logic problem.