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.

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.

Looks like i dun gets a possishun on yer English staff!