Trick question. Within any calendar year, Christmas and New Year’s Day always fall on different days:

If the prisoners knew that both switches were off at the start, then the solution would be fairly straightforward. One switch (say, the left one) will be used for counting, and the other is a meaningless dummy. One prisoner is chosen as the Counter. Each other prisoner is instructed to flip the left switch from off to on one time, at the first opportunity, but otherwise to flip the dummy switch on each visit. When the Counter enters the room and finds the left switch on, he turns it off; otherwise he too flips the dummy switch. When the Counter has found the left switch turned on 22 times, he knows that all prisoners have visited the room and can safely tell the warden so.

Because the prisoners don’t know the initial states of the switches, this plan must be enhanced (otherwise, if the left switch starts in the on position, then the Counter will make his announcement prematurely). The solution is to ask each prisoner to flip the left switch from off to on twice in total, rather than once. The Counter will wait until he’s seen the left switch turned on 44 times altogether, and then tell the warden that all prisoners have visited the room.

(At first I wondered why they couldn’t just stick with the first scenario but add 1 to the expected count to allow for the possibility that the left switch had started in the on position. But this won’t work: If the Counter is waiting until he’s found the left switch turned on 23 times, and if it started in the off position, then he’ll wait forever, because each of the 22 other prisoners will turn it on only once.)

1. Ng5!, and the other knight will discover mate.

From O.A. Brownson Jr., Chess Problems, 1876.

Rotating the figure 60° counterclockwise around B carries A to R and P to C. And rotating it 60° clockwise around A carries B to R and Q to C. So AP = BQ = CR.

From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.

1. Ra4+! Bxa4 2. b4#

From Theophilus A. Thompson, Chess Problems, 1873.

No. When viewed from above the three pucks form triangle ABC. With each hit the orientation of the triangle changes, so that sometimes “ABC” reads clockwise, sometimes counterclockwise. The original position was reached after zero hits, so it can be restored only after an even number of hits.

From Arthur Engel, Problem-Solving Strategies, 1998.

I don’t have the olympiad solution, but it seems to me that there must be an equal number of knights and scoundrels. The first half of the interviewees are knights, and the second half are scoundrels.

If there were more than n/2 knights, then they’d crowd out the scoundrels and their greater claims could not be true.

If there were more than n/2 scoundrels, then they’d crowd out the knights and their lesser claims could not be false.

UPDATE: Also, n must be even. If it’s odd, then the middle interviewee can be neither a knight nor a scoundrel, which is impossible. Thanks to those who wrote in to point this out.

POCO is 4595 and MUCHO is 68925.

There are 15 POCOs, so the sum of each column is a multiple of 15, and hence (without the carry from the column to its right) ends in a 5 or a 0. The rightmost column has no carry, so the O in MUCHO stands for 0 or 5. Which is it?

• If it’s 0, that means that the sum of all the Os in the rightmost column is 0, and that there’s no carry into the tens column. If there’s no carry, we need to find a digit for C that when multiplied by 15 produces a two-digit answer (the CH in MUCHO) that starts with that digit (since the hundreds column in the sum is otherwise zero). The only digit that will do that is 1 (1 × 15 = 15). But if CH is 15 then there’s no carry into the thousands column, which means that U must be 0 or 5 (15 × P). And we’ve already assigned 0 to O and 5 to H. So this can’t be right.
• If it’s 5, then the carry into the tens column is 7 (15 × 5 = 75). What is C? Every possibility from 0 to 8 produces conflicts or requires C to have two different values, but C = 9 produces a sum in the tens column of 142 (15 × 9 plus the 7 carry). In that case H = 2 and 14 is carried into the hundreds column. We know that O is 5, so that column now totals 89 (5 × 15 + 14). C is 9 and 8 is carried into the thousands column, where it’s added to 15 Ps. Here every possibility except P = 4 creates duplications or conflicts, so U is 8 and M is 6.

“Because O must equal 5 and C must equal 9 and P must equal 4, I believe the solution is unique,” writes NYU computer scientist Allan Gottlieb, who conducts the puzzle column. “I like these alphabetical number problems; but, cannot recall seeing this one before.”

1. Bc8! Kg4 2. Re4#

From Benjamin Glover Laws, The Two-Move Chess Problem, 1890.

Consider all the possible divisions, and in each case count up the total number of cases E in which a member has an enemy in his own house. The division in which E is minimized must have the required property. If it didn’t, that would mean that some member has at least two enemies in his own house. In that case he’d have one enemy at most in the other house, and could reduce E further by moving to the opposite house. That’s a contradiction, so the required task must be possible.