**5th Conundrum: January 2014: Ringing in the New Year**

Not only was midnight January first the start of a new year, but it also marked a moment when the hour, minute and second hands of a clock were all aligned. OK, the latter isn’t as interesting or special; unlike the first, it happens every day. This leads, of course, to some natural questions: what other interesting clock configurations happen each day?

`Suppose you have a standard analog clock with an hour hand, minute hand, and second hand. Given that the second hand moves continuously, is there a time at which the clock will be configured such that all hands make a 120 degree angle with each other (in other words, given any two of the three hands, they will be at an angle of 120 degrees)? ``If so, how many times and at what time will this occur each day? ``What if the second hand moves discretely, so that during each second it points precisely to the appropriate position on the clock face and jumps to the next position every second? `

Please email solutions to Professor Hanson (Edward.Hanson@williams.edu) or Professor Miller (sjm1@williams.edu).

**4th Conundrum: November 2013: Thanksgiving Challenge**

**As Thanksgiving is rapidly approaching, many turkeys are understandably worried. Several of them have gotten together and convinced humanity to accept the following challenge (rather than settling things with the sword).**

**The turkeys will create a polynomial P(x) such that, no matter what integer k the humans give them, the output P(k) will be an integer. If they can do this with one of the coefficients of P(x) being 1/2013 then ***no* turkeys will be eaten for the rest of 2013.

**Can the turkeys succeed? More generally, if you give them finitely many years (say n+1 years) can they create a similar polynomial which has 1/2013, 1/2014, …, 1/(2013+n) as coefficients?**

**2nd Conundrum: October 2013: We Value Your Input**

Alice and Bob are playing the following game: Alice has a secret polynomial P(x) = a_0 + a_1 x + a_2 x^2 + … + a_n x^n, with non-negative integer coefficients a_0, a_1, …, a_n. At each turn, Bob picks an integer k and Alice tells Bob the value of P(k). Find, as a function of the degree n, the minimum number of turns Bob needs to completely determine Alice’s polynomial P(x). *Hint: the answer is much smaller than you might believe!*

Send solutions to Professor Miller (sjm1@williams.edu) or Stoiciu (Mihai.Stoiciu@williams.edu) or Biro (Michael.J.Biro@williams.edu).

**As Thanksgiving is rapidly approaching, many turkeys are understandably worried. Several of them have gotten together and convinced humanity to accept the following challenge (rather than settling things with the sword):**

**The turkeys will create a polynomial P(x) such that, no matter what integer k the humans give them, the output P(k) will be an integer. If they can do this with one of the coefficients of P(x) being 1/2013 then ***no* turkeys will be eaten for the rest of 2013.

**Can the turkeys succeed? More generally, if you give them finitely many years (say n+1 years) can they create a similar polynomial which has 1/2013, 1/2014, …, 1/(2013+n) as coefficients?**