Conundrums

2011:12: December Conundrum: Stuffing the Trunk

It’s that time of year again!  You’ve blown your savings on the holiday shopping, having purchased a total of 7 gifts for the relatives who will make an appearance during the winter break.  You’ve made it through the exam and final paper push–all that remains is to pack the car and hit the road for

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2010:10: October Conundrum: Calling the Great Pumpkin

Each year, the Great Pumpkin rises out of the pumpkin patch that he thinks is the most sincere.  Well, what could be more sincere than a community garden?  Your entry has planted a community pumpkin patch, and each of you is responsible for caring for all of the pumpkins in some rectangular portion of the

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2010:09: September Conundrum: Panting at the Purple Key Fair

It’s the beginning of a new year, and you’re feeling optimistic:  “This year, I’m going to manage my time well and do all of the extracurricular activities that I want to, in addition to the coursework for my Math/Stat, Comparative Literature, and Japanese triple major!”  In your moment of naivety, you and a fellow cross

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2010:03: March Conundrum!

Imagine you have pennies on each of the spots above and you want to move all the pennies to the spots below the line using only checkers moves (jumps over an adjacent penny to the next spot — left or right and up or down, but not diagonal).  Can this be done?  If a penny

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2010:10: February Conundrum!

Define a sequence {a_n} as follows: a_0 = 2,  a_1 = 3,  a_2 = 6 and, for n ≥ 3, a_n  =  (4+n) a_{n-1} – 4n a_{n-2} + (4n -8) a_{n-3}. Thus, the first several values are 2, 3, 6, 14, 40, 152, 784, and 5168. Find and prove a formula of the form  a_n =

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2010:01: Winter Study Conundrum

Let n  be a natural number greater than 1 and suppose n has k distinct prime factors.  Please prove that  log(n)  ≥  k  log(2). Solution. Congrats to Sean Pegado, the first to submit the following correct solution: If n = p_1^e_1 * p_2^e_2 * …. * p^e_k, we know that the smallest possible value for p_i will

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2009:12: December Conundrum!

The philosophy of yin and yang, much like Santa, divides life into two dual halves:  naughty and nice, dark and light, low and high, etc.  The problem here is to draw one straight line through the yin yang symbol to split the yin and yang each into equal areas… and prove they’re equal!  Good luck

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2009:11: Special Thanksgiving Conundrum with Prize!

News flash: correct solutions (9 edges) were received by the deadline from Donny Huang, Kristen Layden, David Moore, Ralph Morrison, Wei Sun, and Sharon Ron’s brother, all eligible for the prize drawing at colloquium Monday, November 30, 1 pm. Morrison solution. The Turkey prize gift basket went to Donny Huang, presented by Jo Proctor for the Office of

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2009:09: September Conundrum.

The conundrum was to prove that given five points on a sphere you can always find a closed hemisphere that contains four of them.    Thanks to all of you who sent in great solutions!  The first person to send one in was Daniel Phelps whose solution is published below.  This is his third correct

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