MP51: May 2005
If the sequence a_{0}, a_{1}, a_{2}, ... satisfies
(1) a_{1} = 1, and
(2) for all integers m and n with m ≥ n ≥ 0, a_{2m} + a_{2n} = 2(a_{m+n} + a_{m–n}),
determine a_{2005}.
This problem appeared in the team competition sponsored by the Math Association of America's North Central Section, held at Concordia College (Minnesota) November 10, 2001.
MP50: April 2005
The lengths of consecutive sides of a plane quadrangle are 6, 33, 47, and 34. What is the angle between the two diagonals?
(Although these side lengths might look as if they have been chosen at random, they weren't.)
MP49: March 2005
If the real numbers x and y satisfy
find their sum x + y.
We thank Andy Liu of the University of Alberta for this month's problem.
MP48: February 2005
Doublezapping
This month we show how television can be used for educational purposes. Suppose that you have two televisions side by side, Television A and Television B, that are activated by separate remote controls. We call doublezapping the act of simultaneously pressing the channel up or channel down button of A's remote and the channel up or channel down button of B's remote. For instance, from (Channel 8, Channel 6) you can doublezap to any one of (Channel 7, Channel 5), (Channel 7, Channel 7), (Channel 9, Channel 5) or (Channel 9, Channel 7).
For this month's problem you actually have three televisions side by side, each activated by its own (distinct) remote and connected to different cable companies:
Television A is connected to cable company Alpha which has 70 channels A1 to A70,
television B is connected to cable company Beta which has 60 channels B1 to B60, and
television C is connected to cable company Gamma which has 94 channels C1 to C94.
However, you find that there are very few different networks, each of which is duplicated over and over; moreover the networks are exactly the same on all three cable company; in particular, channel A1 is the same as channel B1 and channel C1, and channel A70 is the same as channel B60 and C94. Eventually you find out that it is possible to start from (A1, B1) and doublezap all the way to (A70, B60) in such a way that the program displayed on television A is always the same as the program displayed on television B. Similarly, it is possible to start from (B1, C1) and doublezap all the way to (B60, C94) in such a way that the program displayed on television B is always the same as the program displayed on television C.
The February problem: In these circumstances is it necessarily possible to doublezap from (A1, C1) to (A70, C94) in such a way that the program displayed on television A is always the same as the program displayed on television C.
WARNING: Do not attempt to push the channel down button when a television is on channel 1; similarly, do not attempt to push the channel up button when a television is on the top channel. This would cause the corresponding television to explode, and might void your insurance policy.
MP47: January 2005
As a warmup exercise you can prove
Every way you line up the numbers 0, 1, 2, ..., 9, the resulting 10digit number is a multiple of 3.
It does not matter whether or not we allow the leading digit to be zero – technically the resulting number would only have 9 digits, but it would still be a multiple of 3, which is the important point. There is no need for you to send us the proof – we already have similar results on the Math Central web page; see, for example,
Our problem this month is the version that would be natural for a citizen of the planet Kuku, where the inhabitants have five hand of two million digits each.
Let us build a 70million digit number N by stringing together, in any order, all 107 possible arrangements of 7 digits (using the numbers from 0 to 9 as our digits). In other words, each consecutive 7digit block of N is one of the arrangements from the list 0000000, 0000001, 0000002, ..., 9999999. (There are 10million factorial such numbers N, if we allow them to begin with initial zeros.)
The problem for January: Prove that every one of these very many very big numbers N is divisible by 239.
MP46: December 2004
The Toque Game (for winter months).
Alice and her two friends Bob and Chris are selected to play the "Toque game": They will be seated so that each can see the others. Each will be given a bag containing two toques, one with a white pompom and one with a red pompom. They will then be blindfolded and each will pick a toque at random out of the bag and put it on. When the blindfolds are removed, the players will be able to see the pompom on the other's toque, but not their own. They will not be allowed to communicate by gestures or shouting, and they will have to whisper to a referee standing next to them either "I guess my pompom is white", "I guess my pompom is red", or "I pass". If at least one of them guesses right and nobody guesses wrong, they all win a trip to Moose Jaw, Saskatchewan. Otherwise they just get popcorn. Alice and her friends can devise a strategy before the game to give them better odds. For instance, they could decide that Alice alone will guess and the others will pass; such a strategy gives them a fifty percent chance of winning. Is there a way to do better?
MP45: November 2004
For this month’s problem we travel to the world’s first mathematics museum, theMathematikum in Giessen, Germany. One exhibit there displays 7 lights evenly spaced about a circle. In front of each is a button that, when pushed, changes the state of its light and that of its two neighbors – a push of the button turns off any of the three that are on, and turns on any of the three that is off. If we begin with all lights switched off, what is the least number of buttons to push that will turn them all on? In what order should they be pushed?
Of course, it is natural to ask what happens to your answer when 7 is replaced by n: you have n lights around the table and n buttons to push. What is the least number of buttons that must be pushed to change from n lights al off to n lights all on? Describe how to do it, and provide a convincing argument why nobody can do it in fewer steps.
MP44: October 2004
In Ira Hauptmann's play Partition, which is playing to sellout crowds around the world, Ramanujan tells Hardy that 153 is indeed an interesting number.
"It equals the sum of the cubes of its digits:
153 = 1^{3} + 5^{3} + 3^{3}.
And I know that there are, in all, just five positive base10 numbers
that equal the sum of the cubes of their digits."
that equal the sum of the cubes of their digits."
Show that Ramanujan is correct.
MP43: September 2004
The unit circle is divided into twelve equal parts, and the twelve dividing points are joined to the circle's centre, producing twelve rays. Starting from one of the dividing points a segment is drawn perpendicular to the next ray in the clockwise sense; from the foot of this perpendicular another perpendicular segment is drawn to the next ray, and so on. What is the limit of the sum of the lengths of these segments?

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