MP51: May 2005
If the sequence a0, a1, a2,
... satisfies
(1) a1 =
1, and
(2) for
all integers m and n with m ≥ n ≥ 0,
a2m + a2n = 2(am+n + am–n),
determine a2005.
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.
Solution to May
2005 problem
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.)
Solution to April
2005 problem
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.
Solution to March
2005 problem
MP48: February
2005
Double-zapping
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 double-zapping 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 double-zap 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 double-zap 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 double-zap 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 double-zap 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.
Solution to February
2005 problem
MP47: January
2005
As a warm-up exercise you can prove
Every way you line up the numbers 0, 1, 2, ..., 9, the
resulting 10-digit 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,
http://mathcentral.uregina.ca/QQ/database/QQ.09.00/kelera1.html.
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 70-million 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
7-digit block of N is one of the arrangements from
the list 0000000, 0000001, 0000002, ..., 9999999. (There are 10-million
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.
Solution to January
2005 problem
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, the Mathematikum 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.
Solution to November
2004 problem
MP44: October 2004
In Ira Hauptmann's play Partition, which is playing
to sell-out 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 = 13 + 53 + 33.
And I know that there are, in all, just five positive base-10 numbers
that equal the sum of the cubes of their digits."
Show that Ramanujan is correct.
Solution to October
2004 problem
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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