Puzzles

Published on May 21st, 2013 | by Simi

0

AustMS Gazette Puzzle Corner 32

Welcome to the Australian Mathematical Society Gazette’s Puzzle Corner number 32. Each puzzle corner includes a handful of fun, yet intriguing, puzzles for adventurous readers to try. They cover a range of difficulties, come from a variety of topics, and require a minimum of mathematical prerequisites for their solution. Should you happen to be ingenious enough to solve one of them, then you should send your solution to us.

For each puzzle corner, the reader with the best submission will receive a book voucher, not to mention fame, glory and unlimited bragging rights! Entries are judged on the following criteria, in decreasing order of importance: accuracy, elegance, difficulty, and the number of correct solutions submitted. Please note that the judge’s decision — that is, my decision — is absolutely final. Please email solutions to ivanguo1986@gmail.com

The deadline for submission of solutions for Puzzle Corner 32 is 1 July 2013. The solutions to Puzzle Corner 32 will appear in Puzzle Corner 34 in the September 2013 issue of the Gazette.

[latexpage]

Telescoping product

Let $n$ be an integer greater than 1. Simplify

$\frac{2^3 − 1}{2^3 + 1}$ x $\frac{3^3 − 1}{3^3 + 1}$ · · · ×$\frac{n^3 − 1}{n^3 + 1}$

Tangent intersections

Let $\Gamma$1 and $\Gamma$2 be two non-overlapping circles with centres $O$1 and $O$2 respectively. From $O$1, draw the two tangents to $\Gamma$2 and let them intersect $\Gamma$1 at points $A$ and $B$.

Similarly, from $O$2, draw the two tangents to $\Gamma$1 and let them intersect $\Gamma$2 points by $C$ and $D$.

Prove that $AB$ = $CD$


Screen Shot 2013-05-21 at 8.38.29 PM

 

Colour coordination
Submitted by Joe Kupka

I need to hang 20 garments on a clothes line. Each garment requires two pegs. I have 20 green and 20 red pegs. I choose pegs at random. On average, how many garments will have pegs of the same colour?

 

Team tactics 2

In a game show, there are three girls, each wearing a blue or a red hat. Each girl can only see the hats of the other two but not her own. Without any communi- cation between themselves, each girl has to choose a real number and whisper it to the host. At the end, the host will add up the numbers chosen by girls wearing red hats, then subtract the numbers chosen by girls wearing blue hats. The girls win if the final answer is positive.

Before the show, the girls try to devise a strategy to maximise their probability of winning.

(i) What is the maximum probability of winning?
(ii) If the girls were only allowed to choose from {−1, 0, 1}, what is the maximum probability of winning?

Bonus: If there are seven girls instead of three, and each girl can see the hats of the other six but not her own, how do the answers change?

A sequence of sequences

Let $S_1$, $S_2$, . . . be finite sequences of positive integers defined in the following way.

Set $S_1$ = (1). For $n$ > 1, if $S_{n-1}$ = ($x_1$,…,$x_m$) then

$S_n$ = (1,2,…,$x_1$,1,2,…,$x_2$,…,1,2,…,$x_m$,$n$).

For example, the next few sequences are $S_2$ = (1, 2), $S_3$ = (1, 1, 2, 3) and $S_4$ = (1,1,1,2,1,2,3,4).

Prove that in the sequence $S_n$ where $n$ > 1, the kth term from the left is 1 if and only if the $k$th term from the right is not 1. (Hint: Pascal’s triangle.)

 

Screen Shot 2013-02-20 at 11.42.39 AM
Ivan Guo
Ivan is a PhD student in the School of Mathematics and Statis- tics at The University of Sydney. His current research involves a mixture of multi-person game theory and option pricing. Ivan spends much of his spare time playing with puzzles of all flavours, as well as Olympiad Mathematics.

The Puzzle Corner is a regular feature in the AustMS Gazette, puzzles are created by Ivan Guo unless otherwise stated. 

 AustMS

[subscribe2]




Comments are closed.

Back to Top ↑