ERJ Brainteaser: May

Question 4: Colour clash

In a game of two dice with just green or red faces instead of numbers, Arthur wins if he throws and gets two top faces of the same colour. On his throw, Sean wins when the colours are different. They each have an equal chance of winning. If one die has five green faces and 1 red face, how many green faces and how many red faces are there on the other?

Answer: One sign of a good teaser is getting a range of neat but different answers (see below) all leading to the same result: the second dice would have 3 red and 3 blues faces. Very well done, in order of reply,  to: Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Andrew Knox, Rubbond International, The Netherlands; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; and everyone else who had a go.

Solutions

Amparo Botella

Each die has 6 faces. When two dice are thrown, there are 36 equally possible results. For chances to be even, there must be 18 ways of getting the same color on top. Let X be the number of red faces on the second die. We have: 18 = 5X + 1(6 - X)

X = 3

The second die must have 3 red faces and 3 blue faces.

Andrew Knox

Surprisingly, the answer is that the second die has to have 3 green faces and 3 red faces.

There are then four possible outcomes:

Sean:

Different colours: 1st green, 2nd red: Chance is 5/6 x 3/6 = 15/36

Different colours: 1st red, 2nd green: Chance is 1/6 x 3/6 = 3/36

Arthur:

Same colours: 1st green, 2nd green: Chance is 5/6 x 3/6 = 15/36

Same coulours: 1st red, 2nd red: Chance is 1/6 x 3/6 = 3/36

In this way, both have equal chances of winning (18/36).

Michele Girardi

The solution is 3 red and 3 green faces .

Let's call x the number of green faces on the second dice . The probabilities are  :

1            2                              ;

G    R    G     R                      ;    

5/6 1/6  x/6  (6-x)/6                ;  

GG+RR     : 5/6*x/6 + 1/6*(6-x)/6                ;

RG+GR    :  1/6*x/6 + 5/6*(6-x)/6                ;

putting GG+RR = RG+GR   we get              :

5/6*x/6 + 1/6*(6-x)/6 = 1/6*x/6 + 5/6*(6-x)/6       ;

5x+6-x = x + 30 -5x                  ;

8x = 24       x = 3                          ;

Stephan Paischer

The second die has 3 green and 3 red faces.

Probability for same colour:

(5/6 * x/6) + (1/6 * (6-x)/6)

Probability for different colour:

(5/6 * (6-x)/6) + (1/6 * x/6)

Where for dice 1:

Green = 5/6

Red = 1/6

And dice 2:

Green = x/6

Red = (6-x)/6

Both probabilities are equal, so you can set up an equation of the two parts.

Solving the equation gives x=3.

John Bowen

A bit of nifty algebra for this week, plus work so apologies for late submission.

There are 3 Green faces

Calculation:

Die 1 has 1 green and 5 red faces

Die 2 is unknown, so let's say number of green faces = G, so number of red faces = [6-G]

Arthur needs [Green + Green] or [Red + Red]

The probability p of which is given by [5/6 x G/6] + [1/6 x [6-G]/6]

Sean needs [Green + Red] or [Red + Green]

The probability of which is given by [5/6 x [6-G]/6] + [1/6 x G/6]

These probabilities are equal so we can simplify to : 

5G/36 +[6-G]/36 = [30-5G]/36 + G/36

Multiply both sides by 36 and rearrange to give 8G = 24 so G = 3

Question 3: Reverse engineering

Apply the same mathematical function to each of the following numbers to get a sequence that reads the same way backwards as it does forward:

31, 3, 5, 11, 23, 13

Answer: This teaser generated some interesting mathematical contortions, alongside the official answer: to square each number (see below). Very well done, in order of reply to: Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, The Netherlands; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Mehmet Koral, Erhardt-Leimer representative for Turkey, managing director, C&C Endüstriyel Dan??manl?k, E?itim ve Mümessillik Ltd, Göztepe, Istanbul, Turkey; and everyone else who had a go.

Solutions

Amparo Botella

3135112313+3000= 3135115313

John Bowen

If each term is squared,then written down, it will read the same both ways:

961, 9, 25, 121, 529, 169

Andrew Knox

Answer: Square each number to get: 

961, 9, 25, 121, 529, 169 

Stephan Paischer 

Each number needs to be squared in order to read the same way backwards and forward: 

961, 9, 25, 121, 529, 169

Mehmet Koral

My approach is related to the Palindromic numbers, so we have a set of 1 digit, 2 digit palindromic numbers in the series given.

All numbers in base 10 (and indeed in any base) with one digit are palindromic, so there are ten decimal palindromic numbers with one digit:

{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

There are 9 palindromic numbers with two digits:

{11, 22, 33, 44, 55, 66, 77, 88, 99}.

A better final solution I developed now is as follows;

Given. : 31, 3, 5, 11, 23, 13

Answer : 31, 23, 5, 3,11, 3, 5, 32, 13

Michele Girardi

Since all these numbers are odd, if you apply the function  "remainder of the division by 2"  you'll get the sequence  1,1,1,1,1,1, which reads the same way backwards as it does forward .

The solution may also be the function

Y = 7,269 - 2,559 X + 0,3126 X*X - 0,01614 X*X*X + 0,000382 X*X*X*X - 0,000003 X*X*X*X*X

that transforms the numbers in the sequence  1,2,0.5, 0.5,2,1 .

That can be derived as follows :

let's put

f(31) = f(13) = a

f(3) = f(23) = b

f(5) = f(11) = c

with the help of excel :

-  let's choose a,b,c so that the data can be interpolated from a nice curve , for example 1, 2, 0.5  

- insert a trendline choosing a polynomial of grade 5, that will fit exactly the data

Question 2: Romanesque

Using only a single ‘line’, how can you change the Roman numeral IX (9) into the number 6?

Answer: Very well done to the following readers who identified that putting S (a curved line) in front of IX gives the word SIX: Michael Easton, director, Globus Group, Trafford Park, Manchester, UK; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, The Netherlands; Jose Padron, laboratory analyst, Waterville TG Inc., Waterville, Québec, Canada; Mehmet Koral, Erhardt-Leimer representative for Turkey, managing director, C&C Endüstriyel Dan??manl?k, E?itim ve Mümessillik Ltd, Göztepe, Istanbul, Turkey.

Not content with that, however, two of our top Brainiacs also offered the following:

Mehmet Koral

A mathematical perception of the question raised gives the answer as follows; When the I is written as a single vertical line, it can be seen perceived as number 1. The letter x also mathematically symbolises multiplication. Therefore, the answer is IX6=6(1X6=6), 1 multiplied by 6 equals 6.

Jose Padron

Solution 1 Writing Roman Numeral IX then placing a mirror to obtain the inverted image XI and then drawing a wide line to cover the half of the numeral.

Solution 2: Writing Roman Numeral IX then turning it 180° degrees to obtain XI and then drawing a wide line to cover the half of the numeral.

Question 1: Ring fence

Police are on the trail of a notorious, international criminal who has stolen sets of rings in a series of thefts: starting in Athens followed by Turin, Beijing, and most recently Vancouver. Having studied this pattern of activity, detectives have worked out where the thief is likely to strike next and have stationed officers across the city. Which city?

Answer: Most readers were left completely puzzled by this teaser, but not our regular Brainiacs, who quickly detected that the police should be stationed in London, the thief targeting a sequence of Olympic cities: Athens, Summer, 2004; Turin, Winter, 2006; Beijing, Summer, 2008; Vancouver, Winter, 2010; then  London, Summer, 2012

Very well done, in order of reply, to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, The Netherlands; David Mann, key account manager, SPC Rubber Compounding, UK; Jose Padron, laboratory analyst, Waterville TG Inc., Waterville, Québec, Canada; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; and everyone else who had a go.