ERJ Brainteaser - October

Question 4: On a roll

Gail rolls three dice. What is the probability that the numbers showing total 7?

Answer: Almost 7% (see Solutions below). Well done to this week’s high rollers: John Bowen BSc [Hons], CEng, MIMMM, consultant, Bromsgrove, UK; Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; Bharat B Sharma, technical director, Rajsha Chemicals Pvt. Ltd, (TWC Group), Vadodara (Guj), India; Jose Padron, project manager / chargé de projet, E2Metrix Canada Inc., Sherbrooke, Quebec, Canada; Andy Longdon, technical sales manager, Ceetak Ltd, Bedford UK;  Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Hans-Bernd Luechtefeld, consultant, Germany; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; and everyone else who had a go.

SOLUTIONS

John Bowen

There are 6 x 6 x 6 = 216 possible outcomes from 3 rolls.

The possible combinations to give 7  are:

(1, 1, 5) and its permutations: (1, 5, 1), (5, 1, 1). There are 3 outcomes.

(1, 2, 4) and its permutations: (1, 4, 2), (2, 1, 4), (2, 4, 1), (4, 1, 2), (4, 2, 1). There are 6 outcomes.

(1, 3, 3) and its permutations: (3, 1, 3), (3, 3, 1). There are 3 outcomes.

(2, 2, 3) and its permutations: (2, 3, 2), (3, 2, 2). There are 3 outcomes.

Adding these up, the total number of favorable outcomes is 15.

So, the probability is 15/216 or 0.0694 or 6.94%.

Kamila Staszewska

Rolling three dice gives 6x6x6=216 possible outcomes,

To get exactly 7 there is 15 combinations: (1,1,5), (1,2,4), (1,3,3), (2,2,3), (2,3,2), (2,4,1), (3,1,3), (3,2,2), (3,3,1), (4,1,2), (4,2,1), (5,1,1).

Probability is 15/216=6.94%.

Bharat B Sharma

The probability is 15/216 or 5/72

Total number of possible outcomes 6 possible results  (1, 2, 3, 4, 5, or 6) x 3 dices = 6

Favorable Required results  (combinations of three numbers that sum to 7)   --  (1,1,5  or  1,2,4,  or  1,3,3   or 2,2,3

Probability of Combination (1, 1, 5):  three possible manners  (1, 1, 5), (1, 5, 1), and (5, 1, 1). This is  factorial 3 by factorial 2 == 3

Probability of Combination (1, 2, 4): Possible manners are factorial 3 = 6  (1, 2, 4), (1, 4, 2), (2, 1, 4), (2, 4, 1), (4, 1, 2), (4, 2, 1).

Probability of Combination (1, 3, 3): three possible manners  : (1, 3, 3), (3, 1, 3), and (3, 3, 1). This is  factorial 3 by factorial 2 == 3

Probability of Combination (2, 2, 3): three possible manners  : (2, 2, 3), (2, 3, 2), and (3, 2, 2). This is  factorial 3 by factorial 2 == 3

The total number of favourable combinations for getting a sum of 6 in rolling three dices are  3+6+3+3 = 15    out of possible 216 combinations.

So, the probability is  15 / 216 (or 5/72).

Andy Longdon

Q4: On a roll

With three dice, the number of outcomes is 6 x 6 x 6 = 216 variations

Of those, 15 would result in a total of 7: 

1,1,5,,,1,2,4,,,1,3,3,,,1,4,2,,,1,5,1

2,1,4,,,2,2,3,,,2,3,2,,,2,4,1

3,1,3,,,3,2,2,,,3,3,1

4,1,2,,,4,2,1

5,1,1

So probability is 15/216, or 5 times in 72 rolls. Or approximately 7%. Or approximately once in every 15 rolls.

Amparo Botella

Each die has 6 faces, so with 3 dice, there are:

6 x 6 x 6 = 216 total possible combinations.

If we need all combinations of three numbers (each 1–6) add up to 7, we have the possible combinations:

1+1+5 / 1+5+1 / 5+1+1 (3 combinations)

1+2+4 / 1+4+2 / 2+1+4 / 2+4+1/ 4+1+2 / 4+2+1 (6 combinations)

1+3+3 / 3+1+3 / 3+3+1 (3 combinations)

2+2+3 / 2+3+2 / 3+2+2 (3 combinations)

The total of combinations adding up 7 is

3 + 6 + 3 + 3 = 15

The probability to get 7 with 3 dices is:

From 217 we have 15

From 100 we have x

X = 6,912

6.91% probability.

Hans-Bernd Luechtefeld

Rolling three dice means you're dealing with 216 possible outcomes (Just as one die has six outcomes and 2 dices have 6^2 = 36 outcomes, the probability of rolling three dice has 6^3 = 216 outcomes.)

The range of possible sums expands, now from 3 to 18:

There is one possible way three dice can total 3

3 ways for 4

6 for 5

10 for 6                                                                       

#

15 for 7

21 for 8

25 for 9

27 for 10

27 for 11

25 for 12

21 for 13

15 for 14

10 for 15

6 for 16

3 for 17

1 for 18

So, probability is 15/216 or around 6,94%.

Andrew Knox

Answer: 0.0694444

There are four ways to throw a total of 7:

1. a) 1 1 5  (in any order)
2. b) 1 2 4  (in any order)
3. c) 2 2 3  (in any order)
4. d) 3 3 1  (in any order)

In case b) there are 6 permutations:

4   2   1

4   1   2

2   4   1

2   1   4

1   2   4

1   4   2

In cases a), c) and d), there are only 3 permutations for each, i.e.

5   1   1

1   5   1

1   1   5 

(and similarly for 2   2   3 and 3   3   1)

So the chance of throwing 7 in total in the case of a), c) and d) is in each case only 1/6 x 1/6 x 1/6 x 3 = 3/216, in the case of b) it is 1/6 x 1/6 x 1/6 x 6 = 6/216

Adding all four possible outcomes is 6/216 + 3/216 + 3/216 + 3/216 = 15/216 = 0.0694444

Jose Padron

The probability of a sum = 7, with three dice                      

1. a) number of possibilities when rolling three dices: 6 numbers each dice   

6 x 6 x 6 = 216 possibilities             

1. b) number of possible combinations to sum 7 = 15 

1	2	4
2	4	1
4	1	2
1	4	2
4	2	1
2	1	4
3	3	1
3	1	3
1	3	3
2	2	3
2	3	2
3	2	2
1	1	5
1	5	1
5	1	1

Thus, the probability of having 7 with three dice is: 15/216 = 0.06944 = 6.94%

 

New teaser on Monday.

Mercury + Venus = 0

Earth + Mars = 3

Mercury + Neptune = 16 (corrected from 14).

Neptune + Jupiter = ?  

Answer: It was more of a ‘small step’ than a ‘giant leap’ for our Brainiacs to work out the connection (see Solutions below) here. Well done to: Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain;Jose Padron, project manager / chargé de projet, E2Metrix Canada Inc., Sherbrooke, Quebec, Canada; Andy Longdon, technical sales manager, Ceetak Ltd, Bedford UK; Eur Ing John Bowen BSc [Hons], CEng, MIMMM, consultant, Bromsgrove, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Kho Irani, Lion Elastomers, USA; John Coleman, membership manager, Circol ELT, Dublin, Ireland; Sudi Sudarshan, principal consultant, Global Mobility Strategies, USA; Peter D. Talbot, research scientist, Chem-Trend LP, Howell, MI, USA. 

SOLUTIONS

Kamila Staszewska

This relates to number of moons each planet has:

Mercury + Venus = 0+0=0

Earth + Mars = 1+2=3

Mercury + Neptune = 0+14=14

Neptune + Jupiter = 14+95=109 (although most recent count is 97 - then 14+97=111).

Amparo Botella

The answer is 113, as:

Mercury has 0 moons and Venus has 0, so Mercury + Venus = 0.

Earth has 1 moon and Mars has 2, so Earth + Mars = 3.

Mercury (0) + Neptune = 14 → corrected to 16, so Neptune = 16.

Therefore Neptune (16) + Jupiter = ?

As of the latest counts, Jupiter has 97 confirmed moons, so Neptune + Jupiter = 16 + 97 = 113.

Andy Longdon

Answer is the total number of moons these planets have, although Neptune has 16, so Mercury + Neptune should = 16, not 14

Mercury and Venus don’t have any.

Earth has 1, Mars has 2, total = 3

Mercury has 0, Neptune has 16, total = 16

Neptune has 16, Jupiter has 97, total = 113

Jose Padron

Hello, it seems that the addition is the numbers of moons of each planet, in that way:

Mercury + Venus = 0 (no moons)

Earth + Mars = 3 (1+2)

Mercury + Neptune = 16 (according to the NASA website Neptune has officially 16 moons instead of 14)

Neptune + Jupiter = 16 + 95 = 111 (moons officially known for each planet by NASA)

However, in National Geographic kids web site, it says Jupiter has 70 moons. So, It could be: 86 (subject to discussion with a pint of beer)

John Bowen

Good morning Patrick, this is to do with planets and their moons.

Earth has 1, Mars 2 so total is 3 and so on.

Jupiter has 97 and you say Neptune has 14, so total is 111

Andrew Knox

Answer: 79

Question appears to ask how many moons do each pair of planets have together, although there appear to be differences of opinion as to the correct numbers:

Mercury and Venus: None

Earth and Mars: ( 1 + 2) = 3

Mercury and Neptune: (0 + 13) = 13

Neptune and Jupiter: (13 + 66) = 79

Kho Irani, Lion Elastomers, USA

This summation is based on the number of moons of each planet.

Neptune has 16 moons and Jupiter has an astounding 97 moons.

16+97 = 113

John Coleman

Mercury + Venus = 0

Earth + Mars = 3

Mercury + Neptune = 14 (correction 16...)Neptune + Jupiter = ? 

This teaser relates to the sum of the identified moons among the named planets.

Therefore, Neptune + Jupiter = 16 + 97 = 113 moons.

Sudi Sudarshan

My answer: 109

It is the sum of the number of known moons of the two planets.

Peter Talbot

The equations in this puzzle relate to the number of moons orbiting the named planet. Mercury and Venus both do not have any moons, therefore 0+ 0 = 0. Earth has 1 moon and Mars 2, therefore 1+2 = 3. Mercury, we already know has no moon. Neptune has either 14 or 16, depending on authoritative source.  Therefore 0+14 (16) = 14 (16). Jupiter has 95 moons (though some sources say 67) and Neptune 14 (or 16). Therefore, in this case, the sum of moons can be as low as 81 or as high as 111.

New teaser on Monday.

…, Hu Jintao, Queen Elizabeth II, Michel Temer, Emperor Naruhito,  ?

Answer: Special mention this week to Hans-Bernd Luechtefeld, consultant, Germany, who was quickest out of the blocks with his answer. He was followed soon after by: Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Jose Padron, project manager / chargé de projet, E2Metrix Canada Inc., Sherbrooke, Quebec, Canada; John Coleman, membership manager, Circol ELT, Dublin, Ireland; John Bowen, consultant, Bromsgrove, UK; Sudi Sudarshan, principal consultant, Global Mobility Strategies, USA. Well done to all and everyone else who had a go.

SOLUTIONS

Hans-Bernd Luechtefeld

We talk about the representatives who have opened the Olympic Summer Games in its countries. So, # 5 is Emmanuel Macron, who opened the 33. Summer Games in Paris in 2024.

Kamila Staszewska

The question is who officially opened Summer Olympic Games: Hu Jintao - China 2008 , Queen Elizabeth II - UK 2012 , Michel Temer - Brazil 2016, Emperor Naruhito - Tokio 2020, Next Olympics were France 2024 and Emmanuel Macron is the answer.

Andrew Knox

Answer: Emanuel Macron.

This is a list of the heads of state who opened the Olympic games from 2008 to 2024

Jose Padron

The next person in the list is Emmanuel Macron as the dignitary in the XXXIII Summer Olympiad Opening Ceremony in Paris. Hu Jintao (2008) , Queen Elizabeth II (2012), Michel Temer (2016), Emperor Naruhito (2021), Emmanuel Macron (2024), Donald Trump (2028), KIng Charles III (2032) (expected)...

John Coleman

These are the heads of state that opened the most recent Summer Olympic Games. Therefore, President Emmanuel Macron of France is the missing name at the end of the series as he opened the Paris games on 26 July 2024.

John Bowen

These names are those of the heads of state of the countries holding the summer Olympic games in 2008, [Beijing, China] 2013 [London, UK], 2016 [Rio, Brazil] The previous one was in Greece, in 2004 and officially opened by President Komstantin Stephanopoulos.

Sudi Sudarshan

My response to this week's Brainteaser: Emmanuel Macron. The list shows the name of the national leader who inaugurated the summer Olympic games starting with the 2008 games.

New teaser on Monday.

ABC, BCE, CEH, EHM, _

Answer: While the official answer is HMU, some Brainiacs came up with interesting alternatives (see Solutions below). Very well done to: Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; Eur Ing John Bowen B.Sc.[Hons], CEng, MIMMM, consultant, Bromsgrove, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Andy Longdon, technical sales manager, Ceetak Ltd, Bedford UK; Peter D. Talbot, research scientist, Chem-Trend LP, Howell, MI, USA; Hans-Bernd Luechtefeld, consultant, Germany; Sudi Sudarshan, principal consultant, Global Mobility Strategies, USA; and everyone else who had a go.

 

SOLUTIONS

Amparo Botella

A=1, B=2, C=3, E=5, H=8, M=13, …

Each third letter of the three letters has the correspondent alphabet number of the sum of the first+second:

ABC → (A)1+B()=C(3)

BCE → (B)2+(C)3=(E)5

CEH → (C)3+(E)5=(H)8

EHM → (E)5+(H)8=((M)13

Next triple is 8+13=21 →

H (8) + M (13) = U (21).

So the next term is HMU.

Kamila Staszewska

ABC - start of the sequence - two last letters become the first two of next.

BCE  (1st prime is 2, C+2=E)

CEH (2nd prime is 3, E+3=H)

EHM (3rd prime is 5, H+5=M)

The next is HMT (4th prime is 7, M+7=T)

John Bowen

I think this would be H,M,R

Taking the first letter, we add 1,1,2,3 in Fibonacci manner, so next is +3 which is  H

Second letter we add 1,2,3, so next is 4 which is M

Third letter we add 2,3,4, so next is 5 which is R

Andrew Knox

Answer: HMV

Pattern appears to be that the last two letters of each group of 3 become the first two letters of the next set of 3 letters, and for the last letter of each group,  in the first case the next letter in the alphabet is missed out, then the next 2 letters, then the next 4, and then (one presumes) the next 8 letters.

So the letters skipped are D, FG, IJKL, NOPQRSTU, leaving V to follow HM.

Andy Longdon

Answer: HMU

The last two letters of the previous sequence become the first two letters of the next. The third letter is found by the position/number in the alphabet (A=1 ….. Z=26) resulting from the sum of the positions of the first two letters:

ABC      1+2=3

BCE      2+3=5

CEH      3+5=8

EHM      5+8=13

HMU     8+13= 21

HM = last two digits of sequence before and U = letter 21 out of 26).

Peter Talbot

If a letter is represented by its numerical order in the alphabet i.e. A=1, B=2, C=3 etc., ABC, BCE, CEH, EHM, _ becomes 123, 235, 358, 5813.

It quickly becomes apparent that the third letter of the three letter composite equals the sum of the first two letters

i.e. A+B=C, B+C=E, C+E=H, E+H=M or 1+2=3, 2+3=5, 3+5=8 & 5+8=13.

Therefore, the next set of letters is represented by 8+13 = 21 or HMU.

Hans-Bernd Luechtefeld

The next one is: HMU

Assume

A = 1

B = 2

C = 3

etc.

ABC => 123

BCE => 235

CEH => 358

EHM => 58 13

The figures for two first letters gives the third letter: 1+2 = 3 resp. A + B = C, etc.

And the.next letter triple begins with the last two letters of the letter triple before => in our case H and M, the 'adding up' H and M (resp. 8 + 13 gives 21), resulting in U (= 21).

Sudi Sudarshan

The ordinal numbers in the English alphabet of the third letter in each set is the sum of that of the first two letters.

Also for the second set onwards, the first two letters are the last two letters of the prior set.

ABC = 1,2,3   1+2=3

BCE = 2,3,5   2+3=5

CEH = 3,5,8   3+5=8

EHM = 5,8,13 5+8=13

8+13=21

So, the next series comprises of the 8th, 13th and the 21st letter i.e. HMU.