ERJ Brainteaser - April

Question 5: Cannot be series (#7)

1 1 2,  2 2 0,  3 3 9,  _?_

Answer: A nice little one to round off April, and perhaps not quite as tricky as it seemed (see Solutions below). Very well done to: Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; Bharat B Sharma, technical director TWC Group (Techno Waxchem & Rajsha Chemicals), Kolkata, India; John Coleman, membership manager, Circol ELT, Dublin, Ireland; John Woods, materials technology consultant, Hanau, Germany; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; and everyone else who had a go.

SOLUTIONS

Kamila Staszewska

The pattern combines identical digits in order, plus (sic!) four basic arithmetic operations:

1+1=2  [1 1 2];

2-2=0 [2 2 0];

3*3=9 [3 3 9]

So the last in the series is 4:4=1 à [4 4 1]

Bharat B Sharma

(STAR ANSWER – Nice table)

Next sequence of numbers could be  4 4 1

Reasoning-

1st series is 1, 1, 2 (addition of first two digits 1,1)

2nd series is 2,2,0 (subtraction of first two digits 2,2)

3rd series is 3,3,9 (multiplication of first two digits 3,3)

4th series could be 4,4,1 (division of first two digits 4,4)?

1st	2nd	3rd	Operation
1	1	2	Addition	1+1
2	2	0	Subtraction	2-2
3	3	9	Multiplication	3*3
4	4	1	Division	4/4

John Coleman

This series appears to follow the four basic operations of maths, i.e.:

Addition: 1+1 = 2

Subtraction: 2-2 = 0

Multiplication: 3x3 = 9

Therefore, the fourth operation would be division, i.e. 4/4 = 1, so 4 4 1 would be the next in the series.

John Woods

I honestly don’t know the answer but I imagine it is a sequence logical.

112 -> 110 + 2

220 -> 220 - 0

339 -> 330 + 9

And so the next number will be ,

440  - undermined function of 4.

The figures that are the last.

2 - an addition of 2+ or a blend of 1 + I

O - could be subtraction

9- could multiplication of 3

Fiscally I suggest without confidence thr final figure is simple fusion of division of 4 by itself

Missing Number: 441

ALTERNATIVELY

Andrew Knox

Answer: 469 (one possibility)

With little to go on (just three numbers) let's assume it is in fact a simple series. Taking the differences between the numbers we get 108, 119, ....

The second difference gives 11 (119 - 108), so adding 11 each time to the 1st difference would generate 339+130=469. The following numbers would then be 469+141=610, etc.

Not very elegant but it is a series which at least doesn't involve any kind of non-mathematical manipulation.

Amparo Botella

I found the following relation:

1+1=2

2-2=0

3x3=9

So, the next number could be the fourth mathematical sign (+, -, x, /)

7:7=1

So next in the series could be 771.

New teaser on Monday

4.0026

6.938

10.806

14.0064

___?___

Answer: Quite a tough teaser this week, so extra well done to everyone who worked out sodium (Solutions below) as the answer: Andrew Knox, Rubbond International, Ohé en Laak, The Netherland; Dr Kate Burns, senior technologist and regulatory officer, Prisma Colour Ltd, Birch Vale, High Peak, Derbyshire, UK; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Bharat B Sharma, technical director TWC Group (Techno Waxchem & Rajsha Chemicals), Kolkata, India; Hans-Bernd Luechtefeld, consultant, Germany; John Bowen, BSc [Hons], CEng, MIMMM, consultant, Bromsgrove, UK; Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; John Woods, materials technology consultant, Hanau, Germany.

SOLUTIONS

STAR ANSWER - Andrew Knox

Don't try this at home, or anywhere else!

These are elements whose atomic numbers are ascending prime numbers, and what is actually listed in the question is their corresponding atomic mass. So, the next number will be 22.9897, which is the metal sodium. I have fond memories of throwing quite big lumps of pure sodium into the swimming pool at school...

Dr Kate Burns
The series is the atomic masses of the prime-numbered elements - 

Atomic Number	Element	Atomic Mass
2	Helium	4.0026
3	Lithium	6.938
5	Boron	10.806
7	Nitrogen	14.0064
11	Sodium	22.9898

Amparo Botella

These numbers are the atomic weights of elements whose atomic numbers are prime numbers:

4.0026 → Helium (atomic number 2, prime)

6.938 → Lithium (3, prime)

10.806 → Boron (5, prime)

14.0064 → Nitrogen (7, prime)

The next prime is 11, which corresponds to Sodium (Na).

So, the answer is  the atomic weight of Sodium which is 22.989

Bharat B Sharma

ANSWER:            22.9898   and                    26.9815 (next number)

Detail Analogy--

Thanks for giving the Clue – “PRIME” PUZZLE

The numbers are quite distant so these are not mathematical.

Scientifically, such numbers can be atomic mass (in amu)

If we see the periodic table, these numbers given in the sequence are amu of elements having PRIME Z (Atomic Numbers  -- - namely 2, 3,5,7) 

Accordingly, the next two PRIME Atomic Numbers will be Z=11 (Sodium) and Z=13 (Aluminium) and the answer would be atomic mass of these two PRIME ATOMIC NUMBERS.

Z=2 (Helium, He):             4.0026 u

Z=3 (Lithium, Li):              6.94  (rounded off of 6.638)u

Z=5 (Boron, B):                  10.81(rounded off of 10.806) u

Z=7 (Nitrogen, N):            14.007 (rounded off of 14.0064)u

Z=11 (Sodium, Na):         22.990  ((rounded off of 22.9898)u

Z=13 (Aluminium, Al):   26.982 u (rounded off of 26.9815)u

Hans-Bernd Luechtefeld

Next number is  22.9897.

The list shows the atomic mass of elements, following the aromic numbers (BUT only elements with prime numbers 2,3,5,7 are shown, with leaving out helium = 1). Next prime number is 11, for the element SODIUM. The atomic weight is 22,9897.

John Bowen

The numbers given are the atomic masses of the first 4 prime numbered elements [2,3,5,7].  Next is 11, Sodium, atomic mass 22.980.

Kamila Staszewska

These are atomic mases of elements what has ‘prime’ atomic numbers:

2 - Helium (He) 4.0026 u

3 - Lithium (Li) 6.938 u

5 - Boron (B) 10.806 u

7 - Nitrogen (N) 14.0064 u

Next are:

11- Sodium (Na) 22.9898 u

13 - Aluminium (Al) 26.9815 u

17- Chlorine (Cl) 35.4532 u, and so on.

John Woods

Three clues have been given :

Primes -> notion of Prime Numbers

Elementary -> Chemical Elements

Sodium -> confirmation of Chemical Element and a Prime Number of the Atomic Number of Sodium.

4.0026     He 2

6.938        Li 3

10.806      B 5

14.0064    N 7

The data presented shows clearly that the numbers correspond to the atomic weights of four elements with prime atomic numbers.

Therefore, the next one in the series is Sodium, atomic number 11 ( the next prime number which has an atomic weight of 22.989 76928 or rounded up to 22.990).

New teaser on Monday.

Kabul, Tirana, Algiers, _?_...

Answer: These are the capital cities of countries worldwide listed in alphabetical order, starting with the letter "A": so, the next city in the series is Andora La Vella.  Well done, in order of reply, to: Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Bharat B Sharma, technical director TWC Group (Techno Waxchem & Rajsha Chemicals), Kolkata, India; John Bowen, BSc [Hons], CEng, MIMMM, consultant, Bromsgrove, UK; Dr Kate Burns, senior technologist and regulatory officer, Prisma Colour Ltd, Birch Vale, High Peak, Derbyshire, UK; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Hans-Bernd Luechtefeld, Consultant, Germany; John Coleman, membership manager, Circol ELT, Dublin, Ireland; John Woods, materials technology consultant, Hanau, Germany.

STAR ANSWER: Neat table

Andrew Knox

AFGHANISTAN		KABUL
ALBANIA		TIRANA
ALGERIA		ALGIERS
ANDORRA		ANDORRA LA VELLA
ANGOLA		LUANDA
ANTIGUA & BARBUDA		SAINT JOHN'S
ARGENTINA		BUENOS AIRES
ARMENIA		YEREVAN
AUSTRALIA		CANBERRA
AUSTRIA		VIENNA
AZERBAIJAN		BAKU New teaser on Monday.

12 Nov 2010; 9 Aug 2007; 6 May 2004; __?__.

Answer: As easy as 1,2,3 once you recognised that these dates can give the numerical sequence: 12, 11, 10 // 9, 8, 7 // 6, 5, 4 // 3, 2, 1 (or 3 Feb 2001). Well done to: Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; John Bowen, BSc [Hons], CEng, MIMMM, consultant, Bromsgrove, UK;  Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; John Woods, technical consultant, Hanau Klein-Auheim, Germany; Hans-Bernd Luechtefeld, Consultant, Germany; Dr Kate Burns, senior technologist and regulatory officer, Prisma Colour Ltd, Birch Vale, High Peak, Derbyshire, UK; and (for their interesting alternative answers) Bharat B Sharma, technical director TWC Group (Techno Waxchem & Rajsha Chemicals), Kolkata, India; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain. 

SOLUTIONS

Andrew Knox

Answer: 3 Feb 2001

These dates can be abbreviated to read as 12-11-10, 9-8-7, 6-5-4, 3-2-1 (or 03-02-01 if preferred).

Taken together they form a simple sequence from 12 down to 1

John Bowen

Next in the series is 3rd February 2001

Each date is represented as follows: 12,11,10; 9,8,7; 6,5,4; so next will be 3,2,1

Kamila Staszewska

If the dates are presented in numerical format: 12/11/10, 09/08/07, 06/05/04 so the last missing date is 03/02/01 (3 Feb 2001).

John Woods

My proposed solution follows the following steps:

1. Transformation the dates into numerical form ,

12 November 2010 becomes 12.11 .10

9 August 2007 becomes 09.08 07

6 May 2004 ->  06.05.04

If we apply this and continue the spiral sequence downwards , it is logical to imagine that the next date is 03.02.01  , or the

3 February 2001 my proposal as the solution but I confess to being lazy and not looking up Saint’s days , Port Vale’s victories or other remote possibilities of an alternative solution.

Hans-Bernd Luechtefeld, Consultant

Days, months and years follow the row from 12 to 1, so the answer is 3 February 2001.

Dr Kate Burns

When written in dd/mm/yy format, the dates given are:

12/11/10

09/08/07

06/05/04

So the final one is 03/02/01 (counting down), which is 3rd February 2001.

ALTERNATIVELY

Bharat B Sharma

Answer - The next date would be 2 February 2001.

If we see closely, the trend is that be it date or month or year, everything is getting reduced by 3 (subtracting 3 years, 3 months, and 3 days v/s. earlier date).

If we reduce  

12 Nov 2010  If we reduce 3  years, 3 months, and 3 days, we get -- 9 Aug 2007

9 Aug 2007   If we reduce 3  years, 3 months, and 3 days, we get -- 6 May 2004

6 May 2004   If we reduce 3  years, 3 months, and 3 days, we get -- 2 Feb 2001

 Amparo Botella

The answer is 1-4-2001

As if you add the day, month and year, the result is a decrease of 9

12+11+2010 = 2033

9+8+2007 = 2024

6+5+2004 = 2015

As the year decreases in 3 years every time, the next year will be 2001, so the date must be:

1+4+2001 = 2006

This makes the sum of all factor the following series: 2033 -9 = 2024 – 9 = 2025 – 9 = 2006

New teaser on Monday.

 Moulded rubber seals are produced in pairs - one green, the other black - and transferred to containers holding more than 250 but less than 350 of the mouldings in total.

However, one seal has been lost from a container so that, if two seals were randomly drawn out, the probability of them being the same colour would be exactly 50%.

If there are more green seals than black seals, can you work out mathematically the colour of the missing seal?

Answer: Well done to everyone who had a go at this somewhat confusing (apologies) teaser – for starters, black would clearly be the colour of the missing seal. However, using very clever maths, Bharat B Sharma, technical director TWC Group (Techno Waxchem & Rajsha Chemicals), Kolkata, India was able to pinpoint the official answer. For their commendable replies, well done also to: Dr Kate Burns, senior technologist and regulatory officer, Prisma Colour Ltd, Birch Vale, High Peak, Derbyshire, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; and Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland.

SOLUTIONS

Bharat B Sharma

Suppose – G is Green seals and B is Black Seals.

                Total seal in container is T = G+B  and probability (in drawing two seals) is each 0.5 (50%) for Black and Green

                {G(G-1)+B(B-1)}/{T(T-1)} = ½

                2{G(G-1)+B(B-1)}} = T(T-1)

2G2 – 2G + B2- 2B = (G+B)2 – (G+B)

                G2 – 2GB + B2=G+B

                (G-B)2 = T  -- Remaining seals must be a perfect square

Since one of the seals is lost, so the total will not be even (it will be odd).

Squares between 250 and 250 are squares of 16, 17, and 18 and only square of 17 is odd (289) (others are 256 and 324).

With 289 is only odd number, the total that would have been including the missing seal – should have been 289+1=290 (or 145 each of black and green seal).

We have two equations now:

G+B =289

G-B =17

Therefore 2G = 289+17    or G=153

And B= 289-153 or B= 136

Given that only one ball is lost during filling of container , we should be having originally 153 pairs of Green and Black. We have two scenario-

#1 If green is lost, meaning Green = 152 , making black as 153 (Black >Green).

#2 If Black is lost, meaning Black = 152 , making Green as 153 (Green > Black)

In our solution, Green seals are more (153) than Black (136)  (2nd condition)

Therefore, scenario #2 is more mathematically valid meaning

A BLACK SEAL WAS LOST DURING TRANSFERRING IN CONTAINER!!

Dr Kate Burns

If the seals are produced in pairs, one is missing and there are more green seals, then the missing seal must be black.

To go a step further, I set out to prove mathematically that in this case the probability of picking two of the same colour was 50%.

If the number of green seals (ng) is one greater than the number of black seals (nb), ng=nb+1. The total number of seals is 2nb+1

Probability of picking two green seals P(g,g) = nb+1/(2nb+1) * nb/2nb = (nb+1) / 2(2nb+1)

Probability of picking two black seals P(b,b) = nb/(2nb+1) * nb-1/2nb = (nb-1)/2(2nb+1)

Probability of picking the same colour twice = P(g,g) + P(b,b) = (nb+1)/ 2(2nb+1) +  (nb-1)/2(2nb+1) = 2nb/ 2(2nb+1)

This is close to 50%, but is not exactly 50% as stated in the question. As the number of total mouldings is between 250 and 350, the actual chance of both seals being the same colour ranges from 49.801% to 49.857%.

Andrew Knox

Answer: If the seals are produced in pairs, and only one seal from this container has been lost, and given there are more green seals than black, then the colour of the lost seal must be black.

In order for two seals to be withdrawn from this container and the chance to be EXACTLY 50% that they are of the same colour, the first of the two seals selected must be the opposite colour to the lost seal, i.e. the first to be selected must be green, leaving an equal number of both colours before the second seal is withdrawn.

The statement "However, one seal has been lost from a container so that, if two seals were randomly drawn out, the probability of them being the same colour would be exactly 50%" is however not true. This is only the case if the first seal removed is green. If the first seal removed is black, there will be two more green seals remaining n the container than black seals, whereby the chance the second seal selected also being black will be only (n-2)/(2n-2) x 100 percent, where n is the original number of black or green seals that should have been in the container before one was lost.

Kamila Staszewska

There are more green and black so G>B

P of drawing two seals of the same colour  = 50%

P=((G(G-1))/2 + (B(B-1))/2)/((G+B)*(G+B-1)/2)=1/2

Solving this will give (G-B)^2= G+B

The difference d forces G-B to be positive whole number d=G-B

If the green is missing then the original number is G+1 and B

Original difference: (G+1)-B=(G-B)+1=d+1

Since d>0, d+1>=2 which is not possible from losing only ONE seal.

If the black  is missing then the original numbers were G and B+1

Putting in the original difference equation: G-(B+1)=(G-B)-1=d-1

For d to be positive whole number: d-1>=0

So G>=B+1-1=B which is possible so the missing seal was black.

New (more straightforward) teaser on return from Easter holiday on 7 April.