ERJ Brainteaser Results

List of Brainteaser questions over the past couple of months, along with the answers and readers supplying correct replies. The number of correct answers, as well as speed and quality of reply will help decide our Brainiac of the Month. 

 

AUGUST 2019

 

Question 3: Find the formula

If you use a certain formula on 13, you end up with 7. Under the same formula, 2352 becomes 16, 246 becomes 14, 700 turns into 16, and 1030 becomes 14. What would 9304 become?

Answer: Just two readers came up with the answer for this tricky teaser – 19. But never mind the quantity, look at quality of the excellent replies (below) from: Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; and Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany. Very well done to both, and many thanks to everyone else who had a go.

Bharat B Sharma:

Hans-Bernd Lüchtefeld

Formula: Convert the number to binary; add 1 for every zero, add 2 for every 1:

 

Question 2: EU get it?

Which country comes next in the following list?:

Greece

Netherlands

Belgium

France

(Clues given during the week: It is to do with numbers, codes…)

Dog and bone

Answer: It is always good to have a clear winner, and this time around it is a big congratulations to Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands who came up with the correct answer first thing on Monday morning. This related to a numeric sequence of international telephone dial codes in these countries: with Spain next on the list:

Greece 30, Netherlands 31, Belgium 32, France 33, Spain 34

Also ‘ringing in’ with the correct solution – just a little after Andrew –  were: David Mann, Polymer Business Development, France; Yoganand Nannapaneni, Mascot Systems Private Ltd, Mumbai, India; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; John Bowen, consultant, Bromsgrove, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Fariha Rashid, marketing analyst, Kraton Polymers LLC, Houston, Texas, USA.  Well done to all and everyone else who had a go – especially those from outside the EU, as this was a slightly Europe-centric question.

 

Question 1: Counting cards

Tire technician Simon is trying his hand at blackjack on a two-deck table in Las Vegas. The dealer rolls through 3 hands in which Simon counts 17 face cards and 29 non face cards. On the fourth hand he sees 4 face cards and 9 non-face cards, of which he has a seven and a five and the dealer has an ACE showing.

What are the odds that Simon will bust if he takes one card?

Answer: The odds of Simon busting with one card is 24.4% as worked out by just two readers. Extra well done to: Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA.

As Andrew Knox explained:

The odds that Simon will bust taking one card is 11/45.

This is assuming Simon considers the 10’s as one of the face cards, i.e. 10, J, Q & K are the face cards (the so-called ‘X cards’ as they all have a value of 10) and he considers the Ace thru 9 as the non face cards.

The shoe starts off with 2 decks, i. e. 104 cards, of which 72 non- and 32 face cards.

The first 4 hands deal 38 non- and 21 face cards, leaving 34 non- and 11 face cards in the shoe.

Simon has a 5 and a 7, totalling 12, so he busts (> 21) only if dealt a 10, J, Q or K, i.e. one of the face (X) cards.

The odds of this happening with 45 cards (of which 11 face cards) in the shoe is 11/45.

The fact that the dealer’s first card is an Ace is not relevant.

ps If the face cards are taken strictly as J, Q or K, then Simon cannot work out his chances exactly, as he hasn’t been counting the 10’s.

 

JULY

Having responded correctly to all four questions of the month, Ramasubramanian P of Larsen & Toubro Ltd in India, deservedly wins the Brainiac of the Month title.

Question 4: Missing number

Find the missing number in the table above.

 Answer: 55 – a relatively easy one which saw many new entrants.

Well done to Lars Linnemann, R&D manager, Genan A/S, Viborg, Denmark; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Sebastian Barbe, group operations manager Hans W. Barbe Chemische Erzeugnisse GmbH, Germany; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Sajiva Manju; David Cao; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; John Bowen, consultant, Bromsgrove, UK; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands;  Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA;  Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada;  Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Ricardo Azcarate; Serafina Vulcano; Fariha Rashid, marketing analyst, Kraton Polymers LLC, Houston, Texas, US. Berny Bolanos, PE Curing & Tire Room, Bridgestone Costa Rica

Question 3: Fast car?

In a tire test, a race car is driven along a straight horizontal track with constant acceleration. There are three check points A, B and C, in that order, on the road, where AB = 22 m and BC = 104 m. The car takes 2 seconds to travel from A to B and 4 seconds to travel from B to C. What is the acceleration of the car, and the speed of the car at the instant it passes A?

Answer: Acceleration of the car = 5 m/s² and speed at point A = 6 m/s  (see solution from below). Well done to: France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; John Bowen, consultant, Bromsgrove, UK: Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India.

Bonus question: What is the minimum number of serves you would need to take, to be a victor in a three-set match of doubles tennis? The answer was, quite simply, 12 serves.

Fast car? solution from Ramasubramanian P:

Speeds at A,B,C are Ua, Ub, Uc

Acceleration is constant a.

Constant Accleration Formulae:

V = U + aT

D = UT + (1/2)aT^2

Using the above formulae,

Ub = Ua + 2a                —- 1

Uc = Ub + 4a                 —- 2

Uc = Ua + 6a                 —- 3

22 = 2Ua + 2a               —- 4

104 = 4Ub + 8a             —- 5

126 = 6Ua + 18a            —- 6

(6) – 9*(4):  Ua = 6        —- 7

(6) – 3*(4):   a = 5          —- 8

(7)&(8) on (1):  Ub = 16  —- 9

(7)&(8) on (3):  Uc = 36 —- 10

Acceleration of the car : 5 m^2/s: speed at instant A : 6 m/s

 

Question 2: Ye Olde Brainteaser

This was an adaptation of a problem by a well-known puzzler,  Henry Ernest Dudeney (1857 – 1930) – as supplied in The Engineer magazine many years ago.

Last week, the squire and his wife held a party for the villagers. The guests comprised 3 widowers, 3 widows,9 bachelors and boys, 7 eligible maidens and girls, and 7 married couples.

The hosts and all guests kissed everybody else present once, with the following exceptions:

No male kissed another male;

No married man kissed a married woman except his own wife;

The widows did not kiss each other;

The widowers kissed only the widows;

All the bachelors and boys kissed all the maidens and girls twice;

Each kiss between two people counts as one kiss.

How many kisses were there?

Answer: The official answer, from all those years ago, was 472 kisses (see Henry Ernest Dudeney’s solution below). Well done to the following readers who got this correct and to the many others who had a really good go:

Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India.

The table below summarises the position: (males in upper case, females in lower case)
0=no kiss, K=1 kiss, T=2 kisses.

Or as Michele Girardi explained: This is done dividing the people in homogeneous groups and building a table of number of kisses considering the exceptions . The total is given by the sum of the diagonal and one of the quadrants.

 

Question 1: Missing names

Fill in the gaps in the following sequence:

Walker, ___, Herbert, Wilson, Earl, ____.

Answer: A chance to recharge your calculators this week, as we delved deep into the world of politics. The readers below (not the US presidents) correctly identified the sequence as:

George Walker Bush

William Jefferson Clinton

George Herbert W. Bush

Ronald Wilson Reagan

James Earl Carter

Gerald Rudolph Ford

Well done to: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; John Bowen, consultant, Bromsgrove, UK; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan. Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India.

 

JUNE

Everyone who successfully tackled this month’s Q3 and Q4 deserves special mention: there were some really amazing solutions sent in. But for the extremely neat way he tackled both these and other questions, huge congratulations go to Jose Padron our new Brainiac of the Month

 

Question 4: Cycle code

Simon has worked out a new four-digit number for the combination lock on his bicycle. This is coded A, B, C, D, with each letter representing a different digit from 1 to 9. If this number is divisible by 13, BCDA is divisible by 11, CDAB is divisible by 9, and DABC is divisible by 7, what is the original number ABCD?

Answer: Tricky enough but at least no need for supercomputers to work out this week’s answer 3861. Well done to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; David Mann, Polymer Business Development, France; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India.

(Michele Girardi, Scame Mastaf Spa, Suisio, Italy and David Mann came up with an interpretation that gave 7722 as another possible answer.)

We really liked Jose Padron’s table for solving this, amid all the more detailed solutions sent in:

 

Question 3: One to nine

Arrange the digits from 1 to 9 to make a 9-digit number ABCDEFGHI which satisfies the following conditions:

1) AB is divisible by 2;

2) ABC is divisible by 3;

3) ABCD is divisible by 4;

4) ABCDE is divisible by 5;

5) ABCDEF is divisible by 6;

6) ABCDEFG is divisible by 7;

7) ABCDEFGH is divisible by 8;

8) ABCDEFGHI is divisible by 9.

There is only one solution.

Answer: When the going gets tough, the tough really do get going, as evidenced by the fantastic array of solutions to find the correct answer 381654729 to this tough teaser. Well done, in order of reply to: Amparo Botella, Ismael Quesada SA, Spain; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; David Mann, Polymer Business Development, France; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; John Bowen, consultant, Bromsgrove, UK: Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India.

A few examples:

Michele Girardi

This is done more easily with an Excel spreadsheet, using the function that calculates the remainder of a division

CD must be divisible by 4 and not contain 0

E can only be 5

BDFH must  be even numbers 2,4,6,8

Develop all the CD combinations without 0 , double numbers and 2 even numbers

12

16

32

36

52

56

72

76

92

96

Develop combinations with the remaining even numbers for B and the numbers for A for which A+B+C is divisible by 3

Develop the possible combinations for the remaining 2 even numbers between F and H

Filter the numbers for wich ABCDEF is divisible by 3

At this point  there are 15  combinations missing G and I

A        B        C        D        E        F        G       H        I

1        2        3        4        5        6        X        8        X

7        4        1        2        5        8        X        6        X

3        2        1        6        5        4        X        8        X

9        2        1        6        5        4        X        8        X

3        8        1        6        5        4        X        2        X

9        8        1        6        5        4        X        2        X

9        6        3        2        5        8        X        4        X

7        2        3        6        5        4        X        8        X

1        8        3        6        5        4        X        8        X

7        8        3        6        5        4        X        8        X

1        4        7        2        5        8        X        6        X

3        2        7        6        5        4        X        8        X

9        2        7        6        5        4        X        8        X

3        8        7        6        5        4        X        2        X

9        8        7        6        5        4        X        2        X

Try inserting the last 2 missing numbers until ABCDEFG is divisible by 7

Hans-Bernd Lüchtefeld:

E = 5 (condition 4)

B, D, F, H are even numbers (conditions 1, 3, 5, 7), therefore A, C, G, I are 1, 3, 7or 9 (in some order)

CD is divisible by 4 (condition 3, 7) and GH is divisible by 8 (because FGH is divisible by 8 and F is even). Because C and G are odd, D and H must be 2 and 6 in some order

A+B+C, D+E+F, G+H+I are all divisible by 3 (conditions 2, 5)

If D=2, then F=8, H=6, B=4. A+4+C is divisible by 3, therefore A and C must be 1 and 7 in some order, G and I must be 3 and 9 in some order. G6 is divisible by 8, therefore G=9. But neither 1472589 nor 7412589 is divisible by 7

Therefore D=6, and F=4, H=2, B=8. G2 is divisible by 8, therefore G=3 or 7. A+8+C is divisible by 3, therefore one of A and C is chosen from 1 and 7, the other is chosen from 3 and 9

If G=3, then one of A and C is 9, the other is chosen from 1 and 7. But none of 1896543, 7896543, 9816543 and 9876543 is divisible by 7

Therefore G=7, one of A and C is 1, the other is chosen from 3 and 9. From 1836547, 1896547, 3816547 and 9816547, only 3816547 is divisible by 7 (the quotient is 545221)

The result is 381654729.

David Mann

By a combination of logic and excel, I arrived at:381654729

Logic: position 5 has to be 5.

Positions 1,3,7,9 must be odd

Positions 2,4,6,8 must be even

Then I built up the number from left to right using duplicate checks and the ROUNDDOWN function to check for divisibility.

Bharat B Sharma

Detail analogy and interesting conclusion/observation–

Number ABCDEFGHI

Let’s consider E =5 because (to be multiple of 5 as ABCDE to be divisible by 5)  (either 5 or 0 and number diesnot have any 0)

Let the number be ABCD5FGHI. It’s clear that the fifth digit has to be 5.

B,D,F,H are belonging to set of {2,4,6,8}

Remaining A,C,G,I are belonging to set of {1,3,7,9}.

There could be 24×24 possibilities but with conditions , the possibilities reduces drastically)

2C+D has to be divisible by 4,

4D+20 + 4F by 6

2G+H by 4.

Alternate digits must be even, so the rest have to be odd.

C & D have to be “odd, even” (for above criterion) and also to have ABCD  a number divisible/multiple of 4 (so D has to be 2 or 6).

Same applies for number H — must be 2 or 6.

Now even numbers remaining are 4 and 8, and these must be either B or F

ABC ( first 3 digits) to be divisible by 3 . Now there could be 9 possible options. We need to try and workout to identify D as D and similarly to H.

By trial and error, we need to identify 7 digit number (ending with G) to be divisible by 7 and leaving remaining number ending with 8th digit divisible by 8 381654729

Interesting observation—This is how the numbers are displayed on Calculator. Answer — Number is 381654729

Number        Divided by     Value

3        1        3

38      2        19

381     3        127

3816   4        954

38165 5        7633

381654         6        63609

3816547       7        545221

38561472     8        4820184

385614279    9        42846031

 

Question 2: Train game

Mathematician Matt’s train is delayed so he passes the time by generating random numbers from 1 to 9, inclusive, on his calculator. As a game, he tries to work out the probability that the difference between two of these numbers is greater than 5. What is the answer?

Answer: The probability was 12/81 or 14.8% as explained by: Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan.

Well done to all for coming up with the correct answer in a variety of brilliant ways, not least this table by Bharat B Sharma:

 

Question 1: Digit dilemma

192 is one of four three-digit numbers that taken along with its double and triple, ie 192, 384, 576, gives a set with each digit from 1 to 9 – likewise 273. Can you find two other three-digit numbers with this property?

Answer: 219 and 327: as Hans-Bernd Lüchtefeld neatly explained, just swap 19 and 2 resp. 27 and 3. Well done to; John Bowen, consultant, Bromsgrove, UK; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Bharat B Sharma, Sr VP product development & technical Service (elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; David Mann, Polymer Business Development, France; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada: Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India

Special extra mentions to:

Jose Padron for this tidy table

And to David Mann for his inspired use of Excel:

I made a list from 123 to 329. Then columns 2 and 3 are n*2 and n*3. Convert these to text and concatenate. Then use a formula =SUMPRODUCT(LEN($G2)-LEN(SUBSTITUTE($G2,$I$1,””))) to count the instances of each digit from 1 to 9. Finally multiply all these counts together and the required numbers must multiply to give 1. So the 4 sets are:

n        n2      n3

192     384     576

219     438     657

273     546     819

327     654     981

 

MAY

 

 

This time the top award deservedly goes to one of our most consistent high scorers. And for being to the fore in tackling a particularly tough set of teasers, it is big congratulations to:

Industry consultant John Bowen, our new Brainiac of the Month.

 

Question: Fun fractions

What particular feature do the following fractions have in common?

19/95, 26/65, 16/64

Answer: As Jose Padron neatly explained this is a mathematical curiosity, when cancelling the same digit in both denominator and numerator gives the equivalent of the original fraction: 19/95 = 1/5; 26/65 = 2/5; and 16/64 = ¼.

Well done and thanks for your patience, to John Bowen, consultant, Bromsgrove, UK; Jose Padron, material development specialist, Waterville TG Inc., Québec, Canada; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; and, Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; as well as to Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; and Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands, who were on the right track.

 

Also: Chair challenge

For a live webinar discussion, chairman Dave and his 11 fellow industry experts sit at a round-table. How many ways can the group be arranged, if nobody can sit between two older experts?

With apologies all round, let’s call this teaser null and void: too much confusion and difference in interpretation of the wording. A special thanks to John Droogan here for his help in sorting through this muddle.

For what it’s worth, so, below is the answer to the question, which was sourced from another journal. I should have been wary: they only received one correct answer, and that was from the editor’s brother!

There are 1024 ways, up to rotation around the table. To see this, notice that the youngest journalist must sit right next to Dave – there are two possible places for him. Then, the second youngest journalist must sit right next to this group of two. Once again, there are two possible places for him. Continuing like this, we see that for all journalist except for the oldest one, there are two possible spots on the table. Multiplying two to the power of ten out, we get 1024.

 

Question 4: Brian-teaser 2

Brian’s bike was sadly stolen from outside his office, so he had to buy a new one. To secure it, he bought two combination locks each having a five-digit code. For the codes, Brian decided to use the smallest and largest numbers that met the following conditions:

1. Each digit of the number was a prime digit.

2. Each successive pair of digits formed a two-digit number that was not a prime number.

3. Each of the prime digits had to appear at least once in the five-digit number.

What were the codes for Brian’s combination locks?

Answer: Someone should tell Brian that his combinations have been rumbled – 32257 and 35772 – though only by a select few of our sharpest readers. Extra well done, so, to: John Bowen, consultant, Bromsgrove, UK; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Jose Padron, material development specialist, Waterville TG Inc., Québec, Canada; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village, Tamil Nadu, India.

 

Question 3: Biker Brian-teaser

On his regular one-mile commute, Brian took three minutes to cycle to work with the wind directly behind him, and returned home in four minutes, cycling against the same wind-force. How long would his journey have taken if there was no wind?

Answer: Okay, a little bit open-to-interpretation regarding whether this was one-way or return journey, but this Brian-teaser really got even our best Brainiacs on their ‘thinking bikes’ – as shown by the two examples further below.

Brian’s other bike…

Extra well done, so, to all readers who answered either 3.43 minutes (3min 25.7s) or 6.86 minutes (6 min, 51.4s): John Bowen, consultant, Bromsgrove, UK; Michele Girardi, Scame Mastaf Spa, Suisio, Italy;  Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Ashley Fahey, sustainability principal vice president, HERO (LGBTQ ERG), Goodyear Tire & Rubber Co., Akron, Ohio, USA; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village, Tamil Nadu, India.

Solutions:

John Bowen: With no wind, the time would be 3minutes 25.7seconds.

[Method:  Time = distance/speed

Let distance = D, Speed in still air = V and windspeed = W

then With wind speed = V + W, Against wind speed = V – w

Distance is constant, and D = velocity * distabce,

so [V + W]*3 = [V-W]*4, and rearranging, V = 7W, so W = V/7

So with wind, speed =  8V/7, Against wind speed = 6V/7

Time = Distance/Speed, so

3 = D/[8V/7] and 4 = D/[6V-7]

3 = 7D/8V and 4 = 7D/6V

Multiply 3 by 48:

144 = 42D/V

And multiply 4 by 56:

192 = 56D/V

D/V =Time, so when wind is ) and speed is thus V, Time is 144/42 AND 192/56 – both = 3.42857 minutes or 3min 25.7secs]

Jose Padron: Brian would take 6.857 minutes (6 min, 51.4 sec.) would have completed his journey without wind.

Speed with the wind; s+s(wind)=1/3  mile/min

Speed against the wind; s-s(wind)=1/4  mile/min  and,

Solving these 2 equation for s we have: 2s=1/3+1/4=7/12

thus:  s=7/24  mile/min

The whole journey home-work-home is: 2 miles

and the speed with no wind resistance is: 7/24 mile/min

from the equation: t=d/s=(2*24)/7=48/7=6.875 min.

Elapsed time to complete a journey: 6875 min, = 6min. 51.4 sec.

 

Question 2: Missing years

Fill in the missing years in this tricky millennium sequence:

2307, 2417, 2527, 2637, _, _, _.

Answer: Not so tricky after all, as correct answers quickly arrived in from around the world. The key is that the total from adding the first two digits (eg 27) and last two digits (eg 47) equals the number formed by the middle two digits (eg 74).

2307, 2417, 2527, 2637, 2747, 2857, 2967

Well done in order of arrival to: Amparo Botella, Ismael Quesada SA, Spain; Bharat B Sharma, Sr VP product development & technical service (elastomers), Reliance Industries Ltd, Gujarat, India; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Risto Vettenranta, technical specialist, Ham-Re Oy, Neuvoton, Finland; John Bowen, consultant, Bromsgrove, UK: Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Ashley Fahey, sustainability principal vice president, HERO (LGBTQ ERG), Goodyear Tire & Rubber Co., Akron, Ohio, USA; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany.

 

Question 1: Plus or minus 100

Using only a total of 3 symbols and without rearranging the numerals, try to get to a total of 100 by placing + (plus) and – (minus) signs between: 1 2 3 4 5 6 7 8 9

 

Answer: 123 – 45 – 67 + 89 =100

Very well done, in order of reply to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Bharat B Sharma, Sr VP product development & technical service (elastomers), Reliance Industries Ltd, Gujarat, India; David Mann, Polymer Business Development, France; John Bowen, consultant, Bromsgrove, UK; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA. Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Amparo Botella, Ismael Quesada SA, Spain; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Lars Linnemann, R&D manager, Genan A/S, Viborg, Denmark

Special mention also to: Bob Hall, Sr. director global marketing, Kraton Polymers LLC, Houston, Texas; Randa Tharwat, import manager, Nacita Automotive, Cairo, Egypt and Salah Younis (no details given) who got there, but by using a couple of extra symbols.

Also thanks to Jose Padron for showing how to work this out using the matrices in this neat table:

 

APRIL

 

There can be absolutely no dispute about our top performer in April. For showing up first on the grid each time and navigating a tricky set of teasers, particularly Question 4, big congratulations go to:

John Droogan of MegaChem (UK) Ltd, our new Brainiac of the Month

 

Question 4: Mind the gaps

Fill in the gaps in the following sequence?

_, 4, _;  _ 8, 9, 10, _ ;  _, 14, 15, 16, _ ;  _, _, _, _, _.

Answer: The series comprises sets of numbers flanked by prime numbers in ascending order – though some readers got there using a different rationale.

3, 4, 5     7, 8, 9, 10, 11     13, 14, 15, 16, 17        19, 20, 21, 22, 23

We do try to keep our questions as original and as Google-proof as possible. As a result, solutions are sometimes a bit less straightforward than intended – as was the case this week.

However, detecting our ‘prime sandwiches’ straight away was John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK. His reply early Tuesday (day 1) was followed a little bit later by Franz Hochwimmer, sales & support inspection systems, Micro-Epsilon Messtechnik GmbH & Co. KG, Ortenburg, Germany; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, Larsen & Toubro Ltd, Kanchipuram, Tamil Nadu, India; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan. Well done to all, and everyone else who had a go.

 

Question 3: River crossing

A father, mother and their two children are crossing a river. Their boat can only take one adult or two children at a time. What is the minimum number of journeys needed to take the family of four across the river?

Answer: As per the first correct reply in from John Droogan, (see also table below from Jose Padron) it took them 9 journeys.

2 children across > 1 child back > 1 adult across > 1 child back > 2 children across > 1 child back > 1 adult across > 1 child back > (and finally) 2 children across.

Well done to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; David Mann, Polymer Business Development, France; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; John Bowen, consultant, Bromsgrove, UK; Jose Padron, spécialiste en développement de materiaux, Waterville TG Inc., Québec, Canada; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Bob Hall, senior director global marketing, Kraton Polymers LLC, Houston, Texas, USA.

 

Question 2: High five

Complete the following series

8,485, 8,516, 8,586, 8,611, _.

Answer: Apologies to those readers who spent time trying to work this one out as a mathematical series. There was, though, a strong clue in the title, as the figures represent the height above sea level of the world’s five highest mountains. Our answer, so, was 8848m for Mount Everest, known as Sagarmatha in Nepal and Chomolungma in Tibet.

Well done to peak performers: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany (thanks for the photo); David Mann, Polymer Business Development, France; Lars Linnemann, R&D manager, Genan A/S, Viborg, Denmark; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India (thanks for the table); Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; John Bowen, consultant, Bromsgrove, UK.

 

Question 1:  Next number

What number comes next in this series?

100, 500, 1, 50, 1000, 5, ???

Answer: The series derives from Roman numerals placed in alphabetical order:

C (100), D (500), I (1), L (50), M (1000), V (5) and then our answer X (10).

Congratulations maximus go to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Amparo Botella, Ismael Quesada SA, Spain; John Bowen, consultant, Bromsgrove, UK; David Mann, Polymer Business Development, France; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA.

 

MARCH 

 

The top prize this month goes to a clear winner and probably our leading specialist in the field of probability. Well earned congratulations go to: Michele Girardi of Scame Mastaf Spa in Italy, the new Brainiac of the Month.

 

Question 4: Degrees of difficulty

Add the next number to this series?

60, 90, 108, 120, 128.5714*, 135, 140, 144, 147.2727*, _

(*5th and 9th numbers are rounded to four decimal places)

Answer: Well done to: David Mann, Polymer Business Development, France (thanks for the drawing below); John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Amparo Botella, Ismael Quesada SA, Spain; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, Larsen & Toubro Ltd, Kanchipuram, Tamil Nadu, India; Stephan Paischer, head of product management special products,  Semperit AG Holding, Vienna, Austria; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA.

As John Droogan noted, these are the angles inside a dodecagon: so, 10 x 180 / 12  = 150. We also liked Bharat B Sharma’s explanation (below) and similar approaches by Amparo Botella and Ramasubramanian P among others:

Sides Polygon          Sum of all angles      Interior angle l

3          Triangle                       180°                             60°

4          Quadrilateral               360°                             90°

5          Pentagon                    540°                             108°

6          Hexagon                     720°                             120°

7          Heptagon                    900°                             128.5714…°

8          Octagon                      1080°                           135°

9          Nonagon                     1260°                          140°

10        Decagon                     1440°                           144°

11        Hendecagon               1620°                           147.2727…°

12        Dodecagon                 1800°                           150

 

Question 3: Teacher teaser II

Question: A follow-on from last week’s ‘divisive question, we asked: There are 12 boys and 8 girls in the class. The teacher selects three children from the class at random to answer a question. What is the probability that the teacher chooses at least two boys?

Answer: A lot more clear-cut this week though only four correct answers suggest this type of question is proving a bit tricky for some. So extra well done to probably our best probability expert Michele Girardi, Scame Mastaf Spa, Suisio, Italy as well as to Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany, Bharat B Sharma, Sr V.P. product development & technical service (elastomers), Reliance Industries Ltd., India and P Ramasubramaniam , Larson & Toubro Ltd,  India, who provided the following detailed analysis.

 

Solution:

Probability of the teacher choosing at least two boys is the sum of probability of choosing all three boys and probability of choosing 2 boys and 1 girl.

Probability of choosing all 3 boys : P(BBB) = (12/20)*(11/19)*(10/18) = 0.193

Probability of choosing 2 boys and 1 girl: P(BBG,BGB,GBB) = (12/20)*(11/19)*(8/18) + (12/20)*(8/19)*(11/18)+(8/20)*(12/19)*(11/18) = (1056*3 )/6840 = 0.463

Probability of choosing at least 2 boys is : 0.193 + 0.463 = 0.656

Using formula:

P(at least 2 boys in a selection of 3) = (12C2 x 8C1 + 12C3 x 8C0)/20C3

= ((12×11/2) x (8/1) + (12x11x10/6)x1)/(20x19x18/6)

= (528 + 220)/1140 = 748/1140 = 0.656

 

Question 2: Teacher teaser I

In introducing herself. a new maths teacher tells her class that she has three children, before asking: if at least one of the children is a boy, what is the probability that she has three sons?

Answer: There were several different interpretations of this question, but the actual answer is 1/7 (or 14.3%), as all possible ways of the teacher having the three children must be considered – (ie BBB, BBG, BGB, GBB, GBG, GGB, BGG, but not GGG).

Extra well done to Michele Girardi, Scame Mastaf Spa, Suisio, Italy, who was the only one to get the correct answer, neatly explaining: There are 8 possible combinations of children, only one has no boys, and only one has 3 boys. So, the probability is 1/(8-1)

 

Tougher bonus question: Dividing time

If the digits 1 through to 9 are randomly arranged to make a number, what is the probability that that number is divisible by18?

Answer: 4/9 or 0.444. Well done to: Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Amparo Botella, Ismael Quesada SA, Spain; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, Larsen & Toubro Ltd, Kanchipuram, Tamil Nadu, India.

Again, Michele Girardi offered a detailed explanation: 18 = 2*3^2, so a number divisible by 18 must be divisible by 2 and 9. Numbers divisible by 2 must terminate by an even number, numbers are divisible by 9 if the sum of their digits is since: 1+2+3+4+5+6+7+8+9 = 45 (= 9×5), all the numbers created by this random arrangement are divisible by 9. Only the ones terminating by 2, 4, 6, 8, are also divisible by 2 and consequently by 18.  The probability of this event is 4/9 = 14.3%. Then the total number of permutations is 9!,there are 8! permutations for each of the 4 even final digits, so the probability is 4*8! /9! = 4/9.

 

Question 1: Factorial five, plus one

Complete the following set of factorials:

5! = 5 x 4 x 3 x 2 x 1=120, 4! = 4 × 3 × 2 × 1 = 24; 3! = ?; 2! = ?; 1! = ?; 0! = ?.

Answer:

5! = 5 x 4 x 3 x 2 x 1=120; 4! = 4 × 3 × 2 × 1 = 24; 3! = 3 x 2 x 1 = 6; 2! = 2 x 1 = 2; 1! = 1 x 1 = 1; 0! = 1

Well done to all the following readers who avoided tripping over the final step: realising that 0! is not zero: Andrew Beasley, product analyst, Hankook Tyre UK Ltd, UK; David Mann, Polymer Business Development, France; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada: John Bowen, consultant, Bromsgrove, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Ramasubramanian P, L&T Rubber Processing Machinery, India; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria.

Among the many neat explanations provided were Andrew Beasley’s concise definition of n! as n! = n*(n-1)! for n > 0 and this longer one from Ramasubramanian P:

Mathematically, factorial of a positive integer n is the product of all the integers less than or equal to n and greater than or equal to 1. The significance of ‘factorial’ is that it represents the number of ways ‘n’ elements of a set can be distinctly arranged in a sequence. For example, for a set with three elements, say A,B,C the number of ways this set can be distinctly arranged is 6: ABC, ACB, BCA, BAC, CAB, CBA. Similarly, two elements can be arranged in two distinct ways. A single element can be arranged in only one distinct way.  And a set with no elements can be arranged in one distinct way, which is again a NULL set.  Hence 0! = 1 too.

 

FEBRUARY

 

Great to have a first-time winner of our top award and so well deserved for managing an extra tricky series of teasers this month, including Question 1, which seemed to throw so many of our top players. Huge congratulations go to Ramasubramanian P of Larsen & Toubro in India, our new Brainiac of the Month.

 

Question 4: Country file

Starting in Turkey and moving west across Europe, find the missing city in the following sequence:

Ankara, _, Oslo, Andorra la Vella.

Answer: These are the names (in English) of capital cities beginning and ending with the same letter. The missing city, so, was Warsaw.

Expertly well done to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; David Mann, Polymer Business Development, France; Ramasubramanian P, L&T Rubber Processing Machinery, Larsen & Toubro Ltd, Tamil Nadu, India. A special mention also to Michele Girardi Scame Mastaf Spa, Suisio, Italy for his help in pointing out a slight blip with the question on Monday.

Bonus question: More soup

As we had two valid answers for Question 3, we ran this on into this week, asking readers to identify what came next in the ‘officially’ correct sequence ABC, BCE, CEH, EHM, _

Answer: HMU came next in the sequence, as David Mann explained: gaps of 5 and 8 according to the Fibonnaci series.

Well done to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; David Mann, Polymer Business Development, France; Ramasubramanian P, L&T Rubber Processing Machinery, Larsen & Toubro Ltd, Tamil Nadu, India; Amparo Botella, Ismael Quesada SA, Spain. A special mention also to Hans-Bernd Lüchtefeld of PHP Fibers GmbH in Germany, who came quite close with his answer (pictured across).

 

Question 3: Alphabet soup

Find what comes next in the following sequence:

ABC, BCE, CEH, _

Answer: As sometimes happens, a valid, alternative answer has emerged for this week’s Brainteaser. So well done to the following readers who worked out the ‘officially’ correct sequence: ABC, BCE, CEH, EHM: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; John Bowen, consultant, Bromsgrove, UK; Michael Easton, sales & marketing director, Globus Group, Manchester, UK; Michele Girardi, Scame Mastaf Spa, Suisio, Italy;

And, equally well done to readers who provided the alternative valid answer EHL: Amparo Botella, Ismael Quesada SA, Spain; Bharat B Sharma, Sr V.P. product development & technical service (elastomers), Reliance Industries Ltd, Vadodara Manufacturing Division, Petrochemicals, Vadodara, Gujarat, India; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Ramasubramanian P, L&T Rubber Processing Machinery, Larsen & Toubro Ltd, Tamil Nadu, India; Varun Sureka, Hartex Rubber Pvt. Ltd, Hyderabad, India.

This sequence was explained concisely by Amparo Botella: ABC, BC(D)E, CE(FG)H, EH(IJK)L

 

Question 2: Flight time

A projectile is fired at a horizontal velocity of 150 metres/second. It hits a target at a horizontal distance of 30 metres. Assuming negligible air resistance effect, what is a) the flight time of the projectile? and b) its vertical displacement?

Answers: Well done to the following readers who answered part (a) correctly ie 0.2 seconds: John Bowen, consultant, Bromsgrove, UK; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; P. Ramasubramanian, Larsen & Toubro Ltd, Rubber Processing Machinery, Tamil Nadu, India: David Mann, Polymer Business Development, France; Olaf Mayer-Mader, Abteilung BA-M, product management and business development PA, Festo AG & Co. KG, Denkendorf, Germany.

And even more well done to those readers who answered the more tricky part (b) correctly ie 0.196 metres (0.2m also accepted): John Bowen; Thierry Montcalm and David Mann.

John Bowen’s solution neatly shows how it’s done:

a] Flight time = 30/150 = 0.2 secs

b] Vertical displacement: we use s = ut + 1/2.f.tsqd = 0 [fired horizontally so no initial vertical component] + 1/2 * 9.8*0.2*0.2 [f = acceleration due to gravity, 9.8m/sec/sec] = 0.196 metres.

 

 

Question 1: City sequence

Fill in the missing cities in the following sequence:

Reykjavík, _, Oslo Montevideo , _, Wellington.

Answer: Just two correct replies, so extra well done to: P Ramasubramanian, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village, Kanchipuram, Tamil Nadu, India; and Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan. Our other Brainiacs really do need to get out more.

As P Ramasubramanian nicely explained: This is a sequence of national capital cities sorted by latitudes.  Northernmost national capital is Reykjavic (capital of Iceland @ 64.14 N) and the southernmost national capital is Wellington (capital of New Zealand @ -41.28 S).

So, the sequence is:

Reykjavic – capital of Iceland (64.14 N), Helsinki – capital of Finland (60.17 N), Oslo – capital of Norway (59.9 N) …  Montevideo – capital of Uruguay (34.88 S), Canberra – capital of Australia (35.3 S), Wellington – capital of New Zealand (41.28 S).

 

JANUARY

 

When the going got tough, Brainiacs around the world really got going, and this month’s award could have gone to three or four top contestants. Both for dealing so expertly with all four questions, I am sure everyone will join in congratulating JOHN BOWEN, our first BRAINIAC OF THE MONTH for 2019

 

Question 4: Next number B

Find the next number in the following series:

123, 354, 897, 1875, 10626, ?

Answer: Many thanks to all our readers who had a go at this one – not everyone got to the correct answer, 16887. Extra well done, so, in order of reply to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Bharat B Sharma, Sr VP product development & technical service (elastomers), Reliance Industries Ltd, Gujarat, India; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village, Kanchipuram, Tamil Nadu, India; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; John Bowen, consultant, Bromsgrove, UK: Andrew Beasley, product analyst, Hankook Tyre UK Ltd, UK; Michael Easton, sales and marketing director, Globus Group, Manchester, UK; Amparo Botella, Ismael Quesada SA, Spain; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan.

Lots of neat workings out, but here’s John Droogan’s reply – within minutes of the Monday send-out – to show how it’s done:

123 > 231 then add 123 gives 354

354 > 543 then add 345 gives 897

897 > 978 then add 897 gives 1875

1875 >8751 then add 1875 gives 10626

10626 > 06261 then add 10626 gives 16887.

 

Question 3: Football focus

In tests for the next World Cup, a football is modelled as a particle and air resistance is ignored. A player kicks the ball from a height of 0.6m above the ground, propelling it vertically upwards at a speed of 10.5m/s. From the modelling, what is the greatest height above the ground reached by the ball, and calculate the length of time the ball is more than 2m above the ground.

Answer: Said it might get tougher and it did with only three correct replies to this week’s teaser, from: John Bowen, consultant, Bromsgrove, UK: Bharat B Sharma, Sr VP product development & technical service (elastomers), Reliance Industries Ltd, Gujarat, India; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA. Their workings out of the correct answers 6.22 metres and 1.86 seconds are each given below:

John Bowen

This problem uses the basic Laws of Motion:

i] To calculate height reached – ie when the ball stops in its vertical trajectory we use

v sqd – u sqd = 2fs, where f = acceleration due to gravity [ in this case -9.81m/sec/sec] and s = distance, u = initial velocity, v = final velocity, so

0 – 110.25 = -2*9.81*s

s = 5.62, so total height = 5.62 + 0.6 = 6.22m

ii] to determine the time above 2m we need to solve a quadratic of the form s = ut + 1/2f*t sqd where t = time in seconds; s = 1.4 [as it starts at a height of 0.6m]

so 1.4 = 10.5*t – 1/2*9.81*t sqd

solving for t gives values of 1.99 and 0.143, so the ball is above 2m height for 1.85 seconds

Greatest height above the ground reached by the ball = 6.225m

Total time the ball is more than 2m above the ground.= 1.857 Sec. (gravity considered as 9.8m/sq.second)

Bharat B Sharma  

1  Initial height = 0.6                                                               

v2 = u2 + 2as          v=0                                                      

u=10.5m/s                                           

(10.5)2= 2×9.8xs     a= (-9.8m/sec2)                                             

s= (10.5×10.5/2×9.8 )+0.5 m(Initial height)                                     

s=5.625m+0.6m   =  6.225m                                                                            

2. Time require to keep ball above 2 m from ground (upward 2m to 6.225m and return)

Distance covered  4.225 m upward and 4.225 m down ward                               

Going up  time        Final speed = 0, s= 4.225 m, a =-9.8m/sq.sec==                   

Initial speed at 2 m = sq root of 2×9.8×4.225= sq rt of 82.81 m/sec

                    =9.1 m/sec                                                     

Time taken to trave 4.225 m= (V=u+at) =0=9.1-9.8t)                      

                    = time = 9.1/9.8= 0.9286 sec                                               

Going down = distance covered 4.225m, Initial speed is 0 and a=9.8m/sec2

v sqare=u sq+2as    u=0, a=9.8m/sq.sec, s=4.225)                         

v sq=  82.81                                         

v        9.1m/sec                                             

Time for return journey will also be 0.9286sec                                

Total time the ball will be above 2 meter height =2*0.92857=  1.857 sec.

Paul Knutson

Greatest height = 6.22 m.  Time above 2 m is 1.86s.  All derived from formula y=v0*t-.5*g*t^2

 

Question 2: Next number A

 Find the next number in the following sequence: 10, 9, 17, 50, 199, _

Answer: Clues were offered but not really needed as correct replies rolled in, including Michele Girardi’s a(i+1)=a(i)*i-1 making the sequence: 1,10; 2,9; 3,17; 4,50; 5,199; 6, 994. Well done in order of reply to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Andrew Beasley, product analyst, Hankook Tyre UK Ltd, UK; John Bowen, consultant, Bromsgrove, UK: Dave Stuckey, Dow, South Wales, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Bharat B Sharma, Sr VP product development & technical service (elastomers), Reliance Industries Ltd, Gujarat, India; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Michael Easton, sales and marketing director, Globus Group, Manchester, UK; Jon Cutler, materials development manager, Trelleborg Sealing Solutions, Tewkesbury, UK; Martin Jones, customer service manager, network logistics & transport, DHL Supply Chain UK; Amparo Botella, Ismael Quesada SA, Spain; Thierry Montcalm, R&D manager, Soucy Techno Inc., Canada; Trey Thies, growth strategist, engineered performance products, Milliken & Co., Spartanburg, South Carolina, USA; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; David Mann, manager rubber technology, SI Group, Béthune, France.

 

Question 1: Rubber connect

Find the rubber connection in: England, Peru, Denmark, Malaysia.

Answer: A relatively easy one to start the year, as long you were not thrown off track by ‘Malaysia’ – things might get tougher as we go along. Well done to the following readers who identified EPDM in the country names: David Mann, manager rubber technology, SI Group, Béthune, France; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; John Bowen, consultant, Bromsgrove, UK: France Veillette, chef environnement, usine de Joliette, Bridgestone Canada Inc. Canada; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands.

 

Brainteaser Awards and Results 2018