# ERJ Brainteaser: March 2019

25 Sep 2019

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**.

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: 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**

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/

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.*