# ERJ Brainteaser: September

31 Aug 2020

It was well past midnight before our judging panel concluded hours of heated debate over this month’s award. Eventually Q2 and Q5 were key to deciding between six candidates. So, top honours went to **Stephan Paischer** of Semperit and **Michele Girardi** of Scame Mastaf – as well as to, as a newcomer, **Natalie Prescott** of Prescott Instruments for her impressive solution to Q1. Well done to all three new joint holders of the **Brainiac of the Month** title.

How many five-digit numbers are there that do not contain the numbers 3 and 5, and are multiples of 4? (It cannot start with 0, ie. 01234 is not a 5 digit number)

**Answer**: As the solutions below explain he answer is 9408. Very well done to, in order of correct reply: **Stephan Paischer**, head of product management special products, Semperit AG Holding, Vienna, Austria; **Michele Girardi**, Scame Mastaf Spa, Suisio, Italy; **Jose Padron**, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; **John Bowen**, rubber industry consultant, Bromsgrove, UK; **Hans-Bernd Lüchtefeld**, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany. And, as this was quite a tricky teaser, we should also give an honourable mention to the following readers who came very close and/or were on the right track: **Laurent Simonet**, Michelin, France; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; **Michael Easton**, sales and marketing director, Globus Group, Manchester, UK.

*Solutions:*

*John Bowen
To be divisible by 4 there is a choice of 21 possible combinations of the last 2 digits if 5 and 7 are excluded
Possible first digits are 1to 9 excluding 5 and 7 , ie 7
Possible second and third digits are 8 [ie 0 to 9 less 5 and 7]
And there are 21 possible combinations for [fourth and fifth] digit.
So possible numbers is given by 1st x 2nd x 3rd x [4th + 5th] = 7 x 8 x 8 xd 21 = 9408*

*Stephan Paischer
In total, there are 90.000 numbers (five digits), thereof 70.000 without 3 or 5 as the first digit. 25% of them are divisible by 4 (very 4th one).
Of these 17.500, 80% have no 3 or 5 in the 2nd digit, the same for the 3rd digit. Of the remaining 11.200 numbers, 4 out of 25 need to be removed (for numbers ending with 32, 36, 52, 56). This leaves 84% of 11200 -> 9408.*

*Michele Girardi
The first digit is a number from 1 to 9 excluded 3 and 5, there are 7 possibilities
The second and third digits are numbers from 0 to 9 excluded 3 and 5, there are 8 possibilities- the 4th and 5th digit must form a multiple of 4, there are 25 possibilities, but we must exclude 32,36, 52 and 56, so the possibilities are 21.
The numbers are therefore 7x8x8x21 = 9408*

*Jose Padron
There are 9408 5-digits numbers with the mentioned rules, no number 3, no number 5 and multiples of four.
Well, using an Excel worksheet,
1. Starting with 10000 and concluding with 99999
2. Eliminating all number containing 3 and 5
3. Making the rest of the number divide by four
4. Eliminating all fractional results and keeping only integer numbers.
The result is 9408 different numbers.*

New teaser on Monday

What number comes next in the sequence:

61, 691, 163, 487, 4201, ?

**Answer**: The sequence consists of the prime numbers which, when their digits are reversed, are perfect squares. So next in the series is 9631.

So, it’s very well done, in order of reply, to: **John Bowen**, rubber industry consultant, Bromsgrove, UK: **Stephan Paischer**, head of product management special products, Semperit AG Holding, Vienna, Austria; **Michael Easton**, sales and marketing director, Globus Group, Manchester, UK; **David Mann**, key account manager, SPC Rubber Compounding, UK; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; and – with a welcome back to a former Brainiac champion – **Hans-Bernd Lüchtefeld**, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany.

New teaser on Monday

What is the next number in this series?

10, 9, 17, 50, 199, ?*.*

**Answer**: With a clue in the title, easy if you know how – 994. Well done to: **Duncan Thomson**, general manager, Wuxi Elbe Polymer Technology Ltd, China; **Michael Easton**, sales and marketing director, Globus Group, Manchester, UK; **Andrew Knox** Rubbond International, Ohé en Laak, The Netherlands; **Stephan Paischer**, head of product management special products, Semperit AG Holding, Vienna, Austria; **John Bowen**, rubber industry consultant, Bromsgrove, UK; **Jose Padron**, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; **Daniel Willrich**, redakteur, büro Köln, AutoRäderReifen-Gummibereifung, Köln, Germany; **David Mann**, key account manager, SPC Rubber Compounding, UK; **Dr. Katharina Gottfried**, technical manager, Arlanxeo Deutschland GmbH, Köln, Germany; **Carlo Nolli**, sales manager, PMG SpA, Cenate Sotto – BG, Italy; **Sunil Sampat**, CEO, Sonata Rubber Pvt Ltd, Mumbai, India.

*Some neat solutions, including these examples:*

*David Mann
If we label the numbers A to F,
A = 10
B = (A x 1) - 1
C = (B x 2) - 1
D = (C x 3) - 1
E = (D x 4) - 1 = 199
So, F = (E x 5) - 1 = 994*

*Daniel Willrich
The next number in the series should be 994 (5x199-1). Starting from the 9 it’s the number x2-1, then x3-1 and x4-1.*

New teaser on Monday

After a production-line problem, the QC department has found that 5% of the tires being made at a factory are faulty. To track down the fault technician Terry collects samples of tires in batches of 20, which are randomly selected. What is the probability that a batch contains no faulty tires?

**Answer**: The probability of readers coming up with the answer to a fairly tricky stats question seems not so high, with only a select trio here correctly replying 35.8%. Extra well done, so, to: **Michele Girardi**, Scame Mastaf Spa, Suisio, Italy; **Andrew Knox** Rubbond International, Ohé en Laak, The Netherlands; **Stephan Paischer**, head of product management special products, Semperit AG Holding, Vienna, Austria.

*Two nice solutions shown:*

*Michele Girardi
Since there is no indication on production size we assume the binomial distribution .
The probability that one sample of 20 contains no faulty tires is:
(0,95)^20 = 0,358 = 35,8%*

*Andrew Knox
Probability that a random sample of 20 tyres contains no faulty tyres is 0.35849 (rounded off) 5% of the tyres are faulty.
So the chance of picking a faulty tyre is 1/20 and chance of picking a good tyre is 19/20 or 0.95.
So, the chance of picking a good tyre 20 times in a row is is 0.95 x 0.95 x....20 times = approx 0.35849*

*New teaser on Monday*

Riley rolls three dice. What is the probability that the numbers (or number of dots) showing total 5?

**Answer**: The probability is 2,78% or 6/216 (1/36), ie six possible combinations over 216 total combinations. Very well done to: **John Bowen**, rubber industry consultant, Bromsgrove, UK; **Michele Girardi**, Scame Mastaf Spa, Suisio, Italy; **Natalie Prescott**, product manager, dynamic testing, Prescott Instruments Ltd, Tewkesbury, Glos., UK; **Andrew Knox** Rubbond International, Ohé en Laak, The Netherlands; **Stephan Paischer**, head of product management special products, Semperit AG Holding, Vienna, Austria;

*Solution*

*Some neat working out on this one, but our judges particularly liked Natalie Prescott’s detailed explanation:*

*With three dice, the probability of any outcome is 1/6 x 1/6 x 1/6 = 1/216*

*For a sum of 5, we must have a group of either [1, 1, 3] or [1, 2, 2]*

*Next we work out the permutations for each group, i.e. how many different ways this can be thrown:*

*[1, 1, 3], [1, 3, 1], [3, 1, 1], [1, 2, 2], [2, 1, 2], [2, 2, 1] = 6 different ways*

*This can also be worked out using the combination function C(n, r). E.g. for [1, 1, 3] there are: C(3, 1) x C(2,2) = 3 x 1 = 3 ways*

*Therefore, there are 6 different ways to throw three dice that sum to 5*

*The probability of this is 6 x 1/216 = 1/36 = 2.78%*

*Note on combination function: for [1, 1, 3] the 3 can go in any of the three spaces, and the two 1’s can go in the two remaining spaces. *

*Because the two 1’s are the same, there is only one way to place them in those two spaces. Therefore there are 3 x 1 = 3 ways to throw [1, 1, 3].*.

New teaser on Monday