27 Sep 2019
SEPTEMBER 2019
This month's title for Brainiac of the month goes jointly to four people who received special mention for sending 'fantastic' workings out for our very tricky Q3: Ramasubramanian P, Andrew Knox, Michele Girardi and Jose Padron
Q4: Simon's steps
Answer: 49, although those who answered 48 could be correct, as pointed out by some readers. Well done to Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; John D Burrows, textile consultant, France; John Bowen, consultant, Bromsgrove, UK; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Fariha Rashid, marketing analyst, Kraton Polymers LLC , Houston, Texas; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Ricardo Azcarate; 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; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada;
Andrew Knox:
If M = middle step, then top step is M+8-12+1+(9*3)= M+24. So there are 24 steps up from the middle, and 24 steps down from the middle, total 48 steps.
For the calculation, the middle step doesn't count. There is no middle step on any odd number of steps, only on even numbers of steps.
Question 3: Pie ‘r squared
Answer: Another tasty teaser, with the answer working out at 10,39 and 35. Very well done to Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village, Kanchipuram, Tamil Nadu, India; John D Burrows, textile consultant, France; Fariha Rashid, marketing analyst, Kraton Polymers LLC, Houston, Texas, USA; Neha Kaushik, SRF TTB-C, India; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada – and everyone else who had a go.
Special mentions for some fantastic workings out sent in by: Ramasubramanian P, Andrew Knox (shown on wallpaper). Michele Girardi (who pointed out that this can be solved using the Erone/Heron formula area ^2 = p*(p-a)*(p-b)*(p-c), where p = (a+b+c)/2 (semiperimeter)) and Jose Padron for the following:
From data
Shortest side is 10 cms x
longest side is 39 cms y = x+29
third side is 35 cms z = 2x+15
triangle perimeter is 84 cms P = x+y+z = x+x+29+2x+15 = 4x+44
triangle area = 168 cms A = 2P
From Heron's formula, when height is not known
half perimeter s=½*(x+y+z) = 84/2 = 42
triangle area A' = [s(s-x)(s-y)(s-z)]½ = 168
Question 2: Piece of cake
Pete has always had the same number of candles as his age on each of his birthday cakes. If he has had exactly one birthday cake for each year of his life and blown out a total of 861 candles, can you quickly work out how old Pete is?
Among the neat working out was:
Michele Girardi: The general formula for the sum of numbers 1..n is n(n+1)/2 the equation n(n+1)/2 = 861, giving the solution n=41…
Andrew Knox: Sum of a simple arithmetic sequence of n terms is n/2(2a + (n-1)d), where a is the first term (here 1), and d is the common difference (here also 1), and n=Pete's age. This simplifies to:
Sum(1:n) = n/2(2+(n-1)), or Sum(1:n)*2 = n*(n+1)
So, 861 *2 = 1722 = 41*42, so n +41
Bharat B Sharma: Series is 1+2+3+4+………n (all natural numbers) Given --- 1+2+3+……n = n/2 (n+1) = 861 candles
Equation to solve n2 +n= 1722
n2 +n-1722 =0 and (n+42)(n-41)=0
So n= (-42) or 41
Ramasubramanian P: Let Pete’s age be X. For X birthdays he would have blown 1+2+3+…+X candles.
So, sum(1:X) = 861
ð X*(X+1)/2 = 861
ð X*(X+1) = 1722
ð X = 41
Question 1: You cannot be series!
What comes next in this sequence: 2, 6, 42, 1806, _?
Answer: Great to see many in such fine form this week – sign of a well-earned holiday. The answer as explained by several refreshed readers was:
2*(2+1) = 6; 6*(6+1) = 42; 42*(42+1) = 1806; and 1806*(1806+1) = 3263442
Very well done in order of reply to: Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; John Bowen, consultant, Bromsgrove, UK: David Mann, Polymer Business Development, France; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; 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; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada: John D Burrows, consultant, France; Amparo Botella, Ismael Quesada SA, Spain; Yoganand Nannapaneni, Mascot Systems Private Ltd, Mumbai, India; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Mario Swaanen, technical operations manager, Gradient Compounds Netherlands BV, Hilversum, The Netherlands; Ashley Fahey, sustainability principal vice president, Goodyear Pride Network, Goodyear Tire & Rubber Co., Akron, Ohio, USA; Yoganand Nannapaneni, Mascot Systems Private Ltd, Mumbai, India;
AUGUST
For his extra-sharp reply to Question 2 and excellent performance throughout August, Andrew Knox of Rubbond International is the very worthy winner of our Brainiac of the Month title.
Question 5: Come fly with me
Answer: Well done to the following readers who gently landed on the correct answer, 75 minutes: Bharat B Sharma, Sr V.P. Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; John Bowen, consultant, Bromsgrove, UK; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; John D Burrows, (details not supplied); Fariha Rashid, marketing analyst, Kraton Polymers LLC , Houston, Texas; 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.
As Bharat B Sharma neatly worked out:
1.25 Hours (or 75 minutes)
Both planes speed== 120mph
Dons’ plane speed—120-5mph== 115 mph
Pete’s plane speed – 120+5mph== 125 mph.
Time when both plane will meet
115xt + 125xt = 300 (total distance)
240t=300
T= 1.25 hours (75 minutes)
Question 4: Tire collection
Frank, a tire enthusiast, has a collection of various types of tires in his garage. When his wife asked how many tires he had, he replied: "Well! If I divide them into two unequal numbers, then 32 times the difference between the two numbers equals the difference between the squares of the two numbers." How many tires does Frank have?
Answer: 32. Well done to Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany. Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; John Bowen, consultant, Bromsgrove, UK; David Mann, Polymer Business Development, France; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; 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; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Yoganand Nannapaneni, Mascot Systems Private Ltd, Mumbai, India; John D Burrows.
The formula as provided by John Burrows:
32 (x-y) = (x2 – y2)
32(x-y) = (x+y).(x-y)
So 32 = x+y
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...)
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
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?
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
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?
(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 24x24 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?
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?
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:
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.
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
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
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.
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.
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.
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
= ((12x11/2) 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 (= 9x5), 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
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.
Question 3: Alphabet soup
Find what comes next in the following sequence:
ABC, BCE, CEH, _
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.
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.
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.
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.
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= 2x9.8xs a= (-9.8m/sec2)
s= (10.5x10.5/2x9.8 )+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 2x9.8x4.225= 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.
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