ERJ Brainteaser: September

Can you start off this series, written to three decimal places?

__, 0.750, 0.833*, 0.875, 0.900, 0.916*.

*Last digit is recurring 

.For his admirable hand-written and drawn answer to (hotly contested) Q1, the top award goes to Rohit Kale of Michelin, our new Brainiac of the Month. 

Answer: This is a series of fractions 1/2, 3 / 4 , 5 / 6, 7 / 8, 9 / 10 and 11 / 12 written in decimal - making the answer 0.500. On the mark this week were: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Rohit Kalé, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., USA: Michele Girardi, Scame Mastaf Spa, Suisio, Italy. Well done to all five, and everyone else who had a go. 
 

Solutions

John Bowen 
When the inverse of these numbers is calculated, they become 1.33,1.2, 1.14,1.11, 1.09  [recurring] which may be written as  fractions 1 1/3, 1 1/5, 1 1/7, 1 1/9,  so the inverse of the  first one would be 1 1/1  or2,  so the first one would be 0.500.

Stephan Paischer
Expressed as fraction, the series goes as follows:
¾, 5/6, 7/8, 9/10, 11/12.
Both numerator and denominator are increasing by 2 for every step, so for the initial step we decrease both by two (from ¾): that gives ½, so 0,500.

Michele Girardi 
Every number  of the series is an odd number divided by the following even number
1/2 =      0,50000
3/4 =                 0,75000
5/6 =                 0,83333
7/8 =                 0,87500
9/10 =     0,90000
11/12 =     0,91667

Rohit Kalé
When you convert all the decimals into fractions, you would see the pattern in the fractions where for each next decimals both the Numerator and the denominator increases by 2.
Hence from 0.750 which converts into ¾, if you subtract 2 from both from Numerator and denominator it comes out to be ½, which is 0.500 as a start of the series. 

New teaser on Monday

 

 

To divide a group of 250 students into more manageable numbers, they were asked to write down a number between 1 and 9 inclusive. Those that wrote down a prime number were asked to leave. Of the remaining 70 students, how many would be expected to have written an even number?

Answer: Sometimes our Brainteaser questions (unintentionally) generate some great debate –  see ‘solutions’ below. The official answer is 42, as neatly worked out by: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; and (‘not’) Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands.  

Many thanks also to John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Giorgio Marzari, sales director, RDC srl, Bareggio (Mi), Italy; and everyone else who contributed to this week’s ‘debate’.

Solutions

Stephan Paischer 
The 70 remaining students have written down non-prime numbers as follows: 1, 4, 6, 8, 9
Assuming even distribution, 3 out of five students have written an even number (4, 6, 8):
70 / 5 x 3 = 42

Michele Girardi
The answer is 42, that can be calculated as written below.
Assuming that 1 isn't a prime number, between 1 and 9 the prime numbers are 2,3,5,7 and the non-primes are 1,4,6,8 9 , of which the even numbers are 4,6,8 , in proportion  3/5 .
So, on the remaining 70 students, 3/5*70=42 would be expected to have written an even number. 

Michele Girardi
This is an addition to the  answer sent in previous email  .
The bonus question may be "is the selection fair ?", since there is a probability over one million that less than 100 students remain , supposing that they vote randomly with probability 5/9 of  success

Andrew Knox 
Answer: The answer is NOT 42!
Asking 250 students to write down a number is not a random process.
If it had been random selection, then statistically only 250 x 4/9 (as 4 of the 9 numbers are prime) = 111 students would have been expected to pick a prime number, whereas 250 - 70 = 180 students actually did. (7, for example may have been a lucky number for many of them).
If the 70 students who picked a non-prime number (1, 9 being odd and 4, 6, 8 being even) were to have selected these numbers randomly, then 70 x 3/5 = 42 would be expected to have chosen an even number. However, this is not a random selection. A disproportionate number of students might, for example, have selected 1 or 9, simply because these numbers were mentioned in their instructions.
So, the answer is not 42, although it could be…

Giorgio Marzari
Statistically, the question is about expectation… but it could also be that none of them has written 9 or may have all written 9.

New (more straightforward) question on Monday.

 

 

Can you find the missing number in this series?

34, 55, 89, 144, _, 377

Answer: Each number is the sum of the two previous ones, as in a Fibonacci sequence, making the answer 233. Very neatly worked out, in order of reply, by: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Carlo Nolli, PMG Compounds, Italy; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada; Rohit Kalé, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., USA: Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Michael Easton, director, Globus Group, Trafford Park, Manchester, UK; Flemming Vorbeck, purchasing manager commercial, NDI Group A/S, Brørup, Denmark; Bart Ezendam, general manager, Celerity DRS GmbH, Recklinghausen, Germany; Mark Bussy, M. Rothschild & Co. (agent representing NR shippers/producers), Spring Lake Heights, New Jersey, USA. Well done to all, and everyone else who had a go.

Solutions

Jose Padron

The missing numbers is 233 following the Fibonacci’s sequence 

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610… etc.

Right now, there are Fibonacci sequences in my garden in flowers like this

Rohit Kalé 

This is a Fibonacci numbers sequence. In 1200 century, it was first created by Italian mathematician Leonardo Fibonacci who popularised the Hindu numeral system, which was adopted by the Arabic world and then by the Western world. 

F0 = 34
F1=55
F2 = 89
Logic Formula : Fn=Fn?2+Fn?1
Hence :  
F2= F0-F1 => 34+55= 89
F3 = F1+F2 => 55+89= 144
F4=F2+F3=> 89+144= 233
F5=F3+F4 => 144+233=377

Michele Girardi 

This  is a part of the  Fibonacci series, where every number is the sum of the previous  two, 

1
1
2 = 1+1
3 = 1+2
5 = 2+3
8 = 3+5
13 = 5+8
21 = 8+13
34 = 13+21
55 = 21+34
89 = 34+55
144 = 55+89
233 = 89+144  <===
377 = 144+233

Andrew Knox

Each number in this series is the sum of the previous two, so:
34 + 55 = 89,
55 + 89 = 144,
89 + 144 = 233,
144 + 233 = 377

New teaser on Monday.

 

From his fishing trawler, skipper Brad electronically measures the angle of elevation of the top of a lighthouse as 15 degrees. He sails 600 metres directly towards the lighthouse and then e-measures the elevation of the top of the cliff as 45 degrees.

What is the distance from the second position of the yacht to the top of the lighthouse and what is the height of the lighthouse?

Answer: It was the quality of the catch, rather than the quantity, that impressed this week as the correct solution was ‘one that got away’ for many readers. Very well done, so, to the following five, who correctly answered 220m and 311m (with point taken about confusion around the wording of this question): John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Rohit Kalé, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., ISA: Michele Girardi, quality manager at 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.

Solutions

Rohit Kalé

Michele Girardi 

This problem is undefined. Calling X the height of the cliff, which is equal to the projected distance, H the height of the lighthouse and D the distance, one can calculate infinite solutions
tg(15)=    0,267949192

    Heigth of lighthouse    distance
X    H=(600+X)*tg(15°)-X    D=Sqroot(X^2+(X+H)^2)
100    87,6                     212,6
110    80,2                     219,8
120    72,9                     227,2
130    65,6                     234,9
140    58,3                     242,7
150    51,0                     250,8
160    43,6                     259,0
170    36,3                     267,3
180    29,0                     275,8
190    21,7                     284,4
200    14,4                     293,2
210    7,0                     302,0
220    -0,3                     310,9

Stephan Paischer

-    Sum of 3 angles of a triangle is 180°, so the angles are 15, 135 and 30°

-    Split of the triangle in 2 triangles with a 90° angle

-    Calculate internal triangle height h with sin(15) = h / 600

-    Calculate triangle side (1st pos – lighthouse) by using h and sinus

-    Calculate last missing side of triangle (2nd pos – lighthouse) as a²+b²=c²

-    Calculate lighthouse hight as sin (45) = h1 / 310

Andrew Knox

If after sailing 600 metres directly towards the lighthouse Brad were instead to have re-measured the angle to the top of the lighthouse again, then the height of the top of the lighthouse above sea level could be easily determined by simple trigonometry. However, not knowing how far the lighthouse is set back from the edge of the cliff, nor knowing anything about the lie of the land behind the cliff face where the lighthouse is standing, all Brad will know I'm afraid is that at an angle of 45 degrees to the top of the cliff, he is getting dangerously close to the shoreline.

If the question were re-phrased to say that he re-measures an angle of 45 degrees to the top of the lighthouse, then if X is the horizontal distance to a point directly below the top of the lighthouse, and Y is the height of the top of the lighthouse above sea level, and we ignore the curvature of the Earth and assume the height of the measuring point is at sea level, then the calculation becomes simply:

Tan 15 = 0.2679 = Y/(600 + X)

Tan 45 = 1 = Y/X, so Y = X.

Substituting, Y/(600 + Y) = 0.2679, so Y = X = 219.56 metres.

So, the original position of the trawler horizontally was 819.56 metres from a point directly below the top of the lighthouse, the top of the lighthouse being 219.56 metres above that point at sea level.

The distance from the second position of the trawler to the top of the lighthouse (assuming that the measuring point is at sea level) is then cos 45 = adj/hyp so hyp = 219.56/.7071 = 310.51 metres to the top of the lighthouse.

New teaser on Monday.