ERJ Brainteaser: August

Question 4: Complete the series

B, C, F,  H, I ….

Answer: Ooops, apologies for a couple of blips that made it trickier than intended to get to the correct answer: an alphabetical sequence of single-letter elements in the periodic table: B, C, F, H, I, K, N, O, P, S, U, V, W, Y. Undaunted, nevertheless, were: Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Andy Longdon, technical sales manager, Ceetak Ltd, Bedford UK; Sudi Sudarshan, principal consultant, Global Mobility Strategies, USA. Well done to all, as well as to John Bowen, consultant, Bromsgrove, UK, for his inventive, alternative solution: 

I think the missing element is O, Oxygen, the only missing single letter element from the first row of the Periodic Table. W, Tungsten has an atomic number of 74, the sum of the first 5 elements given.

New teaser on Monday

Email your answer: correct replies on Friday.

One, three, eleven, ?, ...

Answer: This was easy enough, but only if you recognised that the teaser was about the number of times the letter ‘e’ appears (see Solutions below). Very well done to our select set: Sudi Sudarshan, principal consultant, Global Mobility Strategies, USA; John Bowen, consultant, Bromsgrove, UK; Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Andy Longdon, technical sales manager, Ceetak Ltd, Bedford UK; and everyone else who had a go.

SOLUTION

Sudi Sudarshan

Each member of the set is the next integer expressed in word form that has an additional "e" versus the previous integer

1 one - one "e"

3 three - two "e"s

11 eleven - three "e"s

17 seventeen - four "e"s

John Bowen

I think that next is :  seventeen

As we go :one.three, eleven there are 1,2,3 '  'e's in. each number. The first number with 4 'e's is seventeen.

Kamila Staszewska

One (1 e), three (2 e’s), eleven (3 e’s).

Next number with 4 e’s would be seventeen.

Andrew Knox

Answer: Seventeen

These appear to be a sequence of the English spelling of numbers in which the number of e's increases each time by one, i.e. 1 'e' in one, two 'e's' in three, etc.. The next number in the series after seventeen will need 5 'e's', so will probably need to be as high as one hundred and seventeen.

Amparo Botella

The answer is seventeen..

One – 1 e

Three – 2 e

Eleven – 3 e

Seventeen  - 4 e

Andy Longdon

Or when is a maths problem not a maths problem ….

One, three, eleven, seventeen, one hundred and eleven, one hundred and seventeen

The next in a simple number sequence that when written (in English) has 1 letter e, 2 letter e’s, 3 letter e’s 4 letter e’s (seventeen), 5 letter e’s (one hundred and eleven), 6 letter e’s (one hundred and seventeen) etc…

Next teaser on Tuesday - due to a UK bank holiday.

Fire officer Fred noticed a blaze in the extrusion area of the rubber processing factory. From his list of all 142 employees at work that day, he phones two, alerts them, asks them to phone two more people before getting out of the factory. Luckily, everyone answered the phone immediately and got out safely – even though each phonecall took 30 seconds, and it took each employee 90 seconds to get out of the building. How long did it take to evacuate all 142 workers from the factory?

Answer: The official answer is 6 minutes and very well done to Sudi Sudarshan, principal consultant, Global Mobility Strategies, USA for getting everyone out right on time. But given the scope for interpretation, it's equally well done to: John Bowen, consultant, Bromsgrove, UK; Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan; and Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands. See the excellent Solutions below.

SOLUTIONS

Sudi Sudarshan

Assume the total number of employees of 142 includes Fire Officer Fred.

Here is a chronology of people notified (by Employee Number, assuming Fred is E1)

0.5 min E2 notified by E1

1.0 min E3 notified by E1

             E4 notified by E2

1.5 min E5 notified by E2

             E6 notified by E3

             E7 notified by E4

2.0 min E8 notified by E3

             E9 notified by E4

             E10 notified by E5

             E11 notified by E6

             E12 notified by E7

2.5 min E13 notified by E5

             E14 notified by E6

             E15 notified by E7

             E16 notified by E8

             E17 notified by E9

             E18 notified by E10

             E19 notified by E11

             E20 notified by E12

3.0 min ...

..

..

Number of additional employees (A) and Cumulative (C) notified at the end of

0.0 min - A=1 C=1

0.5 min - A=1 C=2

1.0 min - A=1+1=2 C=4

1.5 min -A=1+2=3 C=7

2.0 min -A=2+3=5 C=12

2.5 min - A=3+5=8 C=20

3.0 min - A=5+8=13 C=33

3.5 min - A=8+13=21 C=54

4.0 min - A=21+13=34 C=88

4.5 min - A=34+21=55 C=143

Interesting observation: The number of additional employees added each 0.5 minutes follows a fibonacci sequence: 1,1,2,3,5,8,13,21,34,55

At the end of 4.5 minutes all employees would have been notified. It will take another 1.5 minutes for the last set of employees to evacuate the building.

So it would take 6 minutes to evacuate all 142 workers from the factory.

Note: Even if there are 142 employees in addition to Fred, it would take the same amount of time as 143 employees can be evacuated in 6 minutes

John Bowen

First we need to know whether Fred took part in the cascaded phone call scenario.  If not then the numbers are as follows:

Round 1:    Number phoned  2:                 Total number phoned       2

            2                                4                                                           6

            3                                8                                                         14

            4                               16                                                        30

            5                               32                                                        62

            6                               64                                                      126

            7                             128[possible]                                       142 - all evacuated

Each round takes 2 minutes in total, so all 142 will take 14 minutes to safely leave the factory.

If Fred called 2 further people then the numbers are as follows:

Round 1:     Number called:  2 + 1          Total  number called  3

           2                                   6                                                9

           3                                 18                                              27

           4                                 54                                              81

           5                               162                                             142 - all evacuated

In this case total evacuation will take 10 minutes,

Andrew Knox

Answer: Exactly 5 minutes.

For Fred to call 2 fellow workers takes 30 seconds.

They then call 4 workers taking another 30 seconds.

So, in each 30 second interval, 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 people will/can be reached, so after 3 minutes Fred will have reached 126 people, and after 3 minutes and 30 seconds will have covered all 142 workers in the factory. Assuming they all call from their own mobile phones while walking, the last 16 workers to be called will need a further 90 seconds to leave the building, making 5 minutes in all. 

Kamila Staszewska

Time=0s -  Fred (1) calls worker no. 2 (30s) and no. 3 (60s) and exits building (2.5min)

To trace how long does it take for all workers to exit the factory we need to focus on the last person notifed:

Time=60s - Worker 3 calls worker no. 7 (3+2*2) (60+60=2min) and exits building (2min +90s = 3.5min)

Time = 120s - Worker 7 calls worker 15 (7+4*2) (120+60=3min) and exits building at 4.5min

Time = 4min - Worker 15 calls worker 31 (15 + 8*2) (4min+60s=5min) and exits building at 6.5min

Time=5min - Worker 31 calls worker 63, 6min and exit at 7.5

Time =6min - worker 63 calls 127, 7 min and exit at 8.5

Time = 7min - worker 142 call is finished at 7.30 and at 8 min he exits the building.

 New teaser on Monday.

Magicians Alan, Brian and Chris each hold five tokens in a hat: each with the same five colours. Each magician takes a token randomly from the other two hats and puts them into his own hat.  What is the probability that Alan, Brian and Chris each end up with five different coloured tokens in their hats?

Answer: There might well be a blip somewhere in the logic of this question – in which case apologies. Nevertheless, while differing from the official answer, this teaser generated splendid efforts from two of our top Brainiacs (see Solutions below). Extremely well done to: Sudi Sudarshan, principal consultant, Global Mobility Strategies, USA; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; and everyone else who had a go.

SOLUTIONS

Official answer

((1/5)*(3/7) + (4/5)*(2/7 + (2/7)*(1/4))) * (1/9)

= (3/35 + (4/5)*(2/7 + 1/14)) * (1/9)

= (3/35 + (4/5)*(5/14)) * (1/9)

= (3/35 + 2/7) * (1/9)

= (13/35) * (1/9)

= 13/315

= 0.0413 approx.

Sudi Sudarshan

My answer to this week's brainteaser: 9/320 or 0.028125

Let the five token colors that Alan(A), Brian (B) and Chris (C) have be p,q,r,s,t

Assume that the tokens are picked by each from the original 5 tokens each has in their hat.

A picks first and selects, say p and q from B and C respectively. It doesn't matter what colors he picks. So probability of picking specific colors he needs is 1

B picks next. He now needs to pick Color p from either A or B, in order to replace the p token A took from him; the other token he picks can be of any color.

Probability of picking p from A = 1/5; probability of picking p from B = ¼ (since A has already taken a token from B). Combined probability is  1/4+1/5 = 9/20

Let us say that the second token that B picks is r (it doesn't matter what color). Probability of B picking any colored token as their second token is 1. Combined probability = (9/20)*1 = 9/20

C has to pick the specific colors q from A and r from C. Since A and B have 4 original tokens, the probability of each of these selections is 1/4. The combined probability is (1/4)*(1/4) = 1/16

The combined probability of all the selections to occur as above in order that each magician has all the five colors as at the beginning = (1)*(9/20)*(1/16) = 9/320 = 0.028125

Andrew Knox

With only three rounds of exchanges of tokens it is imperitive that only the colours taken in the first round are involved in the next two.

In other words, if in round 1, one of each of two colours are chosen, say red and white, then only tokens of these two colours can be exchanged in the next two rounds.

If in the first round two tokens of the same colour are chosen, then only tokens of that colour can be exchanged in the following two rounds.

1. if Brian goes first, and takes say a red token from Alan and a white token from Chris, that event is a certainty with that (or whatever other outcome.

For the second round, if Chris goes next, his starting position would be:

Alan Brian Chris

2 x Red Red

White 2 x White

Green Green Green

Blue Blue Blue

Yellow Yellow Yellow

Chris would need to take one white from Alan (with a chance of 1/4) and one of the two reds from Brian (with chance of 2/7). The chance of this happening is 1/4 x 2/7 = 2/28.

For the third round, Alan's starting position would be:

Alan Brian Chris

Red 2 x Red

2 x White White

Green Green Green

Blue Blue Blue

Yellow Yellow Yellow

In the 3rd round Alan would need to take a white from Brian (with chance of 2/6) and one of the two reds from Chris (with chance of 2/6). The chance of this happening is 2/6 x 2/6 = 4/36.

1. If in the 1st round Brian were to take 2 tokens of the same colour, say red, the starting position for the second round would be:

Alan Brian Chris

3 x Red

White White White

Green Green Green

Blue Blue Blue

Yellow Yellow Yellow

This event is also a certainty, it is simply the result of what the outcome is.

In the second round, Chris would have to take a Red from Brian and one of the other colours from Alan.

The probability of this is 3/7 x 1/4 = 3/28.

For the third round, Alan's turn, Alan's starting position would then be:

Alan Brian Chris

2 x Red Red

White White White

Green 2 x Green (if Green were taken)

Blue Blue Blue

Yellow Yellow Yellow

In the third round Alan would need to take a Red from Brian and one of the two tokens Chris has of the other colour he took from Alan.

The probability of this is 2/6 x 2/6 = 4/36.

So, the chance of getting back to all three having 5 tokens of 5 different colours through two different routes is:

(2/28 + 4/36) + (3/28 + 4/36) or  (0.07142857 + 0.11111111) + (0.107142857 + 0.111111111) = 0.4007936.

So, probability is roughly 0.4008, or 4/10.

New (less tricky) teaser on Monday.