ERJ Brainteaser: March

Question 4: Super sequence

What comes next in this sequence: 2, 6, 42, 1806, _?

Answer: Nice to get safely back on track this week, with some neat solutions (see below) getting to the answer 3263442. Very well done, in order of reply to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, The Netherlands; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; David Mann, key account manager, SPC Rubber Compounding, UK; Jose Padron, laboratory analyst, Waterville TG Inc., Waterville, Québec, Canada;  Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Giorgio Marzari, sales director, RDC srl, Bareggio (Mi), Italy.

Solutions

John Bowen

Next number is : 3263442

Each number is the square of the previous number plus the previous number [or {previous number} x {previous number + 1}

Andrew Knox

Answer: 3,263,442: next number is the square of the previous number added to the previous number. E.g. (6x6)+6=42

Amparo Botella

2x2=4 + 2 = 6

6x6= 36 + 6= 42

42x42=1764 + 42=1806

1806x1806=3261636 + 1806=3263442

David Mann

If 2=x the sequence is x*(1+x) so the next term is 3,263,442

Jose Padron

Starting with 1 multiplied by the next number

1x2=2x3=6x7=42x43=1 806x1807=-3 263 442

Thus, answer is 3 263 442

Michele Girardi

The next number is 3263442 . Each one is the multiplication of the previous by the previous +1

n       n+1   n*(+1)

2       3       6

6       7       42

42     43     1806

1806 1807 3263442

New teaser on Monday.

Ranking by country in March : The Netherlands 70 / UK 50 / Austria, Canada, Spain, Italy 30 / Austria 20.

Question 3: X factor

Can you find the value for X?

(Corrected) 

57, 12, 39

29, X, 47

98, 17, 89

43, 7, 25

Answer: Sincere apologies to all readers for a couple of misleading blips in this teaser - we will work hard to make sure this does not happen again. But special extra bonus points do go to Andrew Knox, Rubbond International, The Netherlands, who still worked out that 11 is the missing number: 

Adding the individual digits of the numbers to the left and right appears to give the number in the middle, were it not for the fact that two numbers (the fouth and second from the end) should then be 89 (not 87) and 7 (not 6).

 We then get:  

 12, 12, 12

11, X, 11

17, 17, 17

7, 7, 7

 So X would then be 11.

New teaser on Monday.

Ranking to date by country in March : The Netherlands 40 / UK 30 / Austria, Canada, Spain 20 / Italy.

Question 2: Time traveller

Science fiction writer Sam is setting his next book in the year 2307, on the basis that the sum of the first two digits and the last two is the same as the middle two digits. He now wants to find the previous year to have this same mathematical property. Can you help?

Answer: The previous year that meets those conditions is 1978, as so impressively worked out (see solutions below) by: Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Jose Padron, laboratory analyst, Waterville TG Inc., Waterville, Québec, Canada. Well done to all and everyone else who had a go.

Solutions

Amparo Botella

The previous number filling this same rule should be 1978 ; 19+78=97

We cannot get this same rule starting by 20 and the addition should be higher than 20 and cannot be starting by 0, and the same with 21 and 22, we can not get a number whose addition will be starting by 1 or 2. The next year in series should be 2417 (24+17=41).

John Bowen

A headache with this one, but here's some logic:

This cannot be in the 21st century, since the middle two digits form a 1-digit number.

In the 20th century, any year is of the form 19xy

. Then the year is only possible if and only if

19+xy=9x⇔19+10x+y=90+x⇔9x+y=71.

Then since 0≤y≤9

we get

9x≤71≤9x+9⇒62≤9x≤71 so x=7.

Substituting x=7

in 9x+y=71 so 63 + y = 71, so we get that y=8

Thus, the previous year is in the 20th century and is 1978.

Stephan Paischer

Jose Padron

The period of years 1000 and 9999 to apply this condition is between 1000 and 9999, because they have four digits.

36 years have this condition; I figured out a list and a graph to show how many years have the same property.

Here are all the years, enjoy them. I liked the harmonic incidence.

They have a noticeable change each millennium and a year gap of 110 years within. ?

The year in the question and the solution are noticeable in the graph with a red dot.

Question 1: Measure by measure

Find the missing number in this tricky series

5.08, 7.62, ___ , 17.78, 27.94…

Answer: More straightforward this week, once you figured out that this was a sequence of prime numbers converted from inches to centimetres - making 12.70 the missing number. Well done, in order of reply, to: Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; David Mann, key account manager, SPC Rubber Compounding, UK; Andrew Knox, Rubbond International, The Netherlands; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Jose Padron, laboratory analyst, Waterville TG Inc., Waterville, Québec, Canada; and everyone else who had a go.