ERJ Brainteaser - January

Official answer:

Question 3: First division

If the digits 1 through to 9 are randomly arranged to make a number, what is the probability that that number is divisible by 18?

Answer: Well done to the following readers who neatly got to 4/9 (0.4444…) as the answer to this tricky teaser: Hans-Bernd Luechtefeld, consultant, Germany; John Bowen, consultant, Bromsgrove, UK; Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan; Sudi Sudarshan, principal consultant, Global Mobility Strategies, USA; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Bharat B Sharma, technical director, TWC Group ( Techno Waxchem Pvt. Ltd. & Rajsha Chemicals Pvt. Ltd), India. Nice try also to everyone else who had a go.

SOLUTIONS

Hans-Bernd Luechtefeld

Any permutation of the digits 1 to 9 will be divisible by 9. So divisibility by 18 depends only on whether or not the last digit is even. If the last digit is 2, 4, 6 or 8 the number is divisible by 18 and if it is 1, 3, 5, 7 or 9 it is not. So 4 favourable outcomes out of 9 ie probability = 4/9.

John Bowen

The rule for divisibility of a number by 18 is that 1] the number must be ven, and 2] the sum of the digits must be divisible by 9.

In our case

1] 4/9 of all numbers possible are even, so the probability here is 0.4444, and

2] when all digits from 1 to 9 is 45, so any number in our set is also divisible by 18,

So the answer to our problem is 0.4444 [recurring]

Kamila Staszewska

Considering using digits 1 through 9 exactly once to make 9-digit number,

To be divisible by 18, number must be divisible of 9 and be even number.

A number is divisible by 9 is sum of its digits divides by 9, so 1+2+3+4+5+6+7+8+9=45 which is divisible by nine so 100% fulfil the first condition.

To be even number, last digit can be 2, 4, 6 or 8.

The probability of having even number is 4/9=0.444%

Amparo Botella

1: The sum of all the digits is 45, so divisible by 9  (1+2+3+4+5+6+7+8+9=45)

Since 45 is divisible by 9, any arrangement of these digits will satisfy divisibility by 9.

2: For the number to be divisible by 2, the last digit must be one of 2,4,6,or 8. These digits represent 444 out of the 999 total digits, so there are 444 valid choices for the last digit.

3: Total combinations

There are 9 (factorial of 9) ways to arrange the digits 1,2,3,…,9. The total number of combinations is:

9×8×7×6×5×4×3×2×1=362.880

4: Favorable combinations

If from 999 just 444 are multiple of 2, from the 362.880 (with three rule), just 161.280 will be multiple of 2

Probability calculation

The probability that a randomly arranged number is divisible by 18 is the ratio of favorable arrangements to total arrangements:

161.280 / 362.880 =   4 / 9

Sudi Sudarshan

My answer to this week's puzzle: 4/9 or 0.444...

Solution:

Any number that is divisible by 18 has the sum of digits divisible by 9 and ends with an even digit.

Any random number with all 9 digits has a sum of digits of 45 and hence is a multiple of 9. If the number also ends with an even digit, it is also divisible by 18. Since all digits have the same probability of being in the last place, and 4 of the 9 digits are even, the probability that a random number with all 9 digits is divisible by 18 is 4 out of 9.

Bharat B Sharma

The answer is 4/9 or 44.444... %

Any permutation of the digits 1 to 9 will be divisible by 9.

Divisibility by 18 depends only on whether or not the last digit is even.

If the last digit is 2, 4, 6 or 8 the number is divisible by 18 and if it is 1, 3, 5, 7 or 9 it is not.

So 4 favourable outcomes out of 9

The probability therefore is = 4/9.

Question 2: Secret combination

For his new bike, Brian has bought two combination locks each having a five-digit code. For the codes, he decided to use the smallest and largest numbers that met the following conditions:

1. Each digit of the number was a prime digit.
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: 32257 and 35772 as so neatly worked out by: Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; John Bowen, consultant, Bromsgrove, UK; John Coleman, membership manager, Circol ELT, Dublin, Ireland; Sudi Sudarshan, principal consultant, Global Mobility Strategies, USA; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands. Also, valid answers from Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain and David Mann, Polymer Business Development consultant, UK, whose ‘prime pair’ interpretation ruled out finding the second number.

SOLUTIONS

Kamila Staszewska

-Each digit is prime: 2, 3, 5, 7

-each pair of digits is non-prime: 22, 25, 27, 32, 33, 35, 52, 55, 57, 72, 75, 77; to allow for digit 3 in the sequence, the code must start with 3.

-each appear at least once: as we are looking for smallest and largest combinations, first will start with 32 and last with 35.

From all possible combinations, the smallest is 32257 and the largest is 35772.

John Bowen

The prime single digits are 2,3,5,7 [1 is not a prime by definition]. Smallest 5-digit number possible within the specified terms is 32257 and the largest is 35772.

Sudi Sudarshan

Solution: The only two-digit number that ends with 3 and is non-prime is 33. So in the 5-digit code, 3 needs to be in the first position for the second condition to be satisfied, and can only be in the second position if it repeats.

The only possible first two-digit combinations are 32, 33, and 35.

The smallest valid number is 32257.

The highest valid number is 35772

Andrew Knox

For the smallest number: 32257

For the largest number:   35752

Reasoning:

The rules dictate that:

1. Each digit must be prime, so 2, 3, 5 and 7, and one digit is to be used twice.
2. Successive pairs of digits must form non-prime numbers, so we can construct a matrix of pairs of digits:

22  23  25  27

32  33  35  37

52  53  55  57

72  73  75  77

From this matrix pairs highlighted in yellow are themselves prime, so not permitted.

It can be seen that:

1) 2 and 5 can follow anything, and 7 anything but 3.

2) 3 can only follow itself, or nothing. So 3 must be the first digit of both combinations. It is assumed from the outset that a combination begining with 33 will be neither the highest nor the lowest number possible for either combination.

So, it is likely that the lowest number will start with 32..., and the highest with 35..., as 37 is not permitted as it is prime.

Filling in the next 3 digits by what would be the best choice, and checking that the resulting pairs are permissible, by filling in the lowest next digit available for the lowest number, we move directly from 32 to 322, then 3225, then 32257, and for the highest number 35, then 357, then 3575, then 35752.

Amparo Botella

The smallest one using all prime digits, and non prime pairs is: 32527

Prime numbers: 2,3,5,7

Non prime pairs: 32, 25, 52, 27

For the largest number, I cannot find the right code, as to include the 3 from the prime number, one of the pairs contained in the code, is prime.

The biggest one is 75327, but 75, 32 and 27 are non prime pairs, but 53 is prime. So no way to find the right combination.

David Mann

I tried a few ways to solve this, finally saving a bit of time by writing out the non prime pairs that weren’t repeats.

12 15

21 25 27

32 35

51 52 57

72 75

So, working up from the lowest beginning with 12… and down from the highest 75…

I could only find one combination that worked ..  35127

Question 1 - Next number

3125, 256, 27, ?

Answer: Happy New Year to all. Our readers came back well refreshed after the holiday break, as evidenced by the replies (see Solutions below) to our first teaser of 2025. Very well done to: Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; David Mann, Polymer Business Development consultant, UK; Simon Winfield, director, MacLellan Rubber Ltd, Wednesfield, Wolverhampton, UK; John Coleman, membership manager, Circol ELT, Dublin, Ireland; Jon Cutler, business development technologist, SPC Rubber Compoundinng, UK; John Bowen, consultant, Bromsgrove, UK; Stephan Paischer, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Sudi Sudarshan, principal consultant, Global Mobility Strategies, USA; David Mann, Polymer Business Development consultant, UK; Bharat B Sharma, technical director, TWC Group- Rajsha Chemicals Pvt. Ltd & Techno Waxchem Pvt. Ltd, Vadodara, India; Andy Longdon, technical sales manager, Ceetak Ltd, Bedford UK; Doorba Lakhchaura, market analyst, Kraton Chemical BV, Almere, the Netherlands.

SOLUTIONS

Kamila Staszewska

Good Morning, and Happy New Year!

3125 = 5^5

256=4^4

27=3^3

2^2=4

The answer is 4.

Andrew Knox

This sequence really starts with 1, and is 1, 4, 27, 256, 3125, 46656, etc..

Each number is its position in the sequence to the power of that position number, i.e. position 3 is 3x3x3 = 27, position 2 is 2x2 = 4.

David Mann

Happy New Year to you.

The numbers are powers as follows:

3125 = 5 ^5

256 = 4^4

27 = 3^3

So the next will be

4 = 2^2

Simon Winfield

Answer is 4 as each number is the integer multiplied by the power of itself, nn

Sequence is:

22 = 2 x 2 = 4

33 = 3 x 3 x 3 = 27

44 = 4 x 4 x 4 x 4 = 256

55 = 5 x 5 x 5 x 5 x 5 = 3125

John Coleman

Membership Manager

The next in the sequence is 4 (22).

(3125 = 55, 256 = 44 and 27 = 33)

Jon Cutler

Business Development Technologist

DDI:    +44 (0) 1373 866020

The answer this week is 4

Not done it for ages but an easy start to the year …..

John Bowen

Happy New Year to you and all my friends in the rubber industry.

Answer is 4 - Numbers given are 5 to the power 5, 4 to the fourth, three to the third, so the next one is 2 squared, = 4, then 1 to the power of 1 = 1

Stephan Paischer

The answer is 4 and 1.

3125 = 5^5

256 = 4^4

27 = 3³

4 = 2²

1 = 1^1

Amparo Botella

Each number in this sequence is a natural number ultiplied by itself the same number of times (I do not know how to explain it in English, like exponentialy by it self) that means:

3125= 5x5x5x5x5 (5 multiplied 5 times)

256= 4x4x4x4 (4 multiplied 4 times)

27=3x3x3 (3 multiplied 3 times)

4= 2x2 (2 multiplied 2 times)

Thus, the next number in the series is 4.

Sudi Sudarshan

Solution: The sequence is of the form n^n, (n-1)^(n-1), (n-2)^(n-2), ...

5^5, 4^4, 3^3, 2^2, 1^1

3025, 256, 27, 4, 1

?Bharat B Sharma

Answer= 4

The series is

5x5x5x5x5= 3125

4x4x4x4 = 256

3x3x3 = 27

2x2= 4

Andy Longdon

5^5 = 3125

4^4 = 256

3^3 = 27

2^2 = 4

1^1 = 1

0^0 = 1 (on a calculator) or #NUM! number error (in Excel).

Next number is 4.

Doorba Lakhchaura

3125, 256, 27,…

(5)^5, (4)^4, (3)^3, (2)^2 à 3125, 256, 27, 4