ERJ Brainteaser: June 2019

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?

Answer: Tricky enough but at least no need for supercomputers to work out this week’s answer 3861. Well done to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; David Mann, Polymer Business Development, France; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; 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.

(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:

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 24×24 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

 

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?

Answer: The probability was 12/81 or 14.8% as explained by: Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; 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; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan.

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:

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?

Answer: 219 and 327: as Hans-Bernd Lüchtefeld neatly explained, just swap 19 and 2 resp. 27 and 3. Well done to; John Bowen, consultant, Bromsgrove, UK; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; 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, Vadodara, Gujarat, India; 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: Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India

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