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