Here each letter represents a different single-digit number and none of them is zero.

If A B C D E multiplied by 4 = EDCBA , what numbers are they?

**Answer**: Quite a full-on question for this early time of year. The magic numbers are 21978/87912 as so expertly explained (see solutions below) by: **Amparo Botella**, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; **John Bowen**, rubber industry consultant, Bromsgrove, Worcs, UK; **Michele Girardi**, quality manager, Scame Mastaf Spa, Suisio, Italy; **Andrew Knox**, Rubbond International, The Netherlands; **Rohit Kalé**, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., USA; **Stephan Paischer**, head of product management special products, Semperit AG Holding, Vienna, Austria; **David Mann**, key account manager, SPC Rubber Compounding, UK. Very well done to all and everyone else who had a go.

For a bit of extra excitement. we will tally correct responses by country each month. For January, the ranking list, to date, is: UK (4) / Austria, Italy, The Netherlands, Spain (3) / Germany, USA (1). Still time for this change next week, with our final teaser for the month.

Solutions

*Amparo Botella *

The explanation is quite too long, as we must make many trials with all the digits, so the result is:

A=2, B=1, C=9, D=7 and E=8

ABCDE X 4 = EDCBA

21978 X 4 = 87912

*John Bowen*

Here are my thoughts:

E×4 will have to end with even digit A. At the same time A**** \cdot 4 must not overflow into six digit number. Obviously A has to be 2 (a=1

is not possible because it is odd).

∗∗∗∗E⋅4

should end with A which is 2. So E must be 8 or 3.

2∗∗∗∗ x 4=E∗∗∗∗;E must be 8 or 9

To meet the above two conditions, E must be 8

2BCD8⋅4=8DCB2

∗∗∗∗8⋅4=32

will have a carry of 3 to tens place and added to 4⋅D,B has to be odd.Also4⋅B(in thousands place) must not give a carry to ten thousands place B has to be 1.

No D x 4+3=∗1;d should be 2 or 7 but 2 is already assigned to A so D must be 7

21C78⋅4=87C12

It is easy to see that forC∗4+3=∗C,C=9 fits well.

21978 x 4=87912 is the initial number

*Michele Girardi*

The answer for ABCDE x 4 = EDCBA is

21978 x 4 = 87912

Explanation

it's ABCDE < 25000, otherwise multiplying by 4 we get 6 digits numbers, so it's A=1 or A=2 BA must be divisible by 4 so it can be only A=2 E X 4 must end by 2, so E= 3 or E = 8

The numbers sought are in the form 2BCD3 or 2BCD8 with BCD ranging from 111, since there are no zero's , to 499, to not exceed 25000 setting up an excel file to calculate the numbers we get the solution . The key is to use the function that extracts text in a certain position

ABCD E ABCDE x 4 E D C B A EDCAB ABCDE x 4 - EDCAB

2 194 8 21948 87792 2 9 7 7 8 29778 7830

2 195 8 21958 87832 2 3 8 7 8 23878 1920

2 196 8 21968 87872 2 7 8 7 8 27878 5910

2 197 8 21978 87912 2 1 9 7 8 21978 0

2 198 8 21988 87952 2 5 9 7 8 25978 3990

2 199 8 21998 87992 2 9 9 7 8 29978 7980 2 200 8 22008 88032 2 3 0 8 8 23088 1080

*Andrew Knox*

The letters ABCDE represent the digits 21978 respectively, whereby 21978 times 4 is 87912.

This problem can be solved purely by random attempts but certain digits can also be determined by deduction:

For example, the first digit must be a 1 or a 2, otherwise multiplying by 4 would result in a 6 digit number.

The last digit therefore must be a 4 or 8.

Multiplied by 4, the last digit would give a product of 16 or 32, whereby the first digit of the original number would be a 6 or a 2.

2 would work as the first digit, but not 6 (as the result would be a 6 digit number), so the last digit must be an 8.

So now we have 2 . . . 8.

The 4th digit n could be any of 1 thru 9, but not 2 or 8 (as the digits are all different). The product of (4 x 10n) + (4x8) gives:

For n=1: 40+32 = 72

n=3: 120+32 = 152

n=4: 160+32 = 192

n=5: 200+32 = 232

n=6: 240+32 = 272

n=7: 280+32 = 312

n=9: 360+32 = 392

As the 2nd digit of the original number has to be 4 or less (to avoid the product having 6 digits), then the 4th digit as shown in the table above can only be a 5 or a 7 (for the 2nd digit to be a 1 or a 3). Also from the table above, it can be seen that 7 as the 4th digit would only work with a number beginning with 21, and 5 as the 4th digit would only work with a number beginning with 23.

So now the two possibilities are 21.78 and 23.58. Trial & error reveals only one solution: 4 x 21978 = 87912.

There were also however some very near hits when testing for the middle digit whereby the product contained the same 5 different digits but two were not in the correct order. Reason enough however to double check the sequence of the letters in reverse in the question!

*Rohit Kalé *

Since 4A < 10 and A is even, A = 2.

Therefore, E = 8. Note that E x 4 must end in 2.

It follows that B <= 2 because 23 x 4 > 89.

B = 1 because the two-digit number B2 must be a multiple of 4.

So far we have found that 21CD8 x 4 = 8DC12.

Observe that D >= 4 and that D8 x 4 ends in 12. Therefore, D = 7. The only remaining value of C is 9

Stephan

A 2

B 1

C 9

D 7

E 8

21978 x 4 = 87912

A: must be 2 or 1 as multiplication by 4 must still give a result of a 5 digit number. 4 x E is even, so A must be 2.

E: 4 x E ends with 2. 3 and 8 are the only options here – 3 is not possible as 20k x 4 = 80 k. So E is 8.

B: B cannot exceed 2 (as 3k x4 = 12k – too much). Only 1 works, as then the second figure ends in 12 (which is dividable by 4). So B is 1.

D and E found by trial and error in Excel.