Question 2: **City series**

Complete the following:

... Mexico City, Sao Paulo. Shanghai, Delhi, ?.

**Answer**: Our question-setter happened to be in **Tokyo** recently and thought ‘wow, this is quite a busy place’. On checking, it appeared that the Japanese capital is considered by the UN to be the world’s biggest city in terms of total urban-area population.

Well done so, in order of reply, to the following readers, who recognised this as a series of the most populous cities worldwide: **John Bowen**, rubber industry consultant, Bromsgrove, Worcs, UK; **David Mann**, key account manager, SPC Rubber Compounding, UK; **Andrew Knox**, Rubbond International, The Netherlands; **Amparo Botella**, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; **Stephan Paischer**, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; **Michele Girardi**, quality manager, Scame Mastaf Spa, Suisio, Italy; **Jose Padron**, laboratory analyst, Toyoda Gosei, Canada; **John Coleman**, membership manager, Circol ELT, Dublin, Republic of Ireland; **Suman Dhar**, deputy general manager- international business at zinc chemicals producer Rubamin Private Ltd, Vadodara, Gujarat, India.

*New teaser on Monday*

Question 1: **Six pack**

Can you find a six-digit number, which when multiplied by 2, 3, 4, 5 or 6 only the same (ie no other) digits appear?

**Answer**: This one was quite a bruiser, requiring some impressive, mathematical acrobatics to get to the official answer 142857. Extra well done, so, to: **Andrew Knox**, Rubbond International, The Netherlands; **John Bowen**, rubber industry consultant, Bromsgrove, Worcs, UK; **Mehmet Koral**, Erhardt+Leimer representative for Turkey, managing director, C&C Endustriyel Danismanlik, Egitim ve Mumessillik Ltd, Istanbul; **Michele Girardi**, quality manager, Scame Mastaf Spa, Suisio, Italy; and everyone else who had a go , including **Amparo Botella**, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; and **David Mann**, key account manager, SPC Rubber Compounding, UK (see Solutions/notes below).

Solutions

*David Mann*

*This one stumped me! I tried logic and excel without success. Even when I found an answer on google the maths was quite complex. **Kudos to anyone who solved it.*

*Andrew Knox*

*Answer: 999999/7 = 142857*

*John Bowen*

*This is one you either know or don't...*

*Answer is 142857*

*142857*2 = 285714,*

*142857*3=428571,*

*142857*4=571428,*

*142857*5=714285,*

*142857*6=857142*

*The solution is a bit long. Here it is:*

*Idea #1. a and the sum of digits of a are divisible by 9. *

*Here is why:*

*3a is divisible by 3 ->*

*sum of digits of 3a is divisible by 3 ->*

*sum of digits of a is divisible by 3 ->*

*a is divisible by 3 ->*

*3a is divisible by 9 ->*

*sum of digits of 3a is divisible by 9 ->*

*sum of digits of a is divisible by 9 ->*

*a is divisible by 9.*

*Idea #2. The first digit of a is 1. If it were 2 or more 6a would have*

*had 7 digits.*

*Idea #3. The last digit is 7 and the digits in the middle are 2,4,5,8.*

*This is the longest part of the proof. Let us assume that the last digit*

*is 1.*

*Then the last digits of products a, 2a, 3a, 4a, 5a, 6a should be*

*1,2,3,4,5,6. a should consist of these 6 numbers, but 1+2+3+4+5+6=21. 21*

*is not divisible by 7, So, the last digit is not 1.*

*Similarly we exclude 2, 3 , 4 , 6, 8, 9 as the last digit. I did not*

*consider 0 or 5, maybe we should hoping to find another solution.*

*If we suppose that 7 is the last digit, we get the following last digits*

*of products: 7, 4, 1, 8, 5, 2. Their sum is 27. Which is divisible by 9.*

*Now let us consider the second digit. It can not be 2, otherwise the*

*first digit of 3a would have been 3. 3 is not in our set of digits.*

*We then try 4 as the second digit. The first two digits of 3a are then*

*42 (43 is impossible because of digit 3. So, the third digit is 2. The*

*remaining digits to place are 5 and 8.*

** **

*Mehmet Koral*

*In approaching to develop a solution we consider using a permutation property. So,all we have to do is work with the reciprocal of a natural number for now.*

* We are requested to multiply that number by 1,2,3,4,5 and 6.*

*Note that none of the reciprocals of the numbers from 1 to 6 would work in this case, as at least one of these numbers would give 1 when multiplied by their own. *

*Then we consider number 7. *

*As we know that multiples of 1/7 that are not integers, will contain permutations of decimal expansion of 142857. 1/7 will contain a permutation of its digits when multiplied by any number from 1 to 6.*

*e.g. *

*1/7 =0,142857*

*(1/7)x 2=0,285714*

*(1/7)x 3=0,428571*

*(1/7)x4 =0,571428*

*(1/7)x 5=0,714285*

*(1/7)x6 =0,857142 *

** **

*Amparo Botella*

*The six number digit is 111111*

*111111 x 2 = 222222*

*111111 x 3 = 333333*

*111111 x 4 = 444444*

*111111 x 5 = 555555*

*111111 x 6 = 666666*

*I do not know whether I understood properly what you mean, but this is what I think it is. Too easy though!!*

** **

*Michele Girardi*

*The answer is 142857, I could't find alternatives to brute force to find it .*

*Excluding for the moment 0 as a possible digit, the number must be between 123456 and 999999/6=166666.*

*In excel, create a column with numbers from 123456 to 166666 and their multiplications by 2,3,4,5,6. *

*Concatenate the six columns in one string of 36 characters (CONCATENATE formula). *

*Separate the single digits (text to columns command ), that will create 36 columns . Since there are always the same 6 digits, each one must appear 6 times .*

*Create 9 columns where you count the occurrences of the digits 1 to 9 in the range of 36 colums (COUNTIF formula )*

*Filter the rows where, for every digit, the occurrence is 6 . 3 must be excluded because after setting 1 and 2 there are no occurrences of 6 . After setting 5 the only result is 142857.*