ERJ Brainteaser: February

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.