ERJ Brainteaser: January 2022

Question 4: You cannot be series 0122?

Which number comes next in this series?

93, 105, 111, 114, 120, _? 

Answer: Next in this series is 123: adding the sum of the digits of the previous number (1+2 + 0 = 3) to that same number (120). Well done, in order of correct reply to: Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, The Netherlands; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Jose Padron, laboratory analyst, Waterville TG Inc., Waterville, Québec, Canada; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Rohit Kalé, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., USA; David Mann, key account manager, SPC Rubber Compounding, UK; Mehmet Koral, Erhardt - Leimer representative for Turkey: managing director, C&C Endüstriyel Dan??manl?k, E?itim ve Mümessillik Ltd, Istanbul, Turkey.

Solutions

Special mention
As well as identifying the series as per above, Michele Girardi also offered an interesting alternative take:
The answer maybe 132 .
In analogy with a dice, where the sum of opposite faces gives always 7, in this special one the sum is always 225

              1           93
                   2                  105
                       3                          111
                       4                          114
                   5                  120
               6          132

SUM            7   7   7  225         225         225

Amparo Botella 
Next number in this serie is 123.
93 + (9+3) = 105
105 + (1+0+5) = 111
111 + (1+1+1) = 114
114 + (1+1+4) = 120
120 + (1+2+0) = 123
 

Stephan Paischer 
You add the cross-sum (= sum of all digits) of the number for each step.
93 + ( 9 + 3) = 105
105 + (1 + 0 + 5) = 111
111 + (1 + 1 + 1) = 114
114 + (1 + 1 + 4) = 120
120 + (1 + 2 + 0) = 123

Andrew Knox
The next number is the previous number plus the sum of the individual digits of the previous number. Thus, 120 +(1+2+0) = 123.

John Bowen
Different possibilities here, but how about this:
Each number differs from the previous one by the sum of its individual digits thus:
93 [+9+3]= 105 [+1+0+5]=111[+1+1+1] =114 [+1+1+4] =120  then[+1+2+0] = 123

Jose Padron

?

Michele Girardi 
Each number is equal to the previous plus its digits:
93
105= 93+9+3
111 = 105+1+0+5
114 = 111+1+1+1
120 = 114+1+1+4
120+1+2+0 =123

Rohit Kalé 
The next number in the series is derived by adding the earlier number with the sum of the digits of the same number.
93+(9+3)=105
105+(1+0+5)=111
111+(1+1+1)=114
114+(1+1+4)= 120
120+(1+2+0)=123

David Mann
One of those where it took a while to see it!
Each term is the sum of the previous term + the sum of its digits. So the next term is 120 + 1 + 2 = 123

Mehmet Koral
When we observe the numbers in the series one after the other closely, the common rule seems to be ;
(The number itself+(sum of the digits of that number)) ----?  the next number in the series;
e.g. for the number 93, the next number in the series is calculated using the rule given above;
         93+(9+3)= 105.
Here, we apply the same rule for the rest of the given numbers in the series, and calculate the questioned number of the series as follows;
105+(1+0+5)=111 ;  111+(1+1+1)=114 ; 114+(1+1+4)=120  ; 120+(1+2+0)= 123.
Hence the questioned number in the series is calculated to be 123.

Top Table

For further excitement. we now tally correct responses by country each month. There are extra points for awards (2) and special mentions (1). For January, so, the international ranking list ended up as: UK, Spain (6) / Italy (5 ) / Austria, Netherlands (4) / USA (2) / Canada, Germany, Turkey (1).

Question 3: Reverse gear

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.

Question 2: Country file?

What do the following countries have in common?

Kuwait, Luxembourg, Mexico, Monaco, Singapore

Might require further information so check for clues during the week.

Answer: While some readers went in the wrong direction, the answer was that these countries all have same name as their capital cities. Well done, in order of correct reply to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Andrew Knox, Rubbond International, The Netherlands; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Rohit Kalé, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., USA.

Question 1: Six of one...

Using mathematical functions/expressions, how can you get 6 from three 0s (zeros)?

Answer: Not only did a top group of readers manage to get to 6, they also came up with a range of clever solutions (see below) to this tricky teaser. Well done, in order of reply, to: Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Jose Padron, Laboratory Analyst, Waterville TG Inc.. Waterville, Québec, Canada; Hans-Bernd Lüchtefeld, PHP Fibers GmbH, Obernburg, Germany; Stephan Paischer, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; and everyone else who had a go.

Solutions

Amparo Botella
We need to use the three zeroes allowed, each with a factorial. That makes 3, and as we need 6, we use again the factorial of 3, so we have
0!+0!+0!=3!=6

Andrew Knox 
Answer: (0! + 0! + 0!)! = 6
(To solve this puzzle you have to know that 0! (zero factoriall) is 1, not 0).
So, 0! + 0! + 0! = 3, and 3! =6. 
So the answer is (0! + 0! + 0!)! = 6

John Bowen
We need some brackets and a bit of Trig and Pure Maths here!
Factorial 3 is 3 x 2 x 1  = 6, represented by 3!
How to get to 3?
Trigonometry tells us that Cosine 0 degrees = 1, So 3 = [{Cos0} + {Cos0} + [Cos0}]
So 6 = {[Cos0} + {Cos0} + {Cos0}]!

Michele Girardi
One solution is 6= (0!+0!+0!)! .
Alternatively, looking at the tables of famous limits, one can find a solution requiring only one zero:
lim x*x*x/(x-sin x)= 6
x->0

Jose Padron 
First, using factorial operator and by definition 0! =1
Thus;
0! + 0! + 0! = 3 and,
3! = 6
Finally, we can solve the question like this:
( 0! + 0! + 0!)! = 6

Hans-Bernd Lüchtefeld
(0!+0!+0!)! = 3! = 6

Stephan Paischer 
The answer is: 
(0!+0!+0!)! = 6
0! = 1    zero factorial = 1
3! = 6    three factorial = 6.