ERJ Brainteaser: June
27 Jun 2022
This time around we have decided to award the Brainiac of the Month title to readers who provided any of the excellent solutions to Q3. Well done to all.
Question 3: Anyone for tennis?
For her tennis practice, Selena’s trainer divides 200 tennis balls into seven numbered boxes. How many balls are in each box if:
Box 1 + Box 2 = 57 balls
Box 2 + Box 3 = 83 balls
Box 3 + Box 4 = 71 balls
Box 4 + Box 5 = 43 balls
Box 5 + Box 6 = 66 balls
Answer: Probably one of our ever best teasers – at least based on quality of answer (see Solutions below) – getting us to: 6 + 37 + 29 + 14 + 57 + 26 + 31 = 200. For their ace performances, it's jolly well played 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; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; P J Swamy, managing director, Fores Elastomech India Pvt Ltd, Waluj, Aurangabad, Maharastra, India; Srikanth Pothapragada, Fores Group, India; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Jose Padron, laboratory analyst, Toyoda Gosei, Waterville, QC, Canada; John D Burrows, France; Siddartha Jajodia, Reliance Trade International Pvt. Ltd, Tripureshwor, Kathmandu, Nepal.
Solutions
John Bowen
To avoid confusion I shall call the boxes A,B,C,D,E,F,G
We know that A+B+C+D+E+F+G = 200  Eqn 1
and A+B = 57  Eqn 2
B+C = 83  Eqn 3
C+D = 71  Eqn 4
D+E = 43  Eqn 5
E+F = 66  Eqn 6
F+G = 43  Eqn 7
Adding Eqn 2 to Eqn 7 inclusive:
A+2B+2C+2D+2E+2F+G = 363 Eqn 8
Subtract Eqn 1: B+C+D+E+F = 163 Eqn 9
Substitute Eqns 3+5 in Eqn9:
83 + 43 + F = 163
F = 37
Substitute for F in Eqn 6: E = 29
Subs for F in Eqn 7, G = 6
Subs for E in Eqn 5, D = 14
Subs for D in Eqn 4, C = 57
Subs for C in Eqn 3, B = 26
Subs for B in Eqn 2, A = 31
Restating number in each box is thus as follows:
1=31
2=26
3=57
4=14
5=29
6=37
7=6
Add as check: sum = 200 QED
Amparo Botella
I have named each box with one letter: a, b, c, d, e, f and g. so the boxes contain: 31, 26, 57, 14, 29, 37, 6 = 200
See the numbers:
a+b+c+d+e+f+g=200 







a+b=57 
b=57a 


b+c=83 
c=83b 
c=83(57a) 
c=26+a 
c+d=71 
d=71c 
d=71(26+a) 
d=45a 
d+e=43 
e=43d 
e=43(45a) 
e=2+a 
e+f=66 
f=66e 
f=66(2+a) 
f=68a 
f+g=43 
g=43f 
g=43(68a) 
g=25+a 




a+(57a)+(26+a)+(45a)+(2+a)+(68a)+(25+a) 





a 
31 


b 
26 


c 
57 


d 
14 


e 
29 


f 
37 


g 
6 



200 


Andrew Knox
Adding together lines 1 + 3 + 5 gives the contents of boxes 1 thru 6 = 57 + 71 + 66 = 194, so box 7 = 200 194 = 6 balls.
Substituting from line 6 back to line 1 we get:
Box 6 = 436 = 37 balls,
Box 5 = 6637 = 29 balls
Box 4 = 14 balls
Box 3 = 57 balls
Box 2 = 26 balls
Box 1 = 31 balls
To be sure: 6 + 37 + 29 + 14 + 57 + 26 + 31 = 200 which is correct.
Stephan Paischer
31, 26, 57, 14, 29, 37, 6
Solution:
X1 + x2 + x3 + x4 + x5 + x6 + x7 = 200
X1 + x2 = 57 x2 = 57 – x1
X2 + x3 = 83 x3 = 83 –(57 – x1) x3 = 26 + x1
X3 + x4 = 71 x4 = 71 – (26 + x1) x4 = 45 – x1
X4 + x5 = 43 x5 = 43 – (45  x1) x5 = 2 + x1
X5 + x6 = 66 x6 = 66 – (2 + x1) x6 = 68 – x1
X6 + x7 = 43 x7 = 43 – (68 – x1) x7 = 25 + x1
Inserting equation 2 to 7 in equation 1:
X1 + 57  x1 + 26 + x1 + 45 – x1 – 2 + x1 + 68  x1 – 25 +x1 = 200
X1 = 31
Inserting the value for x1 in the remodeled equations 2 to 7 gives:
X2 = 26
X3 = 57
X4 = 14
X5 = 29
X6 = 37
X7 = 6
Srikanth Pothapragada
Let the quantities in 7 boxes be defined as B1, B2,B3,B4,B5,B6 and B7.
Inputs given:
B1+B2+B3+B4+B5+B6+B7=200
and
B1+B2 
57 
B2+B3 
83 
B3+B4 
71 
B4+B5 
43 
B5+B6 
66 
B6+B7 
43 
Let Qty in B1 be defined as “X”, then


Resolving each by substitution 
B1= 

X 
B2= 
57B1 
57X 
B3= 
83B2 
26+X 
B4= 
71B3 
45X 
B5= 
43B4 
X2 
B6= 
66B5 
68X 
B7= 
43B6 
X25 
Therefore: X + 57X + 26+X + 45X + X2 + 68X + X24 = 200 è X + 169 = 200 è X = 31
B1= 
X 
31 
B2= 
57X 
26 
B3= 
26+X 
57 
B4= 
45X 
14 
B5= 
X2 
29 
B6= 
68X 
37 
B7= 
X25 
6 
Jose Padron
Box 1 has 31 balls
Box 2 has 26 balls
Box 3 has 57 balls
Box 4 has 14 balls
Box 5 has 29 balls
Box 6 has 37 balls
Box 7 has 6 balls
Here is how
From the conditions :
A+B+C+D+E+F+G=200
A+B=57
B+C=83
C+D=71
D+E=43
E+F=66
F+G=43
🎾 (15Love)
Rewriting all the conditions in function of one variable, let’s say A (like serving an ace
B= 57A
C=26+A
D=45A
E=2+A
F=68A
G=25+A
🎾 🎾 (30Love)
Solving the first condition with A
A+169= 200
A=200169
🎾 🎾 🎾 (40Love)
Thus :
A= 31
B=26
C=57
D=14
E=29
F=37
G=6
31+26+57+14+29+37+6=200
🎾 🎾 🎾 🎾 (Game)
31+26=57
26+57=83
57+14=71
14+29=43
29+37=66
37+6=43
🎾 🎾 🎾 🎾 🎾 (Game SetMatch)
Bonus tip :
To get the tennis icon; just type in your keyboard: ALT+127934 = 🎾
Siddartha Jajodia
Box 1  31
Box 2  26
Box 3  57
Box 4  14
Box 5  29
Box 6  37
Box 7  6
John D Burrows
I cracked this the brute force way with Excel
Box 1 31 balls
Box 2 26 balls
Box 3 57 balls
Box 4 14 balls
Box 5 29 balls
Box 6 37 balls
Box 7 6 balls
Michele Girardi
Box 1 
31 
Box 2 
26 
Box 3 
57 
Box 4 
14 
Box 5 
29 
Box 6 
37 
Box 7 
6 
The problem can be solved by :
 expressing the last box content as the total minus the sum of the others
B7 = 200  (B1+B2 +B3+B4+B5+B6)
 isolating couples of addends and substituting them
B7 = 200  (B1+B2)(B3+B4)(B5+B6)
B1+B2 = 57
B3+B4 = 71
B5+B6 = 66
B7 = 200577166 = 6
 calculating all the others in succession
B6 = 43B7 = 436 = 37
B5= 66B6 = 6637 = 29
etc.
Alternatively, for the lovers of matrix calculus and Excel:
Question 2: Cycle code
After setting the fivedigit code on his bicycle lock, Simon has written out the following reminder:
Added together, the five numbers total 30.
Added together, the third number and the fifth number total 14 .
The fourth number = the second number plus 1.
The first number is one less than twice the second number.
Added together, the second number and the third number total 10.
What is the code?
Answer: 74658, as so wonderfully worked out by our Brainiacs, established and new (see below).
Well done, in order of reply, to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Tim Clayfield, Interface Polymers Ltd, UK; Andrew Knox, Rubbond International, The Netherlands; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Frank Bloemendaal, research & development, Polycomp, JG Vorden, The Netherlands; Jose Padron, laboratory analyst, Toyoda Gosei, Waterville, QC, Canada; P J Swamy, managing director, Fores Elastomech India Pvt Ltd, Waluj, Aurangabad, Maharastra, India; and everyone else who had a go.
Solutions
John Bowen
Let the 5 digits = a, b, c, d, e
Then we can just substitute in the equations we are given thus:
a + b + c + d + e = 30
then c + e = 14
d = b + 1
a = 2b  1
b + c = 10
a + b + d = 16
so 2b  1 + 2b + 1 =16
4b = 16, so b = 4
and a = 7
d = b + 1 so d = 5
b + c = 10 so c = 6
b + c = 14 so e = 8
Restating the above: a =7, b = 4, c = 6, d = 5, e =8.
Check: a + b + c + d + e = 30  QED
Amparo Botella
a 
b 
c 
d 
e 

a+b+c+d+e=30 
2b1+b+10b+b+1+4+b 
4b+14=30 
b=16/4 
b=4 
4 

c+e=14 
10b+e=14 
e=1410+b 
e=4+b 
e=8 
8 

d=b+1 
d=4+1 
d=5 
5 

a=2b1 
2b=a+1 
b=(a+1)/2 
a=81 
a=7 
7 

b+c=10 
c=10b 
c=104 
C=6 
6 

30 
Andrew Knox
The various bits of information can be used to set up the following equations using A to E to represent the 1st to 5th digits of the code:
1) A + B + C + D + E = 30
2) C + E = 14
3) D = B + 1
4) A = 2B  1
5) B + C = 10
Resolving these equations gives:
1)  5): A + D + E = 20 Substituting for 3) gives A + B + E = 19
1)  2): A + B + D = 16 Subtracting A + B + E = 19 gives D  E = 3 or E = D + 3
So, If B = X, then
D = X + 1 (equation 3)
E = X + 4, (from E = D + 3)
A = 2X  1 (equation 4) and
C = 30  (A + B + D + E) = 30  5X  4
If X =1, C = 21 (not possible)
If X =2, C = 16 (not possible)
If X =3, C = 11(not possible)
If X =4, C = 6,
If X =5, C = 1.
We also know C + E =14, so C and E can only be 9 & 5, 8 & 6 or 7 & 7, or vice versa.
So C can only be 6, and X (B) is therefore 4.
A is then 7, B is 4, C is 6, D is 5 and E is 8, and the combination code is 74658
Michele Girardi
Let's call a,b,c,d,e the five digits, the problem can be stated as :
a= 2b1
b=b
c=10b
d=1+b
e=14c = 4+b
a+b+c+d+e=30
substituting :
2b1+b+10b+1+b+4+b = 30
4b = 30104= 16
b=4
a=2*41= 7
b=4
c= 104= 6
d=1+4=5
e=4+4=8
Stephan Paischer
First v = 7
second w = 4
third x = 6
fourth y = 5
fifth z = 8
solution:
v + w + x + y + z = 30
x + z = 14 (10 – w) + z = 14 (using condition 5) z = 4 + w
w + 1 = y
v + 1 = 2w v = 2w – 1
w + x = 10 x = 10 – w
(2w – 1) + w + (10 – w) + (w + 1) + (4 + w) = 30
4w + 14 = 30
w = 4
I get the other solution by inserting w = 4 into the other equations.
Jose Padron
The bike code is : 7 4 6 5 8
Here is how,
According to the rules:
A+B+C+D+E=30 …(1)
C+E=14 …(2) à E=14C à E= 14(10B) à E= 4+B
D=B+1 …(3)
A=2B1 …(4)
B+C=10 …(5) à C=10B
Solving (1) with all digits in function of B
(2B1) +B+(10B) +(B+1) +(4+B) = 30
2B1+B+10B+B+1+4+B = 30
4B+14 = 30
B = (3014)/4 = 16/4 = 4
B = 4
From (4)
A=2B1 =2*41 = 7
A=7
From (5)
C=10B = 104 = 6
C=6
From (3)
D=B+1 = 4+1 = 5
D=5
From (2)
E= 4+B = 4+4 = 8
E = 8
Conclusion from (1):
A+B+C+D+E=30
7+4+6+5+8 = 30
30 = 30
P J Swamy
Let the cycle code be abcde
The Clues
Added together, the five numbers total 30.
 a+b+c+d+e = 30(1)
Added together, the third number and the fifth number total 14 .
 c+e = 14(2)
The fourth number = the second number plus 1.
 d=b+1(3)
The first number is one less than twice the second number.
 a=2b1(4)
Added together, the second number and the third number total 10.
 b+c=10(5)
The Solution
Substituting (2) in (1) we get
a+b+d + 14 =30
a+b+d = 3014= 16(6)
Putting (3) and (4) in (6) we get
2b1 +b+b+1 = 16
4b=16
b=4 (7)digit b  4
Putting (7) in (3)
d= 4+1 = 5(8)digit d  5
Putting (7) in (4)
a= 2*41= 7 (9)digit a 7
Putting (7) in (5)
4+c =10 ie c = 104 =6(10)digit c  6
Putting 7, 8, 9 ,10 in 1
7+4+6+5+e = 30
22+e= 30 ie e = 3022 =8 (11) digit e  8
Michele Girardi
This is the "no algebra" solution, using matrix calculus and Excel
New teaser on Monday.
Question 1: Quick question
What is the sum of the first 75 odd numbers?
Answer: The answer is 5625, as worked out by our top Brainiacs in a range of wonderful ways (see below). Well done, in order of reply, to: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, The Netherlands; Mehmet Koral, ErhardtLeimer representative for Turkey, managing director, C&C Endüstriyel Danışmanlık, Eğitim ve Mümessillik Ltd, Göztepe, Istanbul, Turkey; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Jose Padron, laboratory analyst, Waterville TG Inc., Waterville, Québec, Canada; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; and everyone else who had a go.
Solutions
Stephan Paischer
I proceed like Gauss:
The first of 75 odd numbers is 1, the last 149.
Summing up both, multiplying with 75 and dividing by 2 (as only odd numbers are considered, gives the result.
(1 + 149) x 75 / 2
150 x 75 / 2
75 x 75 = 5625
Andrew Knox
Answer: 75 x 75 = 5625
Reasoning:
Sum of 1st 3 odd numbers = 1 + 3 + 5 = 9 Sum of 1st 5 = 1 + 3 + 5 + 7 + 9 = 25 So, sum of the first “X” odd numbers is always the middle number (=X) x X.
Mehmet Koral
Given;
The list of the first 75 odd numbers is: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99,101, 103, 105, 107, 109, 111, 113, 115, 117,119,121,123, 125,127, 129,131,133135,137,139,141,143,145,147,149.
To calculate the first and the 75th term of this series;
T(n)= a+(n1).d where
a= the first term of the series; it is 1.
d= the difference between the 2nd term and the 1st term; it is 2.
T(1)= 1+(11).2= 1
T(75)=1+(751). 2=1+148=149
Here the first 75 odd numbers are such an arithmetic series that the difference between the two consecutive terms is constant and equals 2.
To calculate the sum of the first n terms formula is the following ;
Sum(n) = (n/2)*(2.a+(n1).d) where
a= the first term of the series; it is 1.
d= the difference between the two consecutive terms (a constant=2).
Sum(75) = (75/2). (2.1+(751).2)
Then, the sum of the first 75 odd numbers becomes equal to the following;
Sum (75) = (75/2). (2.1+(751).2) =(75/2). (2.1+(751).2=(75/2).(2+74.2)
=(75/2).(2+148)=(75/2). (150)= (75).(75)=5625
Jose Padron
Quick answer; the sum of the first 75 odd numbers is 5625
Here is how: 75^2= 75*75 = 5625
John Bowen
The sum of the first n odd numbers is given by s = n.sqd thus:
The sum of n terms of an aririthmetic progression is given by the formula Sn= n/2 × [a + l].
By substituting the values of 'a' and 'l' in the above formula we get,
Sn = n/2 × [1 + (2n  1)]
Sn= n/2 × [2n]
Sn= n × n
Sn= n2
So the sum of the first 75 is 75 x 75 = 5625