ERJ Brainteaser: June

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

Box 6 + Box 7 = 43 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=57-a
b+c=83	c=83-b	c=83-(57-a)	c=26+a
c+d=71	d=71-c	d=71-(26+a)	d=45-a
d+e=43	e=43-d	e=43-(45-a)	e=-2+a
e+f=66	f=66-e	f=66-(-2+a)	f=68-a
f+g=43	g=43-f	g=43-(68-a)	g=-25+a
a+(57-a)+(26+a)+(45-a)+(-2+a)+(68-a)+(-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 = 43-6 = 37 balls,

Box 5 = 66-37 = 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=	57-B1	57-X
B3=	83-B2	26+X
B4=	71-B3	45-X
B5=	43-B4	X-2
B6=	66-B5	68-X
B7=	43-B6	X-25

Therefore:  X + 57-X + 26+X + 45-X + X-2 + 68-X + X-24 = 200  è X + 169 = 200 è X = 31

B1=	X	31
B2=	57-X	26
B3=	26+X	57
B4=	45-X	14
B5=	X-2	29
B6=	68-X	37
B7=	X-25	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

? (15-Love)

Rewriting all the conditions in function of one variable, let’s say  A (like serving an ace

B= 57-A

C=26+A

D=45-A

E=-2+A

F=68-A

G=-25+A

? ? (30-Love)

Solving the first condition with A

A+169= 200

A=200-169

? ? ? (40-Love)

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- Set-Match)

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 = 200-57-71-66 = 6 

- calculating all the others in succession

B6 = 43-B7 = 43-6 = 37

B5= 66-B6 = 66-37 = 29

etc.

Alternatively, for the lovers of matrix calculus and Excel:

 

Question 2: Cycle code

After setting the five-digit 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		2b-1+b+10-b+b+1+4+b	4b+14=30	b=16/4	b=4	4
c+e=14	10-b+e=14	e=14-10+b	e=4+b				e=8	8
d=b+1					d=4+1		d=5	5
a=2b-1	2b=a+1	b=(a+1)/2			a=8-1		a=7	7
b+c=10	c=10-b				c=10-4		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= 2b-1

b=b

c=10-b

d=1+b

e=14-c = 4+b

a+b+c+d+e=30

substituting   :

2b-1+b+10-b+1+b+4+b = 30

4b = 30-10-4= 16

b=4

a=2*4-1= 7

b=4

c= 10-4= 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=14-C à E= 14-(10-B) à E= 4+B

D=B+1                            …(3)

A=2B-1                           …(4)

B+C=10                          …(5) à C=10-B

Solving (1) with all digits in function of B

(2B-1) +B+(10-B) +(B+1) +(4+B) = 30

2B-1+B+10-B+B+1+4+B = 30

4B+14 = 30

B = (30-14)/4 = 16/4 = 4

B = 4

From (4)

A=2B-1  =2*4-1 = 7

A=7

From (5)

C=10-B = 10-4 = 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=2b-1--------------------------------------------------------------------------------------(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 =  30-14= 16-------------------------------------------------------------------------(6)

Putting (3) and  (4) in  (6) we get

2b-1 +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*4-1= 7 ---------------------------------------------------------------------------------------(9)-----digit a- 7

Putting (7) in (5)

4+c =10 ie c = 10-4 =6----------------------------------------------------------------------(10)----digit  c - 6

Putting 7, 8, 9 ,10 in 1

7+4+6+5+e = 30

22+e= 30 ie e = 30-22 =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, Erhardt-Leimer representative for Turkey, managing director, C&C Endüstriyel Dan??manl?k, E?itim 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+(n-1).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+(1-1).2= 1

T(75)=1+(75-1). 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+(n-1).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+(75-1).2)

Then, the sum of the first 75 odd numbers becomes equal to the following;

Sum (75) =  (75/2). (2.1+(75-1).2)  =(75/2). (2.1+(75-1).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