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VIRTUAL CONFERENCE ictcm.com | #ICTCM 32 nd International Conference - - PowerPoint PPT Presentation

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32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE ictcm.com | #ICTCM 32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM BY NADEEM ASLAM INSTRUCTOR OF

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32nd International Conference on Technology in Collegiate Mathematics

ictcm.com | #ICTCM

VIRTUAL CONFERENCE

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32nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE

#ICTCM

Solving Diophantine Equations by Excel

BY NADEEM ASLAM INSTRUCTOR OF MATHEMATICS FLORIDA INTERNATIONAL UNIVERSITY MIAMI, FLORIDA ASLAMN@FIU.EDU JAY VILLANUEVA INSTRUCTOR OF MATHEMATICS MIAMI DADE COMMUNITY COLLEGE MIAMI, FLORIDA JVILLANU@MDC.EDU ICTCM, ORLANDO, FLORIDA. MARCH 12-15, 2020.

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32nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE

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Solving Diophantine Equations by Excel

  • 1. Introduction

1.1 Brief biography of Diophantus 1.2 Framing the Diophantine Equations

  • 2. Methods of Solutions

2.1 By Formula 2.2 By Congruence Equations 2.3 By Excel

  • 3. Examples (5)
  • 4. Conclusions
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32nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE

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1-Introduction

1.1 Diophantine lived in Alexandria ~250 CE. Of 13 books he wrote

  • n Arithmetica, only 6 remain, where he invented a system of

notation complete with unknowns that allowed him to address Algebra problems. The only source of details about his life is an epigram found in a collection called Greek Anthology: “Diophantus passed one sixth of his life in childhood, one twelfth in youth, and

  • ne seventh as a bachelor. Five years after his marriage was born

a son who died four years before his father, at half his father’s age.” How long did Diophantus live? 𝑦 =

! " + ! #$ + ! % + 5 + ! $ + 4

[𝑦 = 84 years] 1.2 A Diophantine equation is an indeterminate equation that admits only integer solutions, of the form 𝑔 𝑦 + 𝑕 𝑧 = ℎ(𝑨), 𝑔, 𝑕, ℎ are polynomials. The linear case: 𝑏𝑦 + 𝑐𝑧 = 𝑑, admits many interesting problems.

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  • 2. Methods of Solutions

2.1 By Formula 𝑏𝑦 + 𝑐𝑧 = 𝑑, 𝑏, 𝑐, 𝑑, 𝑦, 𝑧 are integers Let gcd 𝑏, 𝑐 = 𝑒. If d ∤ 𝑑, then there are no solutions. If d|c, there may be unique or multiple solutions. If (𝑦&, 𝑧&) a solution: 𝑏𝑦& + b𝑧& = 𝑑 ⇒ 𝑏 𝑦& + 𝑐 𝑒 𝑜 + 𝑐 𝑧& − 𝑏 𝑒 n = 𝑑 𝑦 = 𝑦& + 𝑐 𝑒 𝑜, y = 𝑧& − 𝑏 𝑒 n. Methods 2 & 3 will be explained after an example.

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Ex.1: Mr. Professor-Currency-Exchange-Problem

A professor returning from Europe changes her euros and pounds into US dollars, receiving a total of $125.78. If the exchange rate is €1 = $1.31 and £1 = $1.61. How many of euros and pounds did she exchange? Let x = number of pounds and y = number of euros. 1.61𝑦 + 1.31𝑧 = 125.78 ⟹ 161𝑦 + 131𝑧 = 12578. 𝑏 = 161, 𝑐 = 131, 𝑑 = 12578. 𝑒 = 𝑏, 𝑐 = 161 , 131 = 1: 𝑒|𝑑 ∴ ∃ solutions! 𝑏𝑡 + 𝑐𝑢 = 𝑒; 𝑡, 𝑢 are integers. 161𝑡 + 131𝑢 = 1 −−−−→ (1), In order to find s and t we proceed as follow; 161 = 131 O 1 + 30 1 = 3 − 2 . 1 = 3 − 8 − 3.2 . 1 = 3 . 3 − 8 . 1. 131 = 30 . 4 + 11 = 11 − 8.1 . 3 − 8 . 1 = 11 . 3 − 8 . 4 30 = 11 . 2 + 8 = 11 . 3 − 30 − 11 . 2 . 4 = 11 . 11 − 30 . 4 11 = 8 . 1 + 3 = 131 − 30 . 4 . 11 − 30 . 4 = 131 . 11 − 30 . 48. 8 = 3 . 2 + 2 = 131 . 11 − 161 − 131 . 1 . 48 3 = 2 . 1 + 1 = 131 . 59 − 161 . 48 = 161 −48 + 131(59) 2 = 1 . 2 + 0 ⟹ 𝑡 = −48 & 𝑢 = 59

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  • Ex. 1 (Continued)

e 𝑏𝑡 + 𝑓 𝑐𝑢 = 𝑓𝑒 = 𝑑 e 1 = 12578 ⟹ e = 12578 𝑦! = 𝑓𝑡 = 12578 −48 = −603744 𝑧! = 𝑓𝑢 = 12578 59 = 742102 𝑦 = 𝑦! +

" # 𝑜,

𝑧 = 𝑧!−

$ # 𝑜

= −603744 +

%&% % 𝑜,

= 742102 −

%'% % 𝑜

𝑦, 𝑧 ≥ 0 ⟹ 4608.73 ≤ 𝑜 ≤ 4609.33 ∴ 𝑜 = 4609 𝑦 = −603744 + 131 4609 = 𝟒𝟔 𝒒𝒑𝒗𝒐𝒆𝒕 y = 742102 − 161 4609 = 𝟔𝟒 𝒇𝒗𝒔𝒑𝒕

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Method2: Congruence equations

a𝑦 + 𝑐𝑧 = 𝑑 (a) 161𝑦 + 131𝑧 = 12578 161𝑦 ≡ 12578 𝑛𝑝𝑒 131 30𝑦 ≡ 2 𝑛𝑝𝑒(131) We can try 𝑦 = 0,1,2, … , 130: ∴ 𝑦 = 35, 𝑧 = 53 Thus the answer is 𝟒𝟔 𝒒𝒑𝒗𝒐𝒆𝒕 & 𝟔𝟒 𝒇𝒗𝒔𝒑𝒕 Or (b): 131𝑧 ≡ 12578 𝑛𝑝𝑒(161) 131𝑧 ≡ 20 𝑛𝑝𝑒(161) We can try y= 0, 1, … , 160: ∴ 𝑧 = 53 & 𝑦 = 35. Thus the answer is 𝟒𝟔 𝒒𝒑𝒗𝒐𝒆𝒕 & 𝟔𝟒 𝒇𝒗𝒔𝒑𝒕

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Method 3: By Excel

Professor Currency Exchange Problem.

1.61x+1.31y=125.78

Number of Pounds x =

35

Number of Euros y =

53 Objective 125.78 Constraints

Constraint1

78

Constraint2

96

Constraint3 Constraint4

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Ex.2: Dinner Problem

A group dinner costs $96.00. If a lobster order costs $11.00 and a chicken order costs $8.00, how many ordered lobsters and chicken? Let 𝑦 = 𝑚𝑝𝑐𝑡𝑢𝑓𝑠 𝑝𝑠𝑒𝑓𝑠 𝑡 𝑧 = 𝑑ℎ𝑗𝑑𝑙𝑓𝑜 𝑝𝑠𝑒𝑓𝑠(𝑡) 11𝑦 + 8𝑧 = 96, 𝑏 = 11, 𝑐 = 8, 𝑑 = 96 𝑒 = 𝑏, 𝑐 = 11,8 = 1: 𝑒|𝑑 ∴ ∃ solution(s)! 𝑏𝑡 + 𝑐𝑢 = 𝑒; 𝑡, 𝑢 are integers. 11𝑡 + 8𝑢 = 1, to find s, 11 = 8 [ 1 + 3 1 = 3 − 2 [ 1 = 3 − (8 − 3 [ 2) [ 1 8 = 3 [ 2 + 2 = 3 [ 3 − 8 [ 1 = 11 − 8 [ 1 [ 3 − 8 [ 1 3 = 2 [ 1 + 1 = 11 [ 3 − 8 [ 4 2 = 1 [ 2 + 0 = 11 3 + 8(−4) ⟹ 𝑡 = 3, 𝑢 = −4.

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Ex.2 (Continued)

e 𝑏𝑡 + 𝑓 𝑐𝑢 = 𝑓𝑒 = 𝑑 e 1 = 96 ⟹ e = 96 𝑦& = 𝑓𝑡 = 96 3 = 288 𝑧& = 𝑓𝑢 = 96 −4 = −384 𝑦 = 𝑦& + '

( 𝑜,

𝑧 = 𝑧&− )

( 𝑜

= 288 + *

# 𝑜,

= −384 − ##

# 𝑜

𝑦, 𝑧 ≥ 0 ⟹ −36 ≤ 𝑜 ≤ −34.9 ∴ 𝑜 = −36 𝑝𝑠 − 35 ∴ 𝑗 𝑦 = 288 + 8 −36 = 𝟏 𝒑𝒔𝒆𝒇𝒔𝒇𝒆 𝒎𝒑𝒄𝒕𝒖𝒇𝒔. 𝑧 = −384 − 11 −36 = 𝟐𝟑 𝒑𝒔𝒆𝒇𝒔𝒇𝒆 𝒅𝒊𝒋𝒅𝒍𝒇𝒐. Check: 11(0)+8(12)=$96. ∴ 𝑗𝑗 𝑦 = 288 + 8 −35 = 𝟗 𝒑𝒔𝒆𝒇𝒔𝒇𝒆 𝒎𝒑𝒄𝒕𝒖𝒇𝒔. 𝑧 = −384 − 11 −35 = 𝟐 𝒑𝒔𝒆𝒇𝒔𝒇𝒆 𝒅𝒊𝒋𝒅𝒍𝒇𝒐. Check: 11(8)+8(1)=$96.

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Method2: Congruence equations

a𝑦 + 𝑐𝑧 = 𝑑 𝑏 11𝑦 + 8𝑧 = 96 11𝑦 ≡ 96 𝑛𝑝𝑒 8 3𝑦 ≡ 0 𝑛𝑝𝑒(8) We can try 𝑦 = 0, 1, … , 7: ∴ 𝑗 𝑦 = 0, 𝑧 = 12 & ii 𝑦 = 8, 𝑧 = 1. Or (b): 8𝑧 ≡ 96 𝑛𝑝𝑒(11) 8𝑧 ≡ 8 𝑛𝑝𝑒(11) We can try y= 0, 1, … , 10: ∴ 𝑗 𝑧 = 1, 𝑦 = 8 & 𝑗𝑗 𝑦 = 0, 𝑧 = 12.

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Method 3: By Excel

Dinner Problem: 11x + 8y = 96 First Possible Solution: Second Possible Solution:

Number of Lobster Ordered x = Number of Lobster Ordered

8

Number of Chicken Ordered y =

12

Number of Chicken Ordered

1

Objective

96

Objective

96 Constraints:

Constraint1 Constraint1

8

Constraint2

12

Constraint2

1

Constraint3 Constraint3

0.1

Constraint4 Constraint4

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Ex.3: Pigs-Chicken-Problem

In the farm the number of heads and legs of chickens and pigs add to 70. How many chickens and pigs were there? Let 𝑦 = 𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑑ℎ𝑗𝑑𝑙𝑓𝑜 𝑧 = 𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑞𝑗𝑕𝑡 5𝑦 + 3𝑧 = 70 , 𝑏 = 5, 𝑐 = 3, 𝑑 = 70 𝑒 = 𝑏, 𝑐 = 5,3 = 1: 𝑒|𝑑 ∴ ∃ solution(s)! 𝑏𝑡 + 𝑐𝑢 = 𝑒; 𝑡, 𝑢 are integers. 5𝑡 + 3𝑢 = 1, to find s, 5 = 3 K 1 + 2 1 = 3 − 2 K 1 = 3 − ( ) 5 − 3.1 . 1 3 = 2.1 + 1 = 3.2 − 5.1 = 5 −1 + 3(2) 2 = 1.2 + 0 ⟹ 𝑡 = −1 & 𝑢 = 2.

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Ex.3 (Continued)

e 𝑏𝑡 + 𝑓 𝑐𝑢 = 𝑓𝑒 = 𝑑 e 1 = 70 ⟹ e = 70 𝑦1 = 𝑓𝑡 = 70 −1 = −70 𝑧1 = 𝑓𝑢 = 70 2 = 140 𝑦 = 𝑦1 + 2

3 𝑜,

𝑧 = 𝑧1− 4

3 𝑜

= −70 + 5

6 𝑜,

= 140 − 7

6 𝑜

𝑦, 𝑧 ≥ 0 ⟹ 23.33 ≤ 𝑜 ≤ 28 ∴ 𝑜 = 24, 25, 26, 27, 𝑝𝑠 28

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Ex.3 (Continued)

∴ 𝑦 = −70 + 3 24 = 2 𝑞𝑗𝑕𝑡. 𝑧 = 140 − 5 24 = 20 𝑑ℎ𝑗𝑑𝑙𝑓𝑜𝑡. Check: 5(2)+3(20)=70. 𝑝𝑠 ∴ 𝑦 = −70 + 3 25 = 5 𝑞𝑗𝑕𝑡 y = 140 − 5 25 = 15 𝑑ℎ𝑗𝑑𝑙𝑓𝑜𝑡. Check: 5(5) + 3(15)=70. 𝑝𝑠 ∴ 𝑦 = −70 + 3 26 = 8 𝑞𝑗𝑕𝑡. y = 140 − 5 26 = 10 𝑑ℎ𝑗𝑑𝑙𝑓𝑜𝑡. Check: 5(8) + 3(10) = 70. 𝑝𝑠 ∴ 𝑦 = −70 + 3 27 = 11 𝑞𝑗𝑕𝑡. y = 140 − 5 27 = 5 𝑑ℎ𝑗𝑑𝑙𝑓𝑜𝑡. Check: 5(11) + 3(5)=70. 𝑝𝑠 ∴ 𝑦 = −70 + 3 28 = 14 𝑞𝑗𝑕𝑡. y = 140 − 5 28 = 0 𝑑ℎ𝑗𝑑𝑙𝑓𝑜𝑡. Check: 5(14) + 3(0)=70.

n x y N 24 2 20 70 25 5 15 70 26 8 10 70 27 11 5 70 28 14 70

5x + 3y = 70 = N

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Method2: Congruence equations

a𝑦 + 𝑐𝑧 = 𝑑 𝑏 5𝑦 + 3𝑧 = 70 5𝑦 ≡ 70 𝑛𝑝𝑒 3 2𝑦 ≡ 1 𝑛𝑝𝑒(3) ∴ 𝑦 = 2, 𝑧 = 20; 𝑦 = 5, 𝑧 = 15; 𝑦 = 8, 𝑧 = 10; 𝑦 = 11, 𝑧 = 5; 𝑦 = 14, 𝑧 = 0. Or (b): 3𝑧 ≡ 70 𝑛𝑝𝑒(5) 3𝑧 ≡ 0 𝑛𝑝𝑒 5 ∴ 𝑧 = 0, 𝑦 = 14; 𝑧 = 5, 𝑦 = 11; 𝑧 = 10, 𝑦 = 8; 𝑧 = 15, 𝑦 = 5; 𝑧 = 20, 𝑦 = 2.

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Method 3: By Excel

Pigs-Chicken-Problem 5x+3y=70

Number of pigs x =

2

Number of pigs x =

5

Number of pigs x =

8

Number of pigs x =

11

Number of pigs x =

14

Number of chicken y =

20

Number of chicken y =

15

Number of chicken y =

10

Number of chicken y =

5

Number of chicken y = Objective

70

Objective

70

Objective

70

Objective

70

Objective

70 Constraints

Constraint1

2

Constraint5

5

Constraint9

8

Constraint13

11

Constraint17

14

Constraint2

23

Constraint6

18

Constraint10

13

Constraint14

8

Constraint18

3

Constraint3 Constraint7

2.1

Constraint11

5.1

Constraint15

8.1

Constraint19

11.1

Constraint4 Constraint8 Constraint12 Constraint16 Constraint20

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Ex.4: Stamps-Problem

You wish to mail a package, total cost is 83¢. Only 6¢ and 15¢ stamps are available. What combination of stamps can be used to mail the package? Let 𝑦 = 𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 15¢ 𝑡𝑢𝑏𝑛𝑞𝑡 𝑏𝑜𝑒 𝑧 = 𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 6¢ stamps. 15𝑦 + 6𝑧 = 83 , 𝑏 = 15, 𝑐 = 6, 𝑑 = 83 𝑒 = 𝑏, 𝑐 = 15,6 = 3: 𝑒 ∤ 𝑑 ∴ ∃ no solution!? The person has to spend 1¢ extra to mail the package. In that case; 15𝑦 + 6𝑧 = 84 ⟹ 5𝑦 + 2𝑧 = 28; 𝑏 = 5, 𝑐 = 2, 𝑑 = 28. 𝑒 = 𝑏, 𝑐 = 5,2 = 1: 𝑒/𝑑 ∴ ∃ solution(s). 𝑏𝑡 + 𝑐𝑢 = 𝑒; 𝑡, 𝑢 are integers. 5𝑡 + 2𝑢 = 1, to find s, 5 = 2 [ 2 + 1 1 = 5 − 2 [ 2 = 5 1 + 2(−2) 2 = 1.2 + 0 ⟹ 𝑡 = 1 & 𝑢 = −2

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Ex.4 (Continued)

e 𝑏𝑡 + 𝑓 𝑐𝑢 = 𝑓𝑒 = 𝑑 e 1 = 28 ⟹ e = 28 𝑦! = 𝑓𝑡 = 28 1 = 28 𝑧! = 𝑓𝑢 = 28 −2 = −56 𝑦 = 𝑦! +

" # 𝑜,

𝑧 = 𝑧!−

$ # 𝑜

= 28 +

% & 𝑜,

= −56 −

' & 𝑜

𝑦, 𝑧 ≥ 0 ⟹ −14 ≤ 𝑜 ≤ −11.2 ∴ 𝑜 = −14, −13, −12

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Ex.4 (Continued)

∴ 𝑗 𝑦 = 28 + 2 −14 = 0 15¢ 𝑡𝑢𝑏𝑛𝑞𝑡. 𝑧 = −56 − 5 −14 = 14 6¢ stamps. Check: 5(0)+2(14)=28. 𝑝𝑠 ∴ 𝑗𝑗 𝑦 = 28 + 2 −13 = 2 15¢ stamps. y = −56 − 5 −13 = 9 6¢ stamps. Check: 5(2) + 2(9)=28. 𝑝𝑠 ∴ 𝑗𝑗𝑗 𝑦 = 28 + 2 −12 = 4 15¢ stamps. y = −56 − 5 −12 = 4 6¢ stamps. Check: 5(4) + 2(4) = 28.

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Method2: Congruence equations

a𝑦 + 𝑐𝑧 = 𝑑 (a) 5𝑦 + 2𝑧 = 28 5𝑦 ≡ 28 𝑛𝑝𝑒 2 𝑦 ≡ 0 𝑛𝑝𝑒(2) ∴ 𝑦 = 0, 𝑧 = 14; 𝑦 = 2, 𝑧 = 9; 𝑦 = 4, 𝑧 = 4; Or (b): 2𝑧 ≡ 28 𝑛𝑝𝑒(5) 2𝑧 ≡ 3 𝑛𝑝𝑒 5 ∴ 𝑧 = 4 𝑦 = 4; 𝑧 = 9, 𝑦 = 2; 𝑧 = 14, 𝑦 = 0.

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Method 3: By Excel

Stamps Problem 15x + 6y = 83 15x + 6y = 84

  • No. of 15¢ stamps x =
  • No. of 15¢ stamps x =
  • No. of 15¢ stamps x =

2

  • No. of 15¢ stamps x =

4

  • No. of 6¢ stamps y =
  • No. of 6¢ stamps y =

14

  • No. of 6¢ stamps y =

9

  • No. of 6¢ stamps y =

4

Objective

There is no Solution

Objective

84

Objective

84

Objective

84

Constraint1

5

Constraint6

1

Constraint10

3

Constraint14

5

Constraint2

13

Constraint7

14

Constraint11

9

Constraint15

4

Constraint3 Constraint8 Constraint12

1.1

Constraint16

3.1

Constraint4 Constraint9 Constraint13 Constraint17

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Ex.5: Museum-Tickets-Problem

Tickets to the museum are $2.25 for adults and $1.00 for children for a total of 60 seats. One day the total amount collected was $117.25. How many adults and children attended? If we use elimination method, solution by 2 x 2 system of equations. x + y = 60 (1) 2.25x + 1.00y = 117.25 (2) ∴ x = 45.8 adults y = 14.2 children (?!) This means that not all 60 seats were taken!

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Ex.5: Museum-Tickets-Problem (2nd look)

Tickets to the museum are $2.25 for adults and $1.00 for children for a total of 60 seats. One day the total amount collected was $117.25. How many adults and children attended? Let x = number of adults and y = number of children.

2.25𝑦 + 1.00𝑧 = 117.25 ⟹ 225𝑦 + 100𝑧 = 11725 ⟹ 9𝑦 + 4𝑧 = 469,

𝑏 = 9, 𝑐 = 4, 𝑑 = 469. 𝑦 + 𝑧 = 60 𝑒 = 𝑏, 𝑐 = 9 , 4 = 1: 𝑒|𝑑 ∴ ∃ solutions! 𝑏𝑡 + 𝑐𝑢 = 𝑒; 𝑡, 𝑢 are integers. 9𝑡 + 4𝑢 = 1, to find s, 9 = 4 f 2 + 1 1 = 9 − 4 f 2 = 9 1 + 4(−2) 4 = 1 . 4 + 0 ⟹ 𝑡 = 1 & 𝑢 = −2

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EX 5 (Continued)

e 𝑏𝑡 + 𝑓 𝑐𝑢 = 𝑓𝑒 = 𝑑 e 1 = 469 ⟹ e = 469 𝑦1 = 𝑓𝑡 = 469 1 = 469 𝑧1 = 𝑓𝑢 = 469 −2 = −938 𝑦 = 𝑦1 + 2

3 𝑜,

𝑧 = 𝑧1− 4

3 𝑜

= 469 + 8

6 𝑜,

= −938 − 9

6 𝑜

𝑦, 𝑧 ≥ 0 ⟹ −117.25 ≤ 𝑜 ≤ −104.22 ∴ 𝑜 = −117, −116, −115, … , −105

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n x = 469 + 4n y = -938 – 9n Solution verification

  • 117

x = 469 + 4(-117) = 469 -468 = 1 y = -938 – 9(-117) = -938+1053=115 No x + y = 116

  • 116

x = 469 + 4(-116) = 469 -464 = 5 y = -938 – 9(-116) = -938+1044=106 No x + y = 111

  • 115

x = 469 + 4(-115) = 469 -460 = 9 y = -938 – 9(-115) = -938+1035=97 No x + y = 103

  • 114

x = 469 + 4(-114) = 469 -456 = 13 y = -938 – 9(-114) = -938+1026=88 No x + y = 101

  • 113

x = 469 + 4(-113) = 469 -452 = 17 y = -938 – 9(-113) = -938+1017=79 No x + y = 96

  • 112

x = 469 + 4(-112) = 469 -448 = 21 y = -938 – 9(-112) = -938+1008=70 No x + y = 91

  • 111

x = 469 + 4(-111) = 469 -444 = 25 y = -938 – 9(-111) = -938+999=61 No x + y = 86

  • 110

x = 469 + 4(-110) = 469 -440 = 29 y = -938 – 9(-110) = -938+990=52 No x + y = 81

  • 109

x = 469 + 4(-109) = 469 -436 = 33 y = -938 – 9(-109) = -938+981=43 No x + y = 76

  • 108

x = 469 + 4(-108) = 469 -432 = 37 y = -938 – 9(-108) = -938+972=34 No x + y = 71

  • 107

x = 469 + 4(-107) = 469 -428 = 41 y = -938 – 9(-107) = -938+963=25 No x + y = 66

  • 106

x = 469 + 4(-106) = 469 -424 = 45 y = -938 – 9(-106) = -938+954=16 No x + y = 61

  • 105

x = 469 + 4(-105) = 469 -420 = 49 y = -938 – 9(-105) = -938+945=7 Yes x + y = 56

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Method2: Congruence equations

a𝑦 + 𝑐𝑧 = 𝑑 𝑏 9𝑦 + 4𝑧 = 469 9𝑦 ≡ 469 𝑛𝑝𝑒 4 𝑦 ≡ 1 𝑛𝑝𝑒(4) ∴ 𝑦 = 1, 𝑧 = 115 (No x + y > 60); x=5, y=106 (No x + y > 60); ∴ 𝑦 = 9, 𝑧 = 97 (No x + y > 60); x=13, y=88 (No x + y > 60); ∴ 𝑦 = 17, 𝑧 = 79 (No x + y > 60); x=21, y=70 (No x + y > 60); ∴ 𝑦 = 25, 𝑧 = 61 (No x + y > 60); x=29, y=52 (No x + y > 60); ∴ 𝑦 = 33, 𝑧 = 43 (No x + y > 60); x=37, y=34 (No x + y > 60); ∴ 𝑦 = 41, 𝑧 = 25 (No x + y > 60); x=45, y=16 (No x + y > 60); ∴ 𝑦 = 49, 𝑧 = 7 (Yes x + y < 60) Or (b): 4𝑧 ≡ 469 𝑛𝑝𝑒(9) 4𝑧 ≡ 1 𝑛𝑝𝑒(9) Based on the similar reasoning above, we get the unique solution; ∴ 𝑧 = 7 & 𝑦 = 49.

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Method 3: By Excel

Museum Tickets Problem

2.25x + 1.00y = 117.25

Adults 49 Children 7 Revenue 469 Constraint1 52 Constraint2 117 Constraint3 Constraint4 Constraint5 60 Constraint6 56

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  • 4. Conclusions
  • Diophantine equations: 𝑏𝑦 + 𝑐𝑧 = 𝑑 -- are indeterminate

equations (there are more unknowns than there are equations) that admit only integer solutions, started by Diophantus around 250 CE.

  • These equations may be solved by formulae

𝑦 = 𝑦T + 𝑐 𝑒 𝑜, y = 𝑧T − 𝑏 𝑒 n, where n is an integer.

  • A second method is by use of congruences: 𝑏𝑦 + 𝑐𝑧 = 𝑑;

equivalent expressions are: a𝑦 ≡ 𝑐 𝑛𝑝𝑒 𝑑 and b𝑧 ≡ 𝑏 𝑛𝑝𝑒 𝑑 .

  • A final method is by Excel, where we use the tool Solver. We

have found that Solver also works when there are multiple solutions.

  • The advantage of this last method is that it works very fast,

almost instantaneously.

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References

  • 1. J Andrew, 1972. Explorations in Number Theory. CA: Brooks/Cole Pub.
  • 2. RG Archibald, 1970. An Introduction to the Theory of Numbers. OH: Merrill Pub.
  • 3. EP Armindariz & SJ McAdam, 1980. Elementary Number Theory. NY: Macmillan Pub.
  • 4. IA Barnett, 1972. Elements of Number Theory. MA: Prindle, Weber & Schmidth, Inc.
  • 5. DM Bourg, 2006. Excel Scientific and Engineering Cookbook. CA: O’Reilly Media Inc.
  • 6. KH Rosen, 2011. Elementary Number Theory. MA: Addison Wesley Pub.
  • 7. FW Stevenson 2000. Exploring the Real Numbers. NJ: Prentice-Hall, Inc.
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Contact Information

Name: Nadeem Aslam Title: Solving Diophantine Equations using Excel Institution: Florida International University Email: aslamn@fiu.edu Social Media handle/name: zoom