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Symmetry Anyone?
Willy Hereman
After Dinner Talk
SANUM 2008 Conference
Thursday, March 27, 2007, 9:00p.m.
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Tribute to Organizers
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Karin Hunter
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Karin Hunter
Andre Weideman
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Karin Hunter
Andre Weideman
Ben Herbst
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Karin Hunter
Andre Weideman
Ben Herbst
Dirk Laurie
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Karin Hunter
Andre Weideman
Ben Herbst
Dirk Laurie
Stéfan van der Walt
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Karin Hunter
Andre Weideman
Ben Herbst
Dirk Laurie
Stéfan van der Walt
Neil Muller
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Karin Hunter
Andre Weideman
Ben Herbst
Dirk Laurie
Stéfan van der Walt
Neil Muller
and the Support Staff
Behind the Scenes
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A Big THANK YOU !
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Outline
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Symmetry Surrounding Us
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What is Symmetry?
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Father – Daughter Puzzle
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Steiner Trees – Puzzle
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Solving Quadratic, Cubic, Quartic Equations
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The Quintic and the French Revolutionary
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The Seven-Eleven Puzzle
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Modern Applications
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Symmetry Surrounding Us
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Ask people on US campus
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Symmetry Surrounding Us
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Ask people on US campus
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Ask an architect or artist
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Symmetry Surrounding Us
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Ask people on US campus
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Ask an architect or artist
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“Madam, I’m Adam”
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Symmetry Surrounding Us
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Ask people on US campus
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Ask an architect or artist
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“Madam, I’m Adam”
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Ask a mathematician
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What is Symmetry?
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It is all about transformations
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What is Symmetry?
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It is all about transformations
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Wigner: “... the unreasonable effectiveness of
mathematics in the natural sciences”
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“... the unreasonable effectiveness of using
symmetries in mathematics...”
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A simple example: father-daughter puzzle
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Father – Daughter Puzzle
Today, the ages of a father and his daughter add up
to 40 years. Five years from now, the father will be 4
times older than his daughter. How old is each today?
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Father – Daughter Puzzle
Today, the ages of a father and his daughter add up
to 40 years. Five years from now, the father will be 4
times older than his daughter. How old is each today?
Solution:
F : age of the father (today)
D : age of the daughter (today)
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Father – Daughter Puzzle
Today, the ages of a father and his daughter add up
to 40 years. Five years from now, the father will be 4
times older than his daughter. How old is each today?
Solution:
Then,
F : age of the father (today)
D : age of the daughter (today)
F + D = 40
F + 5 = 4 (D + 5)
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Father – Daughter Puzzle
Today, the ages of a father and his daughter add up
to 40 years. Five years from now, the father will be 4
times older than his daughter. How old is each today?
Solution:
Then,
F : age of the father (today)
D : age of the daughter (today)
F + D = 40
F + 5 = 4 (D + 5)
Eliminate F = 40 − D and solve
45 − D = 4 (D + 5)
or 5D = 25
Hence, D = 5 and F = 40 − D = 35.
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Symmetry Reduces Complexity
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Symmetry Reduces Complexity
Solution:
20 + x : age of the father (today)
20 − x : age of the daughter (today)
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Symmetry Reduces Complexity
Solution:
20 + x : age of the father (today)
20 − x : age of the daughter (today)
Then,
25 + x = 4 (25 − x)
or 5x = 75
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Symmetry Reduces Complexity
Solution:
20 + x : age of the father (today)
20 − x : age of the daughter (today)
Then,
25 + x = 4 (25 − x)
Hence, x = 15.
or 5x = 75
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Symmetry Reduces Complexity
Solution:
20 + x : age of the father (today)
20 − x : age of the daughter (today)
Then,
25 + x = 4 (25 − x)
or 5x = 75
Hence, x = 15.
So, father is 20 + x = 35, daughter is 20 − x = 5.
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Steiner Trees
Connecting Cities with Shortest Road System
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D
Three cities in equilateral triangle
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L = 1.87 D
One choice of a road system
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L = 1.73 D
Shortest road system connecting three cities
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D
Four cities in a square
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L = 2.83 D
One choice of a road system
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L = 2.73 D
Shortest road system connecting 4 cities
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Solving Quadratic, Cubic, Quartic Equations
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Quadratic: ax2 + bx + c = 0
Babylonians (400 BC)
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Solving Quadratic, Cubic, Quartic Equations
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Quadratic: ax2 + bx + c = 0
Babylonians (400 BC)
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Cubic: ax3 + bx2 + cx + d = 0
Italians (1525-1545): dal Ferro & Fior, Tartaglia
& Cardano
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Solving Quadratic, Cubic, Quartic Equations
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Quadratic: ax2 + bx + c = 0
Babylonians (400 BC)
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Cubic: ax3 + bx2 + cx + d = 0
Italians (1525-1545): dal Ferro & Fior, Tartaglia
& Cardano
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Quartic: ax4 + bx3 + cx2 + dx + e = 0
Cardano & Ferrari
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Solving Quadratic, Cubic, Quartic Equations
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Quadratic: ax2 + bx + c = 0
Babylonians (400 BC)
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Cubic: ax3 + bx2 + cx + d = 0
Italians (1525-1545): dal Ferro & Fior, Tartaglia
& Cardano
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Quartic: ax4 + bx3 + cx2 + dx + e = 0
Cardano & Ferrari
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Quintic: ax5 + bx4 + cx3 + dx2 + ex + f = 0
The equation that couldn’t be solved!
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Challenge Mathematica to Solve the
Quadratic, Cubic, Quartic, Quintic Equations
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The Quintic and the French Revolutionary
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The Quintic and the French Revolutionary
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Evariste Galois (1830): inventor of group theory,
which quintic equations can (cannot) be solved
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The Quintic and the French Revolutionary
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Evariste Galois (1830): inventor of group theory,
which quintic equations can (cannot) be solved
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Niels Hendrik Abel (1821): general quintic
equation can not be solved analytically
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The Quintic and the French Revolutionary
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Evariste Galois (1830): inventor of group theory,
which quintic equations can (cannot) be solved
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Niels Hendrik Abel (1821): general quintic
equation can not be solved analytically
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Joseph Liouville, Camille Jordan, Felix Klein,
Sophus Lie, ....
The Seven Eleven Puzzle
The sum and product of the prices of four items
is R 7.11
The Seven Eleven Puzzle
The sum and product of the prices of four items
is R 7.11
Hence,
x + y + z + w = 7.11
x y z w = 7.11
The Seven Eleven Puzzle
The sum and product of the prices of four items
is R 7.11
Hence,
x + y + z + w = 7.11
x y z w = 7.11
Solution:
The Seven Eleven Puzzle
The sum and product of the prices of four items
is R 7.11
Hence,
x + y + z + w = 7.11
x y z w = 7.11
Solution:
Prices:
1.20
1.25
1.50
3.16
One Solution Strategy
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Convert to integer problem
x + y + z + w = 711
x y z w = 711 ∗ 106
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One Solution Strategy
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Convert to integer problem
x + y + z + w = 711
x y z w = 711 ∗ 106
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Integer factors: 711 ∗ 106 = 26 ∗ 32 ∗ 56 ∗ 79
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One Solution Strategy
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Convert to integer problem
x + y + z + w = 711
x y z w = 711 ∗ 106
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Integer factors: 711 ∗ 106 = 26 ∗ 32 ∗ 56 ∗ 79
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Thus, x = 79 ∗ 2a1 ∗ 3b1 ∗ 5c1
z = 2a3 ∗ 3b3 ∗ 5c3
y = 2a2 ∗ 5c2
w = 2a4 ∗ 3b4 ∗ 5c4
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One Solution Strategy
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Convert to integer problem
x + y + z + w = 711
x y z w = 711 ∗ 106
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Integer factors: 711 ∗ 106 = 26 ∗ 32 ∗ 56 ∗ 79
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Thus, x = 79 ∗ 2a1 ∗ 3b1 ∗ 5c1
z = 2a3 ∗ 3b3 ∗ 5c3
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With a1 + a2 + a3 + a4 = 6,
c1 + c2 + c3 + c4 = 6.
y = 2a2 ∗ 5c2
w = 2a4 ∗ 3b4 ∗ 5c4
b1 + b3 + b4 = 2,
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One Solution Strategy
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Convert to integer problem
x + y + z + w = 711
x y z w = 711 ∗ 106
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Integer factors: 711 ∗ 106 = 26 ∗ 32 ∗ 56 ∗ 79
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Thus, x = 79 ∗ 2a1 ∗ 3b1 ∗ 5c1
z = 2a3 ∗ 3b3 ∗ 5c3
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With a1 + a2 + a3 + a4 = 6,
c1 + c2 + c3 + c4 = 6.
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Actually, x = n ∗ 79 with n = 1, 2, 3, 4, 5, 6, 8.
y = 2a2 ∗ 5c2
w = 2a4 ∗ 3b4 ∗ 5c4
b1 + b3 + b4 = 2,
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One Solution Strategy
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Convert to integer problem
x + y + z + w = 711
x y z w = 711 ∗ 106
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Integer factors: 711 ∗ 106 = 26 ∗ 32 ∗ 56 ∗ 79
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Thus, x = 79 ∗ 2a1 ∗ 3b1 ∗ 5c1
z = 2a3 ∗ 3b3 ∗ 5c3
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With a1 + a2 + a3 + a4 = 6,
c1 + c2 + c3 + c4 = 6.
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Actually, x = n ∗ 79 with n = 1, 2, 3, 4, 5, 6, 8.
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Using y z w ≤
(y+z+w)3
27
y = 2a2 ∗ 5c2
w = 2a4 ∗ 3b4 ∗ 5c4
b1 + b3 + b4 = 2,
eliminates n = 5, 6, 8.
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Reject x = 79, x = 2 ∗ 79 = 158, and x = 3 ∗ 79 = 237
because 711 − 79 = 632, 711 − 158 = 553, and
711 − 237 = 474 are not five-folds (+ argument)
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Reject x = 79, x = 2 ∗ 79 = 158, and x = 3 ∗ 79 = 237
because 711 − 79 = 632, 711 − 158 = 553, and
711 − 237 = 474 are not five-folds (+ argument)
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Bingo! x = 4*79 = 316 = 79 ∗ 22 ∗ 30 ∗ 50
and y + z + w = 395
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Reject x = 79, x = 2 ∗ 79 = 158, and x = 3 ∗ 79 = 237
because 711 − 79 = 632, 711 − 158 = 553, and
711 − 237 = 474 are not five-folds (+ argument)
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Bingo! x = 4*79 = 316 = 79 ∗ 22 ∗ 30 ∗ 50
and y + z + w = 395
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So, at least one number is a five-fold: either only
one number is (excluded), or all three are
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Reject x = 79, x = 2 ∗ 79 = 158, and x = 3 ∗ 79 = 237
because 711 − 79 = 632, 711 − 158 = 553, and
711 − 237 = 474 are not five-folds (+ argument)
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Bingo! x = 4*79 = 316 = 79 ∗ 22 ∗ 30 ∗ 50
and y + z + w = 395
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So, at least one number is a five-fold: either only
one number is (excluded), or all three are
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Then, y = 5y 0 , z = 5z 0 , w = 5w0 and y 0 + z 0 + w0 = 79
y 0 z 0 w0 = 18000
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Reject x = 79, x = 2 ∗ 79 = 158, and x = 3 ∗ 79 = 237
because 711 − 79 = 632, 711 − 158 = 553, and
711 − 237 = 474 are not five-folds (+ argument)
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Bingo! x = 4*79 = 316 = 79 ∗ 22 ∗ 30 ∗ 50
and y + z + w = 395
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So, at least one number is a five-fold: either only
one number is (excluded), or all three are
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Then, y = 5y 0 , z = 5z 0 , w = 5w0 and y 0 + z 0 + w0 = 79
y 0 z 0 w0 = 18000
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Not all three are five folds.
A single one cannot be a five fold (125 > 79)
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So, one must be a multiple of 25; the other
multiple of 5: y 0 = 25y 00 , z 0 = 5z 00 .
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Then, 25y 00 + 5z 00 + w0 = 79 and y 00 z 00 w0 = 144
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So, one must be a multiple of 25; the other
multiple of 5: y 0 = 25y 00 , z 0 = 5z 00 .
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Then, 25y 00 + 5z 00 + w0 = 79 and y 00 z 00 w0 = 144
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So, y 00 = 1 or 2
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So, one must be a multiple of 25; the other
multiple of 5: y 0 = 25y 00 , z 0 = 5z 00 .
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Then, 25y 00 + 5z 00 + w0 = 79 and y 00 z 00 w0 = 144
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So, y 00 = 1 or 2
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Test either case... conclude that y 00 = 2 is
impossible.
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So, one must be a multiple of 25; the other
multiple of 5: y 0 = 25y 00 , z 0 = 5z 00 .
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Then, 25y 00 + 5z 00 + w0 = 79 and y 00 z 00 w0 = 144
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So, y 00 = 1 or 2
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Test either case... conclude that y 00 = 2 is
impossible.
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Then, y 00 = 1.
Bingo!
y = 125
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So, one must be a multiple of 25; the other
multiple of 5: y 0 = 25y 00 , z 0 = 5z 00 .
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Then, 25y 00 + 5z 00 + w0 = 79 and y 00 z 00 w0 = 144
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So, y 00 = 1 or 2
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Test either case... conclude that y 00 = 2 is
impossible.
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Then, y 00 = 1.
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Finally, we must solve 5z 00 + w0 = 54 and z 00 w0 = 144
Bingo!
y = 125
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So, one must be a multiple of 25; the other
multiple of 5: y 0 = 25y 00 , z 0 = 5z 00 .
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Then, 25y 00 + 5z 00 + w0 = 79 and y 00 z 00 w0 = 144
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So, y 00 = 1 or 2
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Test either case... conclude that y 00 = 2 is
impossible.
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Then, y 00 = 1.
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Finally, we must solve 5z 00 + w0 = 54 and z 00 w0 = 144
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Solve a quadratic equation: z 00 = 6, w0 = 24
Bingo!
y = 125
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So, one must be a multiple of 25; the other
multiple of 5: y 0 = 25y 00 , z 0 = 5z 00 .
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Then, 25y 00 + 5z 00 + w0 = 79 and y 00 z 00 w0 = 144
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So, y 00 = 1 or 2
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Test either case... conclude that y 00 = 2 is
impossible.
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Then, y 00 = 1.
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Finally, we must solve 5z 00 + w0 = 54 and z 00 w0 = 144
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Solve a quadratic equation: z 00 = 6, w0 = 24
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Summary:
x = 316, y = 125, z = 25 ∗ 6 = 150, w = 5 ∗ 24 = 120
Bingo!
y = 125
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Solution:
x = 316 = 79 ∗ 22
y = 125 = 53
z = 150 = 2 ∗ 3 ∗ 52 w = 120 = 23 ∗ 3 ∗ 5
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Solution:
x = 316 = 79 ∗ 22
y = 125 = 53
z = 150 = 2 ∗ 3 ∗ 52 w = 120 = 23 ∗ 3 ∗ 5
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Prices:
1.20
1.25
1.50
3.16
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Solution:
x = 316 = 79 ∗ 22
y = 125 = 53
z = 150 = 2 ∗ 3 ∗ 52 w = 120 = 23 ∗ 3 ∗ 5
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Prices:
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Challenge: Can the 7-11 puzzle be solved using
symmetries (group theory)?
1.20
1.25
1.50
3.16
Modern Applications
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Maxwell’s equations: merging electricity with
magnetism
Modern Applications
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Maxwell’s equations: merging electricity with
magnetism
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Einstein: general relativity
Modern Applications
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Maxwell’s equations: merging electricity with
magnetism
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Einstein: general relativity
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Merging general relativity and quantum mechanics
Modern Applications
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Maxwell’s equations: merging electricity with
magnetism
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Einstein: general relativity
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Merging general relativity and quantum mechanics
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String theory, super string theory
Modern Applications
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Maxwell’s equations: merging electricity with
magnetism
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Einstein: general relativity
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Merging general relativity and quantum mechanics
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String theory, super string theory
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A theory for everything
Literature
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Ian Stewart, Why Beauty is Truth:
The History of Symmetry, Basic Books, The
Perseus Books Group, April 2007, 290 pages.
Podcast series, University of Warwick, 2007
(7 episodes, ∼ 90 minutes total)
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Mario Livio, The Equation That Couldn’t Be
Solved, Simon & Schuster, 2005, 368 pages.
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Thank You!