the puzzling beauty of numbers
TRANSCRIPT
THE P U Z Z L I N G BEAUTY OF NUMBERS
Kiryl Tsishchanka
Department of Mathematics
Knox College
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.1/41
A little known verse of the Bible reads
And he made a molten sea, ten cubits from the
one brim to the other: it was round all about, and
his height was five cubits: and a line of thirty
cubits did compass it about. (I King 7:23)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.2/41
Bible 550?BCE 1 place 3
Al-Kashi 1429 14 places 3.14159265358979
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.3/41
Bible 550?BCE 1 place 3
Archimedes 250?BCE 3 places 3.141
Al-Kashi 1429 14 places 3.14159265358979
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.4/41
Bible 550?BCE 1 place 3
Archimedes 250?BCE 3 places 3.141
Zu Chongzhi 480?AD 7 places 3.1415926
Al-Kashi 1429 14 places 3.14159265358979
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.5/41
Bible 550?BCE 1 place 3
Archimedes 250?BCE 3 places 3.141
Zu Chongzhi 480?AD 7 places 3.1415926
Al-Kashi 1429 14 places 3.14159265358979
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.6/41
Bible 550?BCE 1 place 3
Archimedes 250?BCE 3 places 3.141
Zu Chongzhi 480?AD 7 places 3.1415926
Al-Kashi 1429 14 places 3.14159265358979
Machin 1706 100 places
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.7/41
Bible 550?BCE 1 place 3
Archimedes 250?BCE 3 places 3.141
Zu Chongzhi 480?AD 7 places 3.1415926
Al-Kashi 1429 14 places 3.14159265358979
Machin 1706 100 places
Shanks 1874 707 places (527 correct)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.8/41
Bible 550?BCE 1 place 3
Archimedes 250?BCE 3 places 3.141
Zu Chongzhi 480?AD 7 places 3.1415926
Al-Kashi 1429 14 places 3.14159265358979
Machin 1706 100 places
Shanks 1874 707 places (527 correct)
Bellard 1996 400.000.000.000 places
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.9/41
Wallis:π
2=
2 · 2 · 4 · 4 · 6 · 6 · 8 · 8 · 10 · 10 . . .
1 · 3 · 3 · 5 · 5 · 7 · 7 · 9 · 9 · 11 . . .
Leibniz:π
4= 1 −
1
3+
1
5−
1
7+
1
9−
1
11+ . . .
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.10/41
Wallis:π
2=
2 · 2 · 4 · 4 · 6 · 6 · 8 · 8 · 10 · 10 . . .
1 · 3 · 3 · 5 · 5 · 7 · 7 · 9 · 9 · 11 . . .
Leibniz:π
4= 1 −
1
3+
1
5−
1
7+
1
9−
1
11+ . . .
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.10/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.11/41
π
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.11/41
πTHE P U Z Z L I N G BEAUTY OF NUMBERS – p.11/41
πTHE P U Z Z L I N G BEAUTY OF NUMBERS – p.11/41
π
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.11/41
π
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.11/41
π
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.11/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
TOP TEN LIST
10. e is easier to spell than π.
9. π ≈ 3.14 while e ≈ 2.718281828459045.
8. The character for e can be found on a keyboard, but π can’t.
7. Everybody fights for their piece of the pie.
6. ln π is a really nasty number, but ln e = 1.
5. e is used in calculus while π is used in baby geometry.
4. ’e’ is the most commonly picked vowel in Wheel of Fortune.
3. e stands for Euler’s Number, π does not stand for squat.
2. You don’t need to know Greek to be able to use e.
1. You can’t confuse e with a food product.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.12/41
−1= (−1)1 = (−1)13+1
3+1
3
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.13/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.14/41
−1 = (−1)1 = (−1)13+1
3+1
3
= (−1)16
·2+16
·2+16
·2
= (−1)16
·2(−1)16
·2(−1)16
·2
= (−1)2·16 (−1)2·1
6 (−1)2·16
= [(−1)2]16 [(−1)2]
16 [(−1)2]
16
= 11/611/611/6
= 116+1
6+1
6
= 11/2
=√
1
= 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.15/41
DEFINITION:
Suppose that a is a nonzero real number and that m and n are
integers such that n > 0. Then
am/n = (am)1/n,
provided a1/n exists.
EXAMPLE:
(−1)2·16 (−1)2·1
6 (−1)2·16 6=
[
(−1)2]
16
[
(−1)2]
16
[
(−1)2]
16 ,
since (−1)16 does not exist.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.16/41
DEFINITION:
Suppose that a is a nonzero real number and that m and n are
integers such that n > 0. Then
am/n = (am)1/n,
provided a1/n exists.
EXAMPLE:
(−1)2·16 (−1)2·1
6 (−1)2·16 6=
[
(−1)2]
16
[
(−1)2]
16
[
(−1)2]
16 ,
since (−1)16 does not exist.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.16/41
B C
A
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.17/41
B C
A
D
E
FG
H
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.17/41
B C
A
D
E
FG
H
K L
M
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.17/41
PROBLEM:
A plane is colored blue and red. Is it always possible to find two
points of the same color exactly 1 inch apart?
SOLUTION:
Think of an equilateral triangle with each side exactly 1 inch long.
At least two of its vertices have to be of the same color. This
proves that there have to be two points of the same color exactly 1
inch apart.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.18/41
PROBLEM:
A plane is colored blue and red.
Is it always possible to find two
points of the same color exactly 1 inch apart?
SOLUTION:
Think of an equilateral triangle with each side exactly 1 inch long.
At least two of its vertices have to be of the same color. This
proves that there have to be two points of the same color exactly 1
inch apart.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.18/41
PROBLEM:
A plane is colored blue and red. Is it always possible to find two
points of the same color exactly 1 inch apart?
SOLUTION:
Think of an equilateral triangle with each side exactly 1 inch long.
At least two of its vertices have to be of the same color. This
proves that there have to be two points of the same color exactly 1
inch apart.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.18/41
PROBLEM:
A plane is colored blue and red. Is it always possible to find two
points of the same color exactly 1 inch apart?
SOLUTION:
Think of an equilateral triangle with each side exactly 1 inch long.
At least two of its vertices have to be of the same color. This
proves that there have to be two points of the same color exactly 1
inch apart.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.18/41
PROBLEM:
A plane is colored blue and red. Is it always possible to find two
points of the same color exactly 1 inch apart?
SOLUTION:
Think of an equilateral triangle with each side exactly 1 inch long.
At least two of its vertices have to be of the same color. This
proves that there have to be two points of the same color exactly 1
inch apart.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.18/41
PROBLEM:
A plane is colored blue and red. Is it always possible to find two
points of the same color exactly 1 inch apart?
SOLUTION:
Think of an equilateral triangle with each side exactly 1 inch long.
At least two of its vertices have to be of the same color.
This
proves that there have to be two points of the same color exactly 1
inch apart.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.18/41
PROBLEM:
A plane is colored blue and red. Is it always possible to find two
points of the same color exactly 1 inch apart?
SOLUTION:
Think of an equilateral triangle with each side exactly 1 inch long.
At least two of its vertices have to be of the same color. This
proves that there have to be two points of the same color exactly 1
inch apart.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.18/41
PROBLEM:
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
� � �
� � � �
� � � � �
� � � �
� � �
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
1 2 3
4 5 6 7
8 9 10 11 12
13 14 15 16
17 18 19
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
-22 -8 -17
-23 -20 24 33
-28 19 -88 -9 1
-2 4 -54 -7
0 -42 5
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
-22 -8 -17
-23 -20 24 33
-28 19 -88 -9 1
-2 4 -54 -7
0 -42 5
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
-22 -8 -17
-23 -20 24 33
-28 19 -88 -9 1
-2 4 -54 -7
0 -42 5
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
-22 -8 -17
-23 -20 24 33
-28 19 -88 -9 1
-2 4 -54 -7
0 -42 5
×
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
-22 -8 -17
-23 -20 24 33
-28 19 -88 -9 1
-2 4 -54 -7
0 -42 5
× ×
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
-22 -8 -17
-23 -20 24 33
-28 19 -88 -9 1
-2 4 -54 -7
0 -42 5
× × ×
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
-22 -8 -17
-23 -20 24 33
-28 19 -88 -9 1
-2 4 -54 -7
0 -42 5
× × ×
× × ×× × × ×
× ×
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
× × ×× × × ×
× ×
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
1 2 3
4 5 6 7
8 9 10 11 12
13 14 15 16
17 18 19
× × ×
× × × ×
× × ×
× ×
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
PROBLEM:
A right hexagon is split into 24 triangles.
� � �
� � � �
� � � � �
� � � �
� � �
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.19/41
Math problems?
Call 1 − 800 − (10x)2 −sin(xy)
2.36z
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.20/41
Math problems? Call
1 − 800 − (10x)2 −sin(xy)
2.36z
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.20/41
Math problems? Call 1 − 800
− (10x)2 −sin(xy)
2.36z
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.20/41
Math problems? Call 1 − 800 − (10x)2
−sin(xy)
2.36z
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.20/41
Math problems? Call 1 − 800 − (10x)2 −sin(xy)
2.36z
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.20/41
PROBLEM:
A right hexagon is split into 24 triangles.
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
× × ×× × × ×
× ×
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.21/41
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
×
×
×
×
× × ×
× ×
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.22/41
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
×
×
×
×
× × ×
× ×
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.22/41
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
×
×
×
×
× × ×
× ×
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.22/41
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
×
×
×
×
× × ×
× ×
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.22/41
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
×
×
×
×
× × ×
× ×
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.22/41
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
×
×
×
×
× × ×
× ×
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.22/41
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
×
×
×
×
× × ×
× ×
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.22/41
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
×
×
×
×
× × ×
× ×
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.22/41
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
×
×
×
×
× × ×
× ×
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.22/41
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
×
×
×
×
× × ×
× ×
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.22/41
PROBLEM:
A right hexagon is split into 24 triangles.
22 8 17
23 20 24 33
28 19 88 9 1
2 4 54 7
0 42 5
× × ×
× × ×× × × ×
× ×
19 different numbers are written in 19 nodes. Prove that at least
7 of 24 triangles have the following property: The numbers in its
vertices increase (from the least to the greatest) counterclockwise.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.23/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We now note that
4=2+2 14=3+11 24=3+11 34=3+316=3+3 16=3+13 26=7+19 36=5+318=3+5 18=5+13 28=5+23 38=7+3110=3+7 20=7+13 30=7+23 40=3+3712=5+7 22=5+17 32=3+29 42=5+37
Goldbach’s Conjecture: Every even positive integer greater that 2
can be written as the sum of two primes.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.24/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We now note that
4=2+2 14=3+11 24=3+11 34=3+316=3+3 16=3+13 26=7+19 36=5+318=3+5 18=5+13 28=5+23 38=7+3110=3+7 20=7+13 30=7+23 40=3+3712=5+7 22=5+17 32=3+29 42=5+37
Goldbach’s Conjecture: Every even positive integer greater that 2
can be written as the sum of two primes.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.24/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We now note that
4=2+2 14=3+11 24=3+11 34=3+316=3+3 16=3+13 26=7+19 36=5+318=3+5 18=5+13 28=5+23 38=7+3110=3+7 20=7+13 30=7+23 40=3+3712=5+7 22=5+17 32=3+29 42=5+37
We now note that
4=2+2 14=3+11 24=3+11 34=3+316=3+3 16=3+13 26=7+19 36=5+318=3+5 18=5+13 28=5+23 38=7+3110=3+7 20=7+13 30=7+23 40=3+3712=5+7 22=5+17 32=3+29 42=5+37
Goldbach’s Conjecture: Every even positive integer greater that 2
can be written as the sum of two primes.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.24/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We now note that
4=2+2 14=3+11 24=3+11 34=3+316=3+3 16=3+13 26=7+19 36=5+318=3+5 18=5+13 28=5+23 38=7+3110=3+7 20=7+13 30=7+23 40=3+3712=5+7 22=5+17 32=3+29 42=5+37
We now note that
4=2+2 14=3+11 24=3+11 34=3+316=3+3 16=3+13 26=7+19 36=5+318=3+5 18=5+13 28=5+23 38=7+3110=3+7 20=7+13 30=7+23 40=3+3712=5+7 22=5+17 32=3+29 42=5+37
Goldbach’s Conjecture: Every even positive integer greater that 2
can be written as the sum of two primes.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.24/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We now note that
4=2+2 14=3+11 24=3+11 34=3+316=3+3 16=3+13 26=7+19 36=5+318=3+5 18=5+13 28=5+23 38=7+3110=3+7 20=7+13 30=7+23 40=3+3712=5+7 22=5+17 32=3+29 42=5+37
We now note that
4=2+2 14=3+11 24=3+11 34=3+316=3+3 16=3+13 26=7+19 36=5+318=3+5 18=5+13 28=5+23 38=7+3110=3+7 20=7+13 30=7+23 40=3+3712=5+7 22=5+17 32=3+29 42=5+37
Goldbach’s Conjecture: Every even positive integer greater that 2
can be written as the sum of two primes.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.24/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We now note that
4=2+2 14=3+11 24=3+11 34=3+316=3+3 16=3+13 26=7+19 36=5+318=3+5 18=5+13 28=5+23 38=7+3110=3+7 20=7+13 30=7+23 40=3+3712=5+7 22=5+17 32=3+29 42=5+37
Goldbach’s Conjecture: Every even positive integer greater that 2
can be written as the sum of two primes.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.24/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We now note that
4=2+2 14=3+11 24=3+11 34=3+316=3+3 16=3+13 26=7+19 36=5+318=3+5 18=5+13 28=5+23 38=7+3110=3+7 20=7+13 30=7+23 40=3+3712=5+7 22=5+17 32=3+29 42=5+37
Goldbach’s Conjecture: Every even positive integer greater that 2
can be written as the sum of two primes.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.24/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We also note that
220
+ 1 = 3 is a prime
221
+ 1 = 5 is a prime
222
+ 1 = 17 is a prime
223
+ 1 = 257 is a prime
224
+ 1 = 65537 is a prime
225
+ 1 = 4294967297 is composite!
In fact, one can check that 4294967297 = 641 · 6700417.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.25/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We also note that
220
+ 1 = 3 is a prime
221
+ 1 = 5 is a prime
222
+ 1 = 17 is a prime
223
+ 1 = 257 is a prime
224
+ 1 = 65537 is a prime
225
+ 1 = 4294967297 is composite!
In fact, one can check that 4294967297 = 641 · 6700417.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.25/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We also note that
220
+ 1 = 3 is a prime
221
+ 1 = 5 is a prime
222
+ 1 = 17 is a prime
223
+ 1 = 257 is a prime
224
+ 1 = 65537 is a prime
225
+ 1 = 4294967297 is composite!
In fact, one can check that 4294967297 = 641 · 6700417.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.25/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We also note that
220
+ 1 = 3 is a prime
221
+ 1 = 5 is a prime
222
+ 1 = 17 is a prime
223
+ 1 = 257 is a prime
224
+ 1 = 65537 is a prime
225
+ 1 = 4294967297 is composite!
In fact, one can check that 4294967297 = 641 · 6700417.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.25/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We also note that
220
+ 1 = 3 is a prime
221
+ 1 = 5 is a prime
222
+ 1 = 17 is a prime
223
+ 1 = 257 is a prime
224
+ 1 = 65537 is a prime
225
+ 1 = 4294967297 is composite!
In fact, one can check that 4294967297 = 641 · 6700417.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.25/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We also note that
220
+ 1 = 3 is a prime
221
+ 1 = 5 is a prime
222
+ 1 = 17 is a prime
223
+ 1 = 257 is a prime
224
+ 1 = 65537 is a prime
225
+ 1 = 4294967297 is composite!
In fact, one can check that 4294967297 = 641 · 6700417.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.25/41
DEFINITION:
A prime number is a positive integer greater than 1 that is divisible
by no positive integers other than 1 and itself.
EXAMPLE: The integers 2, 3, 5, 7, 11, 13, 17, 19 are primes.
We also note that
220
+ 1 = 3 is a prime
221
+ 1 = 5 is a prime
222
+ 1 = 17 is a prime
223
+ 1 = 257 is a prime
224
+ 1 = 65537 is a prime
225
+ 1 = 4294967297 is composite!
In fact, one can check that 4294967297 = 641 · 6700417.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.25/41
TRUE OR NOT TRUE?
Numbers 8 and 9 are the only consecutive powers.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.26/41
Cathalan’s Conjecture:
Numbers 8 and 9 are the only consecutive powers.
TRUE (Michalesku, 2002).
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.27/41
Cathalan’s Conjecture:
Numbers 8 and 9 are the only consecutive powers.
TRUE (Michalesku, 2002).
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.27/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.28/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.28/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.28/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.28/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.28/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.28/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.28/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.28/41
THEOREM:
For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF:
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY):
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT):
Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some
x1, . . . , xn ∈ [A, B]. Suppose that f(x1, . . . , xn) attains its
minimal value on [A, B] at some a1, . . . , an ∈ [A, B] with
a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
(
a1+a2
2,a1+a2
2, a3, . . . , an,
)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some
x1, . . . , xn ∈ [A, B]. Suppose that f(x1, . . . , xn) attains its
minimal value on [A, B] at some a1, . . . , an ∈ [A, B] with
a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
(
a1+a2
2,a1+a2
2, a3, . . . , an,
)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some
x1, . . . , xn ∈ [A, B]. Suppose that f(x1, . . . , xn) attains its
minimal value on [A, B] at some a1, . . . , an ∈ [A, B] with
a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
(
a1+a2
2,a1+a2
2, a3, . . . , an,
)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some
x1, . . . , xn ∈ [A, B]. Suppose that f(x1, . . . , xn) attains its
minimal value on [A, B] at some a1, . . . , an ∈ [A, B] with
a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
(
a1+a2
2,a1+a2
2, a3, . . . , an,
)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some
x1, . . . , xn ∈ [A, B]. Suppose that f(x1, . . . , xn) attains its
minimal value on [A, B] at some a1, . . . , an ∈ [A, B] with
a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
(
a1+a2
2,a1+a2
2, a3, . . . , an,
)
�
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.29/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we havea1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
f(a1, a2, a3, . . . , an) =a1 + a2 + a3 + . . . + an
n− n
√a1a2a3 . . . an
f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
=
a1+a2
2+ a1+a2
2+a3+. . .+an
n− n
q
a1+a2
2
a1+a2
2a3 . . . an
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.30/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some
x1, . . . , xn ∈ [A, B]. Suppose that f(x1, . . . , xn) attains its
minimal value on [A, B] at some a1, . . . , an ∈ [A, B] with
a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
(
a1+a2
2,a1+a2
2, a3, . . . , an,
)
�
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.31/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some
x1, . . . , xn ∈ [A, B]. Suppose that f(x1, . . . , xn) attains its
minimal value on [A, B] at some a1, . . . , an ∈ [A, B] with
a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
(
a1+a2
2,a1+a2
2, a3, . . . , an,
)
�
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.31/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some
x1, . . . , xn ∈ [A, B]. Suppose that f(x1, . . . , xn) attains its
minimal value on [A, B] at some a1, . . . , an ∈ [A, B] with
a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
(
a1+a2
2,a1+a2
2, a3, . . . , an,
)
�
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.31/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.32/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
THEOREM:
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.32/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
THEOREM (Extreme-Value Theorem):
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.32/41
THEOREM: For any a1, . . . , an ≥ 0 we have
a1 + . . . + an
n≥ n
√a1 . . . an
PROOF (ELEMENTARY, SHORT): Put
f(x1, . . . , xn) =x1 + . . . + xn
n− n
√x1 . . . xn
Assume to the contrary that f(x1, . . . , xn) < 0 for some x1, . . . , xn ∈ [A, B].
Suppose that f(x1, . . . , xn) attains its minimal value on [A, B] at some a1, . . . , an ∈[A, B] with a1 6= a2. This gives us a contradiction, since
f(a1, a2, a3, . . . , an)>f
„
a1+a2
2,
a1+a2
2, a3, . . . , an,
«
�
THEOREM (Extreme-Value Theorem): If f(x1, . . . , xn) is contin-
uous on a closed and bounded set R, then f has both an absolute
maximum and an absolute minimum on R.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.32/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2 u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2 u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =
∫
2 sin x cos x dx
=
sin x = u
d sin x = du
cos x dx = du
=
∫
2u du
= u2 + C
= sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
1. There is a mistake. Where???
2. −1
2cos 2x = sin2 x. Wrong!
3. −1
2cos 2x +
1
2= sin2 x. Correct.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
1. There is a mistake. Where???
2. −1
2cos 2x = sin2 x. Wrong!
3. −1
2cos 2x +
1
2= sin2 x. Correct.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
1. There is a mistake. Where???
2. −1
2cos 2x = sin2 x. Wrong!
3. −1
2cos 2x +
1
2= sin2 x. Correct.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
1. There is a mistake. Where???
2. −1
2cos 2x = sin2 x. Wrong!
3. −1
2cos 2x +
1
2= sin2 x. Correct.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
1. There is a mistake. Where???
2. −1
2cos 2x = sin2 x. Wrong!
3. −1
2cos 2x +
1
2= sin2 x. Correct.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
1. There is a mistake. Where???
2. −1
2cos 2x = sin2 x. Wrong!
3. −1
2cos 2x +
1
2= sin2 x. Correct.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
We have:∫
sin 2x dx =1
2
∫
2 sin 2x dx
∫
sin 2x dx =
∫
2 sin x cos x dx
=
2x = u
d2x = du
2 dx = du
=
sin x = u
d sin x = du
cos x dx = du
=1
2
∫
sin u du =
∫
2u du
= −1
2cos u + C = u2 + C
= −1
2cos 2x + C = sin2 x + C
1. There is a mistake. Where???
2. −1
2cos 2x = sin2 x. Wrong!
3. −1
2cos 2x +
1
2= sin2 x. Correct.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.33/41
limx→8+
1
x − 8= +∞
limx→4+
1
x − 4= +∞
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.34/41
limx→8+
1
x − 8= +∞
limx→4+
1
x − 4= +∞
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.34/41
limx→8+
1
x − 8= +∞
limx→4+
1
x − 4= +∞
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.34/41
limx→8+
1
x − 8= +∞
limx→4+
1
x − 4=
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.34/41
limx→8+
1
x − 8= +∞
limx→4+
1
x − 4= + 4
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.34/41
limx→8+
1
x − 8= +∞
limx→4+
1
x − 4= +
4
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.34/41
limx→8+
1
x − 8= +∞
limx→4+
1
x − 4= +
4
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.34/41
PROBLEM:
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
1 = 1 7 = 1 + 2 + 4
2 = 2 8 = 8
3 = 1 + 2 9 = 1 + 8
4 = 4 10 = 2 + 8
5 = 1 + 4 11 = 1 + 2 + 8
6 = 2 + 4 12 = 4 + 8THE P U Z Z L I N G BEAUTY OF NUMBERS – p.35/41
1=1 16=16 31=1+2+4+8+162=2 17=1+16 32=323=1+2 18=2+16 33=1+324=4 19=1+2+16 34=2+325=1+4 20=4+16 35=1+2+326=2+4 21=1+4+16 36=4+327=1+2+4 22=2+4+16 37=1+4+328=8 23=1+2+4+16 38=2+4+329=1+8 24=8+16 39=1+2+4+3210=2+8 25=1+8+16 40=8+3211=1+2+8 26=2+8+16 41=1+8+3212=4+8 27=1+2+8+16 42=2+8+3213=1+4+8 28=4+8+16 43=1+2+8+3214=2+4+8 29=1+4+8+16 44=4+8+3215=1+2+4+8 30=2+4+8+16 45=1+4+8+32
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.36/41
a32 − 1 (a − 1)(a31 + a30 + a29 + . . . + a2 + a + 1)11111111
a32 − 1 ≡
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)(a31 + a30 + a29 + . . . + a2 + a + 1)11111111
a32 − 1 ≡
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)(a31 + a30 + a29 + . . . + a2 + a + 1)11111111
a32 − 1
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)(a31 + a30 + a29 + . . . + a2 + a + 1)11111111
a32 − 1 ≡ (a16 − 1)(a16 + 1)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)(a31 + a30 + a29 + . . . + a2 + a + 1)11111111
a32 − 1 ≡ (a8 − 1)(a8 + 1)(a16 + 1)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)(a31 + a30 + a29 + . . . + a2 + a + 1)11111111
a32 − 1 ≡ (a4 − 1)(a4 + 1)(a8 + 1)(a16 + 1)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)(a31 + a30 + a29 + . . . + a2 + a + 1)11111111
a32 − 1 ≡ (a2 − 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)(a31 + a30 + a29 + . . . + a2 + a + 1)11111111
a32 − 1 ≡ (a − 1)(a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)(a31 + a30 + a29 + . . . + a2 + a + 1)11111111
a32 − 1 ≡ (a − 1)(a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)(a31 + a30 + a29 + . . . + a2 + a + 1)11111111
a32 − 1 ≡ (a − 1)(a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 11111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 11111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 11111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 11111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 11111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 11111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 11111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 11111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16+
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 1+ 1111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16+
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 1+ 1111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16+
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16+
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 1++ 1111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16+
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16+
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 1++ 1111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16+
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16+ + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16+
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 11111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
a32 − 1 ≡ (a − 1)a31 + a30 + a29 + . . . + a2 + a + 11111111111111111
a32 − 1 ≡ (a − 1)≡ (a + 1)(a2 + 1)(a4 + 1)(a8 + 1)(a16 + 1)
a32 − 1 ≡ (a − 1)≡ 1 + a + a2 + a1+2 + a4 + a1+4 + a2+4
a32 − 1 ≡ (a − 1) ≡ 1 + a1+2+4 + a8 + a1+8 + a2+8 + a1+2+8
(a − 1) ≡ 1 + a4+8 + a1+4+8 + a2+4+8 + a2+4+8+16
(a − 1) ≡ 1 + a16 + a1+16 + a2+16 + a1+2+16 + a4+16
(a − 1) ≡ 1 + a1+4+16 + a2+4+16 + a1+2+8+16 + a8+16
(a − 1) ≡ 1 + a1+4+8+16 + a2+8+16 + a1+8+16 + a4+8+16
(a − 1) ≡ 1 + a1+2+4+8 + a1+2+4+16 + a1+2+4+8+16
a2n−1 + a2n−2 + a2n−3 + . . . + a2 + a + 1
≡ (a + 1)(a2 + 1)(a4 + 1) . . . (a2n−1
+ 1)
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.37/41
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
GENERATING FUNCTIONS
1
1 − a= 1 + a + a2 + a3 + . . . 1, 2, 22, . . . ⇒ N
= (1 + a)(1 + a2)(1 + a4) . . .
1
(1 − a)2= 1 + 2a + 3a2 + 4a4 + . . . x1 + x2 = N
= (1 + a + a2 + a3 + . . .)2
1
(1 − a)3= 1 + 3a + 6a2 + 10a4 + . . . x1 + x2 + x3 = N
= (1 + a + a2 + a3 + . . .)3
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.38/41
100 = 64 + 32 + 4
= 1 · 64 + 1 · 32 + 0 · 16 + 0 · 8 + 1 · 4 + 0 · 2 + 0 · 1
= (11001001)2
100 = 64 + 32 + 4
= 1 · 64 + 1 · 32 + 0 · 16 + 0 · 8 + 1 · 4 + 0 · 2 + 0 · 1
= (11001001)2
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
ANSWER:
Any, since any decimal number can be converted into the binarysystem.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.39/41
100 = 64 + 32 + 4
= 1 · 64 + 1 · 32 + 0 · 16 + 0 · 8 + 1 · 4 + 0 · 2 + 0 · 1
= (11001001)2
100 = 64 + 32 + 4
= 1 · 64 + 1 · 32 + 0 · 16 + 0 · 8 + 1 · 4 + 0 · 2 + 0 · 1
= (11001001)2
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
ANSWER:
Any, since any decimal number can be converted into the binarysystem.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.39/41
100 = 64 + 32 + 4
= 1 · 64 + 1 · 32 + 0 · 16 + 0 · 8 + 1 · 4 + 0 · 2 + 0 · 1
= (11001001)2
100 = 64 + 32 + 4
= 1 · 64 + 1 · 32 + 0 · 16 + 0 · 8 + 1 · 4 + 0 · 2 + 0 · 1
= (11001001)2
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
ANSWER:
Any, since any decimal number can be converted into the binarysystem.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.39/41
100 = 64 + 32 + 4
= 1 · 64 + 1 · 32 + 0 · 16 + 0 · 8 + 1 · 4 + 0 · 2 + 0 · 1
= (11001001)2
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
ANSWER:
Any, since any decimal number can be converted into the binarysystem.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.39/41
100 = 64 + 32 + 4
= 1 · 64 + 1 · 32 + 0 · 16 + 0 · 8 + 1 · 4 + 0 · 2 + 0 · 1
= (11001001)2
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
ANSWER:
Any, since any decimal number can be converted into the binarysystem.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.39/41
100 = 64 + 32 + 4
= 1 · 64 + 1 · 32 + 0 · 16 + 0 · 8 + 1 · 4 + 0 · 2 + 0 · 1
= (11001001)2
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
ANSWER:
Any, since any decimal number can be converted into the binarysystem.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.39/41
100 = 64 + 32 + 4
= 1 · 64 + 1 · 32 + 0 · 16 + 0 · 8 + 1 · 4 + 0 · 2 + 0 · 1
= (11001001)2
PROBLEM:
Consider the following weights:
1 2 4 8 16 32 64 128 etc.
Which weight can be weighed by them?
ANSWER:
Any, since any decimal number can be converted into the binarysystem.
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.39/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.40/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.40/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.40/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.40/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.40/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.40/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.40/41
THE P U Z Z L I N G BEAUTY OF NUMBERS – p.41/41