Transcript magic rabbit

A Mathematics Teacher’s
Responsibility –
Beyond the Curriculum
o. Univ. Prof. Dr. Alfred S. Posamentier
Dean, The School of Education
The City College
The City University of New York
The Extra Responsibilities
• Identifying the problem
– Most parents are ill-informed about
mathematics:
• They may have had bad experiences with
mathematics.
• They may not like mathematics.
• They may not know the true purpose of learning
mathematics.
• They may not be able to help their children with
mathematics.
The Extra Responsibilities
• Make the parents true partners in the teaching of
mathematics
– Provide them with the mathematics curriculum
– Help them understand the role they are to play at
home to support in-class instruction
– Provide them (if necessary) with the information
they will need (e.g. arithmetic algorithms)
– Have their children bring home examples of
mathematics that demonstrate its power and beauty
The Extra Responsibilities
• To enrich the curriculum
– Acceleration: provide future topics earlier
– Expansion: enlarging a topic beyond the
requirement
– Extension: move to a related topic beyond
the requirement
– Expose the astonishing in mathematics!
The Most Beautiful Magic Square
The “Melencolia I” by Albrecht Dürer (1471 -1528)
The Magic Square
16
3
2
13
5
10
11
8
9
6
7
12
4
15
14
1
Some properties of this magic square:
Sum of all rows, columns and diagonals is 34
The four corner numbers have a sum of 34.
16 + 13 + 1 + 4 = 34
16 3
2 13
5 10 11 8
Each of the four corner 2 by 2 squares has a sum of 34. 9
16+3+5+10 = 34
4
2+13+11+8 = 34
9+ 6+ 4+15 = 34
7+12+14+1 = 34
6
7 12
15 14 1
The center 2 by 2 square has a sum of 34.
10 + 11 + 6 + 7 = 34
The sum of the numbers in the diagonal cells
equals the sum of the numbers in the cells not in the diagonals.
16+10+7+1+4+6+11+13 = 3+2+8+12+14+15+9+5 = 68.
More properties of this magic
square!
The sum of the squares of the numbers in the
diagonal cells equals the sum of the squares
of the numbers not in the diagonal cells.
162 +102 +7 2 +12 +42 +62 +112 +132
= 32 +22 +82 +122 +142 +152 +92 +52 = 748
16
3
2
13
5
10 11
8
9
6
7
12
4
15 14
1
The sum of the cubes of the numbers in the diagonal cells
equals the sum of the cubes of the numbers not in the diagonal cells.
163 +103 +73 +13 +43 +63 +113 +133
= 33 +23 +83 +123 +143 +153 +93 +53 = 9,248
The sum of the squares of the numbers in the diagonal cells
equals the sum of the squares of the numbers in the first and third rows.
162 +102 +7 2 +12 +42 +62 +112 +132  162  32  22  132  92  62  7 2  122  748
The Fabulous Fibonacci
Numbers
Leonardo Pisano
The Rabbit Problem
Beginning 1
First
2
Second
3
Third
5
Fourth
8
Fifth
13
Sixth
21
Seventh
34
Eighth
55
Ninth
89
Tenth
144
Eleventh 233
Twelfth 377
“A certain man had one pair of rabbits together in a certain enclosed place, and one
wishes to know how many are created from the pair in one year when it is the
nature of them in a single month to bear another pair, and in the second month
those born to bear also. Because the above written pair in the first month bore,
you will double it; there will be two pairs in one month. One of these, namely
the first, bears in the second month, and thus there are in the second month 3
pairs; of these in one month two are pregnant and in the third month 2 pairs of
rabbits are born and thus there are 5 pairs in the month; in this month 3 pairs
are pregnant and in the fourth month there are 8 pairs, of which 5 pairs bear
another 5 pairs; these are added to the 8 pairs making 13 pairs in the fifth
month; these 5 pairs that are born in this month do not mate in this month, but
another 8 pairs are pregnant, and thus there are in the sixth month 21 pairs; to
these are added the 13 pairs that are born in the seventh month; there will be
34 pairs in this month; to this are added the 21 pairs that are born in the eighth
month; there will be 55 pairs in this month; to these are added the 34 pairs
that are born in the ninth month; there will be 89 pairs in this month; to these
are added again the 55 pairs that are both in the tenth month; there will be 144
pairs in this month; to these are added again the 89 pairs that are born in the
eleventh month; there will be 233 pairs in this month. To these are still added
the 144 pairs that are born in the last month; there will be 377 pairs and this
many pairs are produced from the above-written pair in the mentioned place at
the end of one year.
You can indeed see in the margin how we operated, namely that we added the first
number to the second, namely the 1 to the 2, and the second to the third and
the third to the fourth and the fourth to the fifth, and thus one after another
until we added the tenth to the eleventh, namely the 144 to the 233, and we
had the above-written sum of rabbits, namely 377 and thus you can in order
find it for an unending number of months.”
a pair of baby (B) rabbits matures in one month to become offspring-producing adults (A), then we can set up the following chart:
No. of Pairs
of Adults
(A)
No. of
Pairs of
Babies (B)
Total
Pairs
1
0
1
1
1
2
2
1
3
3
2
5
5
3
8
8
5
13
July 1
13
8
21
Aug. 1
21
13
34
Sept.
1
34
21
55
Oct. 1
55
34
89
Nov. 1
89
55
144
Dec. 1
144
89
233
Jan. 1
233
144
377
Month
Pairs
Jan. 1
A
Feb. 1
A
Mar. 1
Apr. 1
May 1
June
1
B
A
B
A
A
B
A
B
A
A
B
A
B
A
A
B A
A B A A B A B A A B A A B
This problem generates the sequence of numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377,
The Fibonacci Numbers
1, 1, 2, 3, 5, 8, 13, 21,
34, 55, 89, 144, 233,
377, 610, . . .
Spiral arrangement of the bracts of
a pine cone
Take any four consecutive numbers in the sequence:
3, 5, 8, 13
Find the difference of the squares of the middle two numbers:
8  5  64  25  39
2
2
Then find the product of the outer two numbers:
3 13  39
 Fn1    Fn 
2
2
 Fn 1  Fn  2
The Golden Rectangle
B
l
C
w
A
D
w2  wl  l 2 , or
w
l

l wl
w2  wl  l 2  0.
If we let l  1, then w2  w  1  0.
1 5
5 1 1
w=
=
= ,
Φ
2
2
The Golden Ratio
1

  1
1   2 
Then  2    1  0
1 5
Applying the quadratic formula we get:  
2
1 5

 1.6180339887498948482045868343656
2
 3     2     1   2      1    2  1
 4   2   2    1  1   2  2  1    1  2  1  3  2
 5   3   2  2  1  1  2 2  3  1  2  1  3  1  5  3
 6   3   3  2  12  1  4 2  4  1  4  1  4  1  8  5
 7   4   3  3  22  1  6 2  7  2  6  1  7  2  13  8
  1  0
 2  1  1
 3  2  1
 4  3  2
 5  5  3
 6  8  5
 7  13  8
 8  21  13
 9  34  21
 10  55  34
The Ratios of Consecutive Fibonacci Numbers
Fn 1
Fn
Fn
Fn 1
1
1
 1.000000000
= 1.000000000
1
1
2
 2.000000000
1
1
= 0.500000000
2
3
 1.500000000
2
2
= 0.666666667
3
5
 1.666666667
3
3
 0.600000000
5
8
 1.600000000
5
5
= 0.625000000
8
13
 1.625000000
8
8
= 0.615384615
13
21
 1.615384615
13
13
= 0.619047619
21
34
 1.619047619
21
21
 0.617647059
34
55
 1.617647059
34
34
= 0.618181818
55
89
 1.618181818
55
55
= 0.617977528
89
144
 1.617977528
89
89
= 0.618055556
144
233
 1.618055556
144
144
= 0.618025751
233
377
 1.618025751
233
233
= 0.618037135
377
610
 1.618037135
377
377
= 0.618032787
610
987
 1.618032787
610
610
 0.618034448
987
Fn 1 Fn 1
=
+ 1.
Fn
Fn
There is a single-elimination basketball tournament with
25 teams competing.
How many games must be played in order to get a
winner?
Typical Solution:
Any 12 teams vs. any other 12 teams
tournament.
leaves 12 teams in the
6 winners vs. 6 other winners
tournament.
leaves 6 teams in
3 winners vs. 3 other winners
tournament.
leaves 3 teams in
3 winners + 1 team which drew a bye = 4 teams.
2 teams remaining vs. 2 teams remaining
tournament
1 team vs. 1 team to get a champion!
leaves 2 teams in
Use a chart:
Teams playing
Games played
Winners
24
12
12
12
6
6
6
3
3
3+ 1 bye=4
2
2
2
1
1
The total number of games played is:
12+6+3+2+1=24
Solution
using another point of view:
Consider the losers in the
tournament.
There must be 24 losers to get one
champion.
Therefore there must be 24 games
played
An amazing result:
The sum of the squares of the
digits of a given number
52  5  2  25  4  29  2  9  4  81  85 
2
2
2
2
 8  5  64  25  89
2
2
 82  92  64  81  145  12  42  52  1  16  25  42 
 42  22  16  4  20  22  02  4  42  16 
 12  62  1  36  37  32  7 2  9  49  58 
 52  82  25  64  89
You will always end up with either
89 or 1
A (shorter) version that ends up with 1
• This time begin with the number 23:
23  2  3  4  9  13  1  3  1  9  10 
2
1  0 1
2
2
2
2
2
Consider the sum of the cubes of
the digits of a number
•There are only five numbers that revert back:
11 1
3
153  1  5  3  1  125  27  153
3
3
3
370  3  7  0  27  343  0  370
3
3
3
371  3  7  1  27  343  1  371
3
3
3
407  4  0  7  64  0  343  407
3
3
3
•Others will also form a loop – in more steps
The Ulam-Collatz Loop
• Select any arbitrary number
• If the number is odd, then multiply by 3
and add 1
• If the number is even, then divide by 2
• Surprise: You will always end up with 1
An example of the “3n+1” phenomenon
Consider the number: 18 – follow the path:
18 – 9 – 28 – 14 – 7 – 22 – 111 –
34 – 17 – 52 – 26 – 13 – 40 – 20 –
10 – 5 – 16 – 8 – 4 – 2 – 1
Palindromes
RADAR
REVIVER
ROTATOR
LEPERS REPEL
MADAM I’M ADAM
STEP NOT ON PETS
NO LEMONS, NO MELON
DENNIS AND EDNA SINNED
ABLE WAS I ERE I SAW ELBA
A MAN, A PLAN, A CANAL, PANAMA
SUMS ARE NOT SET AS A TEST ON ERASMUS
Palindromic numbers
Here are some palindromic numbers:
121
1331
12345654321
55555555555
To generate palindromic
numbers:
• Take any two-digit number and add it to its
reversal.
• For example 92 + 29 = 121.
• If you don’t get a palindrome (93 + 39 =
132), then continue the process (132 +
231 = 363).
• Continue till you get a palindrome.
• Caution: 97 requires 6 reversals
98 requires 24 reversals!
The Amazing Number 1089
• 1. Choose any three-digit number (where the unit and
hundreds digit are not the same).
• We will do it with you here by arbitrarily selecting: 825
• 2. Reverse the digits of this number you have selected.
• We will continue here by reversing the digits of 825 to get: 528
• 3. Subtract the two numbers (the larger minus the smaller)
• Our calculated difference is: 825 – 528 = 297
• 4. Once again, reverse the digits of this difference.
• Reversing the digits of 297 we get the number: 792
• 5. Now, add your last two numbers.
• We then add the last two numbers to get: 297 + 792 = 1089
• Their result should be the same as ours even though their
starting numbers were different from ours.
•
If not, then you made a calculation error. Check it.
Let’s look at the first ten multiples
of 1089
• Can you see a
pattern?
1089 1  1089
1089 2  2178
1089 3  3267
1089 4  4356
1089 5  5445
1089 6  6534
1089 7  7623
1089 8  8712
1089 9  9801
Getting into an Endless Loop
• Choose a 4-digit number (not one with all
four digits the same).
• Rearrange the digits to make the biggest
and smallest number.
• Subtract the two numbers.
• With this new number, continue this
process.
• Soon you will get 6,174.
• But keep going!
• What do you notice?
We will (randomly) select the number 3,203
•
•
•
•
•
•
•
•
•
•
•
•
The largest number formed with these digits is: 3320.
The smallest number formed with these digits is: 0233.
The difference is: 3087.
The largest number formed with these digits is: 8730.
The smallest number formed with these digits is: 0378.
The difference is: 8352.
The largest number formed with these digits is: 8532.
The smallest number formed with these digits is: 2358.
The difference is: 6174.
The largest number formed with these digits is: 7641.
The smallest number formed with these digits is: 1467.
The difference is: 6174.
• And so the loop is formed, since you keep on getting 6174 if
you continue
When is the sum of the digits of a
number, taken to a power, equal to
the number?
• Consider these two examples:
81  (8  1)  9  81
2
2
4,913  (4  9  1  3)  17  4,913
3
3
=
(Sum of the Digits)n
Number
81
=
92
512
=
83
4,913
=
173
5,832
=
183
17,576
=
263
19,683
=
273
2,401
=
74
234,256
=
224
390,625
=
254
614,656
=
284
1,679,616
=
364
17,210,368
=
285
52,521,875
=
355
60,466,176
=
365
205,962,976
=
465
Some Beautiful Relationships
11  61  81  15  21  41  91
12  62  82  101  22  42  92
11  51  81  121  26  21  31  101  111
12  52  82  122  234  22  32  102  112
13  53  83  123  2,366  23  33  103  113
11  51  81  121  181  191  63  21  31  91  131  161  201
12  52  82  122  182  192  919  22  32  92  132  162  20 2
13  53  83  123  183  193  15, 057  23  33  93  133  163  203
14  54  84  124  184  194  260, 755  24  34  94  134  164  20 4
More Interesting Relationships
Amazing!
3
3
3
153 = 1 + 5 + 3
4
4
4
4
6
6
6
6
1634 = 1 + 6 + 3 + 4
5
5
5
5
5
54748 = 5 + 4 + 7 + 4 + 8
6
548834 = 5 + 4 + 8 + 8 + 3 + 4
7
7
7
7
7
6
7
7
1741725 = 1 + 7 + 4 + 1 + 7 + 2 + 5
8
8
8
8
8
8
8
8
24678050 = 2 + 4 + 6 + 7 + 8 + 0 +5 + 0
9
9
9
9
9
9
9
9
146511208 = 1 + 4 + 6 + 5 + 1 + 1 + 2 + 0
Notice how the powers reflect the original number.
4624  4  4  4  4
4
6
2
4
1033  81  80  83  83
595968  45  49  45  49  46  48
3909511  5  5  5  5  5  5  5
3
9
0
9
5
1
1
13177388  71  73  71  77  77  73  78  78
52135640  195  192  191  193  195  196  194  190
Friendly Numbers
• A pair of friendly numbers: 220 and 284.
• The divisors of 220 are:
1, 2, 4, 5, 10, 11, 20, 22, 44, 55, and 110.
Their sum is
1+2+4+5+10+11+20+22+44+55+110 = 284.
• The divisors of 284 are:
1, 2, 4, 71, and 142
Their sum is 1+2+4+71+142 = 220.
More pairs of friendly numbers:
1,184 and 1,210
2,620 and 2,924
5,020 and 5,564
6,232 and 6,368
10,744 and 10,856
9,363,584 and 9,437,056
111,448,537,712 and 118,853,793,424
The Monty Hall Problem
(“Let’s Make a Deal”)
There are two goats and one car behind three closed doors.
You must try to select the car.
You select Door #3
1
2
3
Monty Hall opens one of the doors that you did
not select and exposes a goat.
1
2
3
Your selection
He asks : “Do you still want your first choice
door, or do you want to switch to the other
closed door”?
To help make a decision, Consider an extreme case:
Suppose there were 1000 doors
1
2
3
4
997
998
999
1000
You choose door # 1000.
How likely is it that you chose the right door?
1
Very unlikely:
1000
How likely is it that the car is behind one of the other
doors: 1-999?
“Very likely”:
999
1000
1
2
3
997
4
998
999 1000
Monty hall now opens all the doors except one (2-999), and
shows that each one had a goat.
A “very likely” door is left:
Door #1
Which is a better choice?
• Door #1000 (“Very unlikely” door)
• Door #1 (“Very likely” door.)
1
2
3
4
997
998
999
1000
These are all “very likely” doors!
So it is better to switch doors from the initial selection.
Morley’s Theorem for the Angle Trisectors of a Triangle
C
E
F
A
D
B
A Reminder!
It is the teacher’s responsibility
to have their students bring home
information to excite their
parents about mathematics
– so that the parents can then
support the importance of
mathematics in their child’s
education.