Transcript Problem No. 1

Promoting Creativity in
Mathematics by Use of
Non-standard Problems
Sara Hershkovitz & Pearla Nesher
Center for Educational Technology
&
The University of Haifa
Israel
Imagine a standard classroom
of fifth-grade students, receiving the
following nonstandard problems to solve:
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Problem No. 1
How many two-digit numbers, up to
one hundred, have a tens digit that is
larger than the units digit?
Student’s answers:
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What if not…
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Problem No. 2:
How many two-digit numbers, up to
one hundred, have a units digit that
is larger than the tens digit?
Student’s answers:
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Problem No. 3:
Look at the following numbers:
23, 20, 15, 25,
which number does not belong?
Why?
Students' answers:
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Problem No. 4:
100 nuts are divided among 25
children, unnecessarily in equal
portions. Each child receives an odd
number of nuts.
How many nuts does each child
receive?
Students' answers:
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Problem No. 5:
(Elaborated after Paige, 1962)
We made change from one Shekel into
smaller coins: 5, 10, and 50 Agorot
(or cents).
Make change so that you
hold three times as many 10-Agorot
coins as 5-Agorot coins.
Students' answers:
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Problem No. 6:
A witch wants to prepare a frog drink.
She can buy dried frogs only in packages of
five or eight.
Note that the witch has to buy the exact
number of dried frogs she needs.
How many frogs can she buy? What is the
largest number of frogs she cannot buy?
Students' answers:
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What if not….
In a similar witch problem,
the dried frogs come in packages of four
or eight. What is the largest number of
frogs that the witch cannot buy?
back
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Problem No. 7:
Insert different numbers in the blank spaces,
so that the four-digit number received
divides by 3.
__ 1 4 __
Students' answers:
A teacher’s answer:
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Problem No 8:
I have a magic handbag. If I leave some
money in it overnight, I find twice as much
money in the morning, plus one unit.
Once, I forgot some money in the handbag
for two nights, and on the third day I found
51 dollars.
How much money was in the handbag
before the first night?
Students'
answers:
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Discussion
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A. The types of problems:
How do the problems differ?
What are the different types of problems?
Can the types be characterized?
Write a few novel problems for each
problem type.
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B. Promoting creativity:
Creativity, as defined by some researchers
(Guilford 1962, Haylock 1987, Silver 1994),
contains the following components:
Fluency: measured by the total number of
replies.
Flexibility: measured by the variety of
categories given.
Originality: measured by the uniqueness of
a reply within a given sample of replies.
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Do such problems promote creativity?
Which additional types of problems can be
employed for promoting creativity?
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C. Using these problems in the
classroom:
We routinely use these problems in the
classroom, approximately once a week,
without connecting them to ongoing topics
studied in class.
Our goal is to allow students to experience
the following:
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Extending and applying previous
mathematical knowledge.
 Constructing relationships among topics
by integrating various topics within one
problem.
 Encouraging the use of different
strategies: working systematically, while
using naïve strategies.
 Stimulating reflections on student
experiences. Articulating the solution in
various ways: words, or diagrams.

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We emphasize the following merits of class
discussion:
 An opportunity for students to explain how
they think about problems; different
problems provide different ways of
analysis.
 Individual ways of presentation may bring
up disparate relations or different
mathematical points of view.
 Students are encouraged to understand
and assimilate someone else’s strategies.
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
A discussion emphasizes that
mathematics entails active processes,
such as investigating, looking for patterns,
framing, testing, and generalizing, rather
than just reaching a correct answer.

The discussion demonstrates that
mathematical thinking involves more
questions than answers.
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Thank you
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Problem No. 1 - Avi
T(Teacher): How many ……. (posing the
question)?
A(Avi): Mmmm…. (thinking).
T: Do you understand the problem?
A: Yes.
T: Can you find an example for such a number?
A: 52.
T: Ok; so how many two-digit numbers are there?
A: A lot.
.
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T: How many?
A: A lot.
T: How many? Give a number.
A: 8.
T: Tell me what they are.
A: 53, 42, 85, 31, 64, 97, 75, 61, 98;
that’s all
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Problem No. 1 - Ben:
I’m going to find them in the “First-hundred table”:
1
11
21
31
41
51
61
71
81
91
2
12
22
32
42
52
62
72
82
92
3
13
23
33
43
53
63
73
83
93
4
14
24
34
44
54
64
74
84
94
5
6
15 15
25 26
35 36
45 46
55 56
65 66
75 76
85 86
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95 96
7
17
27
37
47
57
67
77
87
97
8
9 10
18 19 20
28 29 30
38 39 40
48 49 50
58 5 9 60
68 6 9 70
78 79 80
88 89 90
24
98 99 100
Now I will count them: 1,2,3,…..45.
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Problem No. 1 - Galit
21, 30, 31, 32, Mmmm.
I see 10, 20, 21, 30, 31, 32 (thinking a bit
before writing down each number.)
Then Galit began writing faster: 40, 41,
42, 43, 44, and deleted the 44.
T: Can you explain what you are doing?
Galit didn’t answer, and began writing quickly:
50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70,
71, 72, 73, 74. 75, 76, 80, 81, 82, 83, 84, 85,
86, 87, 90, 91, 92,93, 94, 95, 96, 97, 98
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Galit raised her head and said: Now I have to
count them: 1,2,3,4 ...
T: I saw that you began writing quickly. What
happened at that point?
G: I saw that for each tens number in the list, I
could write the numbers less one 1,2,3,4,5,..
T: Do you know how to sum the numbers by a
shorter method?
G: 6,7,8,9,10,11,12,………43,44.45.
There are 45 numbers.
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Problem No. 1 - Dana
D: The smallest number is 20, Mmmm; no,
sorry, 10,…. and the largest is …98
98 – 10 = 88; there are 88 numbers.
T: Mmmm.
D: Sorry, between the numbers there are
also 55, and 34 and 28; I have to think.
20, 21 ……..30, 31, 32
T: You forgot 10.
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D: Yes, 10,
20, 21,
30, 31, 32
40, 41, 42, 43
I see;
In ten I have one number,in twenty there are
two numbers, in thirty there are three,
in forty there are four; I can continue up to
ninety where there are nine..
So, 1+2+3+4+5+6+7+8+9=
1+2 is 3, plus 3 is 6, 10, 15, 21, 28, 36,….45.
There are 45 numbers.
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Problem No. 1 - Hadar
1
10
2
20
21
3
30
31
32
4
40
41
42
43
5
50
51
52
53
54
6
60
61
62
63
64
65
7
70
71
72
73
74
75
76
8
80
81
82
83
84
85
86
87
9
90
91
92
93
94
95
96
97
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30
Now I see that there are ten (while pointing
to the arrows), twenty, thirty, forty,
forty five …numbers.
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Problem No. 1 - Vered
Vered wrote in the columns:
21
31
41
51
61
71
81
92
32
42
52
62
72
82
43
53
63
73
83
93
54
64
74
84
94
65
75
85
95
76
86
96
87
97
91
98
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Now I can sum them;
8+7+6+5+4+3+2+1
are
15, 21, 26, 30, 33, 35, 36.
Actually, I see the same unit in each row.
back
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Problem No. 2 - Ziva
This problem is symmetrical to the previous
one, so the answer is the same:
45 numbers.
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Problem No. 2 - Chedva
If I think of the “First-hundred table”,
100 has three digits, so 99 numbers are left;
subtract the previous 45, so there are 55
numbers.
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Problem No. 2 - Mick
Beginning with 10: 12, 13, 14. 15, 16, 17, 18, 19
Beginning with 20: 23, 24, 25, 26, 27, 28, 29
With 30:
34, 35, 36, 37, 38, 39
…………..
…………..
With 80: --with 90:
89
it’s
8+7+6+5+4+3+2+1 =36
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Problem No. 2 - Yoni
I’m thinking of the “First-hundred table”.
100 has three digits, and 99 numbers are left.
1-9 are one-digit numbers, so there are only 90
numbers left.
11-99 do not fit, as well, because they have the
same digits; we have to remove 9 additional
numbers, and then we are left with 81 numbers.
If we remove 45 numbers, from the previous
problem, we end up with 36 numbers.
back
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Problem No. 3 - Adi
20 – It is the only even number.
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Problem No. 3 - Bill
15 - The only number that divides by 3.
20 - The units digit is 0; It doesn’t have units.
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Problem No. 3 - Gur
15 - It is in the 2nd ten and the rest are in the
3rd ten.
20 - The only round number.
This number has more factors
23 - Not a multiple of 5.
25 - The sum of its digits is the largest.
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Reasons for choosing a certain
number as exceptional
15





It is under 20
Its tens digit is 1, and the rest have the digit 2
It is in the 2nd ten and the rest are in the 3rd ten
It is the smallest number
The only number that divides by 3
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20








The only even number
A multiple of 2; divides by 2
The units digit is 0; it doesn’t have units
The sum of its digits is doesn’t fit the series
The number divides by 10
The only round number
The only number divides by 4
The number that has more factors
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23






Not a multiple of 5
Doesn’t divide by 5
Not in the series that adds 5 to each number
The only prime number
Doesn’t appear in the multiplication table
The only number that has the digit 3 in it
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25
 A square number
It is the largest number
 The sum of its digits is the largest
 Is 30 in approximation

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Distribution of
Number Property Categories
15 20 23 25 Total
Category
Iconic properties
Size consideration
Additive series
Sums of digits
Multiples and divisors
75
28
89
1
1
109
24
17
114
18
25
2
27
12
34 105
1
152
1
7
39
8
55
96
1
3
100
177 192 168 38
575
Squares, Prime, Fractions, Negatives
Evenness; Others
Total
6
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Unique Replies
Number The given reason
15
If
you divide all numbers by 20, it is the
only one that becomes a fraction
If
you multiply each number by 100, it
is the only one below 200
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Number The given reason
20
If
you add 2 to this number, it remains
an even number.
If you subtract the digits, it is the only
negative number
It is the only number that has the
number 5 as its quarter (1/4)
It is the average between 15 and 25
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Number The given reason
23
When
subtracted from 30, it doesn’t
divided by 5
When multiplied by 2, it is not a round
number
The only one I cannot find an
exercise for
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Number The given reason
25
The only number that didn’t have a
reason to be exceptional
back
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Problem No. 4 - Aluma
It's impossible;
100 : 25 = 4
That means that each child gets 4 nuts,
but it is an even number.
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Problem No. 4 - Bat-Sheva
It's impossible;
if we give an odd number of nuts to 24
children, together they have an even
number of nuts ;
odd + odd = even.
The 25th child can only get an even number
of nuts.
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Problem No. 4 - Gidi
It's impossible;
all the numbers I try are odd, and 100 is
even.
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Problem No. 4 - Dotan
It's impossible.
Whenever we multiply an odd number by
an odd number, the result is odd.
100 is always even, so it's impossible to
divide it by 25 to get an odd number.
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Problem No. 4 - Hen
21 children get 3 nuts each.
One child gets 7 nuts.
Two children get 15 nuts each.
The last child gets 5 nuts.
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Problem No. 4 - Vivi
There are two possibilities:
I 100 : 25 = 4; in this case each child
gets an even number, and the answer
is wrong.
II Not all children get the same amounts
of nuts; some get 3 nuts and the
others 5 nuts.
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Problem No. 4 - Tzvi
Odd +odd = even.
There are 25 children; we can arrange them
in pairs. So, 24 people have an even
number of nuts together.. The 25th child
doesn’t have a pair, so he must have an
even number.
My conclusion: it is impossible to divide 100
nuts this way.
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Problem No. 4 - Yonit
I tried many combinations, and did not
succeed.
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Problem No. 4 - Ofek
I tried to solve this problem with smaller
numbers:
I divided 20 nuts.
I prepared a table, tried all the numbers, and
did not succeed;
I think it is impossible.
back
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Problem No. 5: Avigail
Me and my friend have tried and couldn’t do
it, so it is impossible.
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Problem No. 5: Bilha
1 coin of 50 = 50
3 coins of 10 = 30
4 coins of 5 = 20
100
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Problem No. 5: Gill
It is impossible,
because for each coin of 5 I have to take
three coins of 10 (5+30=35)
If I multiply this sum I get 35+35=70
and three times: 70+35=105 .
back
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Problem No. 6: Arava
She can buy all the multiples of 8, of 5, and
of (8+5).
She cannot buy the rest of the numbers.
There are endless numbers from both kinds.
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Problem No. 6: Bill
She can buy all the multiples of 5 and all the
multiples of 8;
there are endless numbers she cannot buy.
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Problem No. 6: Gila
She cannot buy these numbers:
1,2,3,4,6,7,9,11,12,14,17,22,23,19,29,31.
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Problem No. 6: Dalia
I removed all the multiples of 5, 8, and 13
with their combinations.
The biggest number she cannot buy is 27.
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Problem No. 6: Hava
She cannot buy:
6,4,14,2,12,22,9,19,11,1,3,7,17,27
She can buy
8,16,24,32,40,48,56,64,72,80,88,96
All the unit digits: 5,0,8,6,4,2,9,3,1,7
The largest amount she cannot buy is 27.
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Problem No. 6: Viki
We began with the numbers she cannot buy, for
example, 1,2,3,4,6…
We tried numbers up to 100.
Then we tried to see how many frogs she can buy.
We figured out: she can buy all the multiples of
5,8,13,
or common multiples as 26 (2*8+2*5).
We decided that she could buy endless amounts
of frogs, as long as there were still frogs in the
shop and she had the money to buy them.
back
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Problem No. 7:
Adva
Ben
2142
2143
Gad
3141
Dalit
0141, 1140, 2142, 3141,
1143
There are many
possibilities, but not
endless.
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Problem No. 7: Hilla
1149
1140
1143
4140
1450
3141
2142
7140
2145
2448
3141
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Problem No. 7: Osnat
1140 2142 3141 4140 5142 6141 7140 8142 9141
1143 2145 3144 4143 5145 6144 7143 8146 9144
1146 2148 3147 4146 5148 6147 7146 8148 9147
1149
4149
7149
back
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Problem No. 7: A teacher
Module1 – 1
Function hh (a As Integer) As Integer
Dim c As Integer
Dim I As Integer, J As Integer
b=a
For i = 1 To 9
For j = 0 To 9
a=b
a = i * 1000 + a + j*1
If a Mod 3 = 0 Then c = c + 1
Next j
A=b
Next i
hh = c
End Function
back
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Problem No. 8:
Adam
Ben
25 dollars.
51-1=50
50:2=25
25X2+1=51
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Problem No. 8: Galia
Each day, 1 + 2 X__= money in handbag
1+ 2 X 8 = 17, 1 + 2 X 17 = 35
1 + 2 X 10 = 21, 1 + 2 X 21 = 43
1 + 2 X 11 = 23, 1 + 2 X 23 = 47
1 + 2 X 12 = 25,
1 + 2 X 25 = 51
that’s it!!!
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Problem No. 8: Dov
51 – 1 = 50,
50:2 = 25
25 – 1 = 24,
24:2 = 12
back
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