Promoting Creativity in Mathematics by Use of Non-standard Problems Sara Hershkovitz & Pearla Nesher Center for Educational Technology & The University of Haifa Israel CET - May, 2004
Imagine a standard classroom of fifth-grade students, receiving the following nonstandard problems to solve: CET - May, 2004 2
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: CET - May, 2004 3
What if not… CET - May, 2004 4
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: CET - May, 2004 5
Problem No. 3: Look at the following numbers: 23, 20, 15, 25, which number does not belong? Why? Students' answers: CET - May, 2004 6
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: CET - May, 2004 7
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: CET - May, 2004 8
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: CET - May, 2004 9
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 CET - May, 2004 10
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: CET - May, 2004 11
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: CET - May, 2004 12
Discussion CET - May, 2004 13
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. CET - May, 2004 14
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. CET - May, 2004 15
Do such problems promote creativity? Which additional types of problems can be employed for promoting creativity? CET - May, 2004 16
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: CET - May, 2004 17
Extending and applying previous mathematical knowledge. n Constructing relationships among topics by integrating various topics within one problem. n Encouraging the use of different strategies: working systematically, while using naïve strategies. n Stimulating reflections on student experiences. Articulating the solution in various ways: words, or diagrams. n CET - May, 2004 18
We emphasize the following merits of class discussion: n An opportunity for students to explain how they think about problems; different problems provide different ways of analysis. n Individual ways of presentation may bring up disparate relations or different mathematical points of view. n Students are encouraged to understand assimilate someone else’s strategies. CET - May, 2004 19
n 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. n The discussion demonstrates that mathematical thinking involves more questions than answers. CET - May, 2004 20
Thank you CET - May, 2004 21
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. . CET - May, 2004 22
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 CET - May, 2004 23
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 CET - May, 2004 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. CET - May, 2004 25
Problem No. 1 - Galit 21, 30, 31, 32, Mmmm. I see 10, 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 CET - May, 2004 26
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. CET - May, 2004 27
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 also 55, and 34 and 28; I have to think. 20, 21 ……. . 30, 31, 32 T: You forgot 10. CET - May, 2004 28
D: Yes, 10, 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. CET - May, 2004 29
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 CET - May, 2004 98 30
Now I see that there are ten (while pointing to the arrows), twenty, thirty, forty five …numbers. CET - May, 2004 31
Problem No. 1 - 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 91 97 98 CET - May, 2004 32
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 CET - May, 2004 33
Problem No. 2 - Ziva This problem is symmetrical to the previous one, so the answer is the same: 45 numbers. CET - May, 2004 34
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. CET - May, 2004 35
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 …………. . 89 With 80: --with 90: it’s 8+7+6+5+4+3+2+1 =36 CET - May, 2004 36
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 CET - May, 2004 37
Problem No. 3 - Adi 20 – It is the only even number. CET - May, 2004 38
Problem No. 3 - Bill 15 - The only number that divides by 3. 20 - The units digit is 0; It doesn’t have units. CET - May, 2004 39
Problem No. 3 - Gur 15 - It is in the 2 nd ten and the rest are in the 3 rd 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. CET - May, 2004 40
Reasons for choosing a certain number as exceptional 15 n n n It is under 20 Its tens digit is 1, and the rest have the digit 2 It is in the 2 nd ten and the rest are in the 3 rd ten It is the smallest number The only number that divides by 3 CET - May, 2004 41
20 n n n n 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 CET - May, 2004 42
23 n n n 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 CET - May, 2004 43
25 n A square number It is the largest number n The sum of its digits is the largest n Is 30 in approximation n CET - May, 2004 44
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 CET - May, 2004 45
Unique Replies Number The given reason 15 n. If you divide all numbers by 20, it is the only one that becomes a fraction n. If you multiply each number by 100, it is the only one below 200 CET - May, 2004 46
Number The given reason 20 n. If you add 2 to this number, it remains an even number. n. If you subtract the digits, it is the only negative number n. It is the only number that has the number 5 as its quarter (1/4) n. It is the average between 15 and 25 CET - May, 2004 47
Number The given reason 23 n. When subtracted from 30, it doesn’t divided by 5 n. When multiplied by 2, it is not a round number n. The only one I cannot find an exercise for CET - May, 2004 48
Number The given reason 25 n. The only number that didn’t have a reason to be exceptional back CET - May, 2004 49
Problem No. 4 - Aluma It's impossible; 100 : 25 = 4 That means that each child gets 4 nuts, but it is an even number. CET - May, 2004 50
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 25 th child can only get an even number of nuts. CET - May, 2004 51
Problem No. 4 - Gidi It's impossible; all the numbers I try are odd, and 100 is even. CET - May, 2004 52
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. CET - May, 2004 53
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. CET - May, 2004 54
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. CET - May, 2004 55
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 25 th child doesn’t have a pair, so he must have an even number. My conclusion: it is impossible to divide 100 nuts this way. CET - May, 2004 56
Problem No. 4 - Yonit I tried many combinations, and did not succeed. CET - May, 2004 57
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 CET - May, 2004 58
Problem No. 5: Avigail Me and my friend have tried and couldn’t do it, so it is impossible. CET - May, 2004 59
Problem No. 5: Bilha 1 coin of 50 = 50 3 coins of 10 = 30 4 coins of 5 = 20 100 CET - May, 2004 60
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 CET - May, 2004 61
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. CET - May, 2004 62
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. CET - May, 2004 63
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. CET - May, 2004 64
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. CET - May, 2004 65
Problem No. 6: Hava She cannot buy: 6, 4, 14, 2, 12, 22, 9, 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. CET - May, 2004 66
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 CET - May, 2004 67
Problem No. 7: Adva Ben 2142 2143 Gad 3141 Dalit 0141, 1140, 2142, 3141, 1143 There are many possibilities, but not endless. CET - May, 2004 68
Problem No. 7: Hilla 1149 1140 1143 4140 1450 3141 2142 7140 2145 2448 3141 CET - May, 2004 69
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 CET - May, 2004 70
Problem No. 7: A teacher Module 1 – 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 CET - May, 2004 71
Problem No. 8: Adam Ben 25 dollars. 51 -1=50 50: 2=25 25 X 2+1=51 CET - May, 2004 72
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!!! CET - May, 2004 73
Problem No. 8: Dov 51 – 1 = 50, 50: 2 = 25 25 – 1 = 24, 24: 2 = 12 back CET - May, 2004 74
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