Problem Solving Non-standard Problems
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Problem SolvingNon-standard Problems
Sara Hershkovitz
Center for Educational Technology
Israel
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Sara Hershkovitz, CET Israel, SEMT05 - Aug 2005
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Imagine a standard classroomof 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:”
12345678910
11121314151517181920
21222324252627282930
31323334353637383940
414243444546474849 50
51525354555657585 960
61626364656667686 970
71727374757677787980
81828384858687888990
9192939495 96 979899100
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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
110
22021
3303132
440414243
55051525354
6606162636465
770717273747576
88081828384858687
9909192939495969798
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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:
2131415161718191
32425262728292
435363738393
5464748494
65758595
768696
8797
98
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Now I can sum them;8+7+6+5+4+3+2+1are
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, 19Beginning with 20: 23, 24, 25, 26, 27, 28, 29With 30: 34, 35, 36, 37, 38, 39..…………..…………
89 With 80--- :with 90 :
it’s 36 =8+7+6+5+4+3+2+1
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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 Category15202325Total
Iconic properties75286109
Size consideration89124114
Additive series11718
Sums of digits25227
Multiples and divisors12341051152
Squares, Prime, Fractions, Negatives1739855
Evenness; Others 9613100
Total17719216838575
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Unique Replies
NumberThe given reason
15If 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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NumberThe given reason
20If you add 2 to this number, it remains
an even number.If you subtract the digits, it is the only negative numberIt 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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NumberThe given reason
23When subtracted from 30, it doesn’t divided by 5When multiplied by 2, it is not a round numberThe only one I cannot find an exercise for
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NumberThe given reason
25The 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 = 4That 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.
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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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Sara Hershkovitz, CET Israel, SEMT05 - Aug 2005
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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=70and three times: 70+35=105.
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Sara Hershkovitz, CET Israel, SEMT05 - Aug 2005
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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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Sara Hershkovitz, CET Israel, SEMT05 - Aug 2005
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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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Sara Hershkovitz, CET Israel, SEMT05 - Aug 2005
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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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Sara Hershkovitz, CET Israel, SEMT05 - Aug 2005
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Problem No. 6: VikiWe 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 :
Adva2142
Gad
3141
Ben2143
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
1143
1146
1149
2142
2145
2148
3141
3144
3147
4140
4143
4146
4149
5142
5145
5148
6141
6144
6147
7140
7143
7146
7149
814281468148
9141
9144
9147
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Problem No. 7: A teacherModule1 – 1Function hh (a As Integer) As IntegerDim c As IntegerDim I As Integer, J As Integerb = 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 ihh = cEnd Function
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Problem No. 8:
Adam
25 dollars.
51-1=50
50:2=25
Ben
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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Sara Hershkovitz, CET Israel, SEMT05 - Aug 2005
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Problem No. 8: Dov
51 – 1 = 50 ,50:2 = 25
25 – 1 = 24 ,24:2 = 12
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