Problem Solving Activity 1 (Irfan Rusmar)

8/7/2019 Problem Solving Activity 1 (Irfan Rusmar)

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NAME : IRFAN RUSMAR

STUDENT ID : 211076913

E 700/ESM 704

PROBLEM SOLVING ACTIVITIES

ACTIVITY 1.1

In school, I usually was solving number problems by using mathematics formula that has learned in

school. The Mathematics teachers, when I was in primary, lower secondary and upper secondary, did

not introduce the problem solving method to solve any Mathematics problems during classroom

activities. They usually ask their students to remember formulas from Mathematics textbooks and use it

to solve any Mathematics problems. For example, in solving this kisses problem. The way that

appropriate to use is Combination formulas, such as,

. However, here I use a different way

from the combination formula in solving this problem. Moreover, I am not depend on the formula to

solve this kisses problem. I use tables, numbers sequence and find the appropriate pattern as the way to

solve the problem, as shown below.

There are 12 people at a New Years Eve party. At midnight they al l exchange kisses. How many kisses

altogether?

What if there were 50 people at the party?

Solution:

Firstly, I use a table, and start from probability of the smallest number of people.

A B

A AA AB

B BA BB

The table reveals that, there are four possibilities kisses if 2 people kiss altogether, AA, BB, AB, and BA.

But, for AA and BB impossible, because it is impossible for a person to kiss himself/ herself, and AB = BA.

So, if there are 2 people kiss altogether, there will one kiss altogether.

For 3 people, I use the same method as the 2 people possibilities.

A B C

A AA AB AC

B BA BB BC

C CA CB CC

In the same way, AA, BB, and CC is impossible, and AB = BA, AC = CA and BC = CB. So, for three people

kisses altogether, there will be three kisses altogether.


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For 4 people, I also use the same method.

A B C D

A AA AB AC AD

B BA BB BC BD

C CA CB CC CD

D DA DB DC DD

In the same way, AA, BB, CC and DD is impossible, and AB = BA, AC = CA, BC = CB, AD = DA, BD = DB and

CD = DC. So, if there are 4 people exchange their kisses altogether, there will be 6 kisses altogether.

By looking at those three possibilities, I found an interesting sequence, as shown in the table below.

By looking at the pattern in table, I find another interesting pattern for the unit of the sequence (Un)

and the number of people (n). There will be n = U(n-1), for example for n = 2 people, it will be consider

as the first unit, or = 2, = 3, = 4, , = 12 people.

Another pattern for many kisses altogether that I got from the sequence is,

S1 = 1

S2 = 1 + 2

S3 = 1 + 2 + 3

Based on the pattern above, I believe that for S11= 1 + 2 + 3 + + 9 + 10 + 11, so that , the sum can also

be written by working backwards from the last term that is to get to the previous term.

S11= 1 + 2 + 3 + + 9 + 10 + 11

S11= 11 + 10 + 9 + + 3 + 2 + 1 +

2 S11= 12 + 12 + 12 + + 12 + 12 + 12

S11 =

Un The number of

people (n)

Exchange Kisses

altogether (Sn)

1 2 1

2 3 3

3 4 6

. . .

. . .

. . .

9 10 ?

10 11 ?

11 12 ?

+2

+3


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S11 = 66 kisses altogether

So, for 12 people, there will be 12 kisses altogether.

Now, what if there were 50 people?

By using the same way, for 50 people, I consider as the 49th unit, so U49 = 50 people.

S49= 1 + 2 + 3 + + 47 + 48 + 49

S49= 49 + 48 + 47 + + 47 + 48 + 49 +

2S49= 50 + 50 + 50 + + 50 + 50 + 50

2S49 = 49 x 50

S49 =

S49 = 1225 kisses altogether.

Actually, this problem can be solved by using combination formula, as shown bellow.

ACTIVITY 1.3

Label the solutions according to the strategies used by the children.

A typical list of problem solving strategies usually includes some or all of the following:

Guess and check, often also referred to trial and error. Make and use a model or diagram Act out the problem Try some simple cases Construct a table Look for a pattern Find a general rule Explain why Break the problem into sub-problems Solve a simpler but related problem Eliminate all possibilities


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According to the figure 1.1, there are three different solutions from the children used to solve the ducks

and cows problem.

The first solution, I label as the make and use a model or diagram strategy. As can be seen, in the

figure shows the students use matchsticks and bottle tops to construct the model of ducks and cows,where ducks have 2 feet and cows have 4 feet, respectively. They use matchsticks to construct the

animals feet, and bottle tops as the bodies of the animals.

The second solution, I tag as the trial and error strategy. This shown by the students explanation, they

told that, they have tried this strategy a few time but it did not work, until they found the appropriate

answer for this problem. Finally, for the third solution, I believe this as the guess and check strategy.

ACTIVITY 1.6

After counting the number of square in a chessboard 8 by 8, I try to do the same way, but for different

form of chessboard. I chose rectangle chessboard (5 by 4). From this chessboard, I would like to find

how many squares could be on the chessboard.


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Solution:

How many rectangles are there on the chessboard?

When I first worked it out, it took such a long time and a lot of work to organize this problem logic.

However, it was valuable, because I learnt a lot about the problem.

Firstly, I noticed that the number of squares in the rectangle is a multiply of the number of squares in

line with many rows are left. For example, for 1 by 1 squares is (5 x 4), and for 2 by 2 squares is (4 x 3),

because there are only three rows are left.

1 by 1 rectangles 2 by 2 rectangles 3 by 3 rectangles

4 by 4 rectangles

1 2 3 4 5

2

3

4

5 squares in each row

and there are 4 rows

5 x 4 = 20 squares



4 x 3 = 12 squares



3 x 2 = 6 squares



2 x 1 = 2 squares


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Then I noticed that the size of square and the number of them each row and column always add up to 6

and 5 (lines in column and row).

Size of square 1 by 1 2 by 2 3 by 3 4 by 4

Number of square 5 x 4 4 x 3 3 x 2 2 x 1

Based on the process of solution, I conclude that the number of squares in rectangle 5 by 4 is 40.

Then I think what if there is a rectangle 6 by 7. Based on the explanation above, I predict it will be always

add up to 7 and 8.

Size ofsquare 1 by 1 2 by 2 3 by 3 4 by 4 5 by 5 6 by 6

Number of

square6 x 7 5 x 6 4 x 5 3 x 4 2 x 3 1 x 2

So, the number of squares on the rectangle (6 by 7) chessboard is

42 + 30 + 20 + 12 + 6 + 2 = 112

6 and 5 6 and 5 6 and 5 6 and 5

7 and 8 7 and 8 7 and 87 and 8 7 and 8 7 and 8

Problem Solving Activity 1 (Irfan Rusmar)

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