A Magic Square

3 CONTENTS Question 1 -A Magic SquareArrange the numbers 1 through 9 into a square subdivided into nine smaller squares like the one shown in Figure A so that the sum of every row, column and main diagonal is the same. Figure A

•

UNDERSTANDING THE PROBLEM

We need to put each of the numbers 1, 2, 3,…, 9 in the small squares, a different number in each squares, so that the sum of the numbers in each row, in each column and in each of the two diagonals is the same. •

DEVISING A PLAN

If we knew the fixed sum of the numbers in each row, column, and diagonal, we would have a better idea of which numbers can appear together in a single row, column or diagonal. Thus our sub goal is to find that fixed sum. The sum of the nine numbers, 1+2+3+…+9, equals 3 times the sum in one row. Consequently, the fixed sum is obtained by dividing 1+2+3+…+9 by 3. Using the process developed by Gauss, we have (1+2+3+…+9) ÷ 3 = (9.10) ÷2 ÷ 3 or 45 ÷ 3=15, so the sum in each row, column, and diagonal must be 15. Next we need to decide what numbers could occupy the various squares. The number in the center space will appear in 4 sums, each adding to 15 (two diagonals, the second row, and the second column). Each number in the corners will appear in three sums of 15. If we write 15 as a sum of three different numbers 1 through 9 in all possible ways, we could then count how many sums contain each of the numbers 1 through 9. The numbers that appear in at least four sums candidates for placement in the center square, whereas the numbers that appear in at least three sums are candidates for the corner squares. Thus our new sub goal is to write 15 in as many ways as possible as a sum of three different numbers from the set {1,2,3,….9}.

4 •

CARRYING OUT THE PLAN

The sums of 15 can be written systematically as follows: 9+5+1 9+4+2 8+6+1 8+5+2 8+4+3 7+6+2 7+5+3 6+5+4 Notice that the order in each sum is not import. Hence, 1+5+9 and 5+1+9, for example are counted as the same. Notice that 1 appears in only two sums, 2 in three sums, 3 in two sums, and so on. Table A summarizes this pattern. NUMBER NUMBERS OF SUMS CONTAINING THE NUMBER

1

2

3

4

5

6

7

8

9

2

3

2

3

4

3

2

3

2

Table A The only number that appears in four sums is 5, hence 5 must be center of the square. Because 2,4,6, and 8 appear three times each, they must go in the corners. Suppose we choose 2 for the upper left corner. Then 8 must be in the lower right corner. Now we could place 6 in the lower left corner or upper right corner. The magic square can now completed, as shown in Figure A- i

2

2 5

6 5

8

Figure A- i

8

2 7 6 9 5 1 4 3 8

5 •

LOOKING BACK

We have seen that only number among the given numbers that could appear in the center. However, we had various choices of corner, and hence it seems that the magic square we found is not the only one possible. Another way to see that 5 could be in the center square is to consider the sums 1+9, 2+8, 3+7, 4+6 as shown in Figure B. We could add 5 to each to obtain 15.

1

2

3

4

5

6

7

8

9

|_____15______| |____________15____________| |__________________15__________________| |________________________15________________________|

Figure B Conclusion: There are two-type strategy solutions for question 1, first strategy is identifying a Sub goal strategy and the second strategy is using by table strategy. I think that identifying a sub goal strategy is suitable for use in Question 1 and it helps me to solve this question easily and quickly rather than using by table strategy because it takes much time to solve the problem.

6 Question 2

-Fruits for DonationPerak mangoes were donated for a Teacher Appreciation lunch. Two mangoes were too ripe to use. Five groups of five students cut up the remaining fruit. Each student cut up five mangoes. How many mangoes were donated? •

UNDERSTANDING THE PROBLEM

We need to know how many mangoes were donated and how we need to find it base on what the clue given. Two mangoes were too ripe to use. We know that five groups which have five students in one group. Lastly, all students should be cut up five mangoes for each student.

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DEVISING A PLAN

Find how many mangoes were cut up. Add the number of mangoes that were not used.

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CARRYING OUT THE PLAN

How many mangoes were cut up

= students × mangoes × groups =5×5×5 = 75

How many mangoes were donate

= cut up + 2 unused = 75 + 2 = 77 mangoes were donate

7

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LOOKING BACK

Another way to see that 77 mangoes were donated for a Teacher Appreciation lunch by looking with draw the picture. Imagine that we have five groups and each group have 5 students. Then, every student should cut up 5 mangoes. We draw it just like shown below:

s s s s s

s s s s s

s s s s s

s s s s s

s s s s s

= 5 mangoes S = student

Conclusion: There are two-type strategy solutions for Question 2, first strategy is work backward strategy and the second strategy is using by draw the picture strategy. I think that work backward strategy is suitable for use in Question 2 and it helps me to solve this question easily and quickly rather than using by draw the picture strategy because it takes much time to solve the problem.

8 Question 3

-CheckerboardUsing the existing lines on the checkerboard shown below, how many squares are there?

• UNDERSTANDING THE PROBLEM To count how many squares on the checkerboard.

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DEVISING THE PLAN

Solve this problem by counting the number of squares. However, this is tedious process. Simplifying the problem into smaller number of squares and looking for a pattern will help you to solve this problem quickly.

9 •

CARRYING OUT THE PLAN

It can make be easier by count squares by squares as shown below: 1 square 4 squares

9 squares

16 squares 25 squares

36 squares

49 squares

64 squares

All-in = 1+4+9+16+25+36+49+64= 204 squares

10 •

LOOKING BACK

You can see that if the size of the grid is n by n, then the total number of squares is obtained by adding the squared number from 12 to n2 . Therefore we can use ‘look for a pattern’ as another strategy. Looking for patterns is a very important strategy for problem solving, and is used to solve many different kinds of problems. Sometimes you can solve a problem just by recognizing a pattern. But often you will have to extend a pattern to find a solution. Making a number table often reveals patterns, and for this reason it is frequently used together with this strategy. Look at the numbers given in the sequence. Try to find the relationships between consecutive numbers. Look for a pattern to find the answer.

1

4 +3

9 +5

16 +7

25 +9

36 +11

49 +13

64 +15

That mean it can also be solve by make it in n2 solution as shown below 12+22+32+42+52+62+72+82= 204 squares.

Conclusion: There are two-type strategy solutions for Question 3, first strategy is simply the problem strategy and the second strategy is looking the pattern strategy. I think that looking the pattern strategy is suitable for use in Question 3 and it helps me to solve this question easily and quickly rather than using by simply the problem strategy because it takes much time to solve the problem.

11 OWN QUESTION

Replace the letters with digits in such a way the computation is correct. Each letter may represent only one digit. MA MA + MA EEL

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UNDERSTANDING THE PROBLEM

Each of the letters in M, A, E, and L must be replaced with the digits 1 until 8 to get the correct sum. •

DEVISING A PLAN

Use guess and check strategy. When the alphabet A is replaces by one of the digits, then A+A+A must be L or 10+L. Since 4+4+4=12, 6+6+6=18, 7+7+7=21, 8+8+8=24, there are four possible values for A, namely 4, 6, 7, or 8. When the alphabet M is replaced by one of the digits, then M+M+M must be EE or 10+EE. Since 7+7+7=21+10 it can be 7+7+7=22. So, we already get suitable the digits for M and the answer for E. •

CARRYING OUT THE PLAN

We can’t replace A with 4 because it seems like E. Also, we can’t replace with 7 because it seems like M and we can’t replace with 8 because it will bring the answer more than 19. If A = 6, then A+A+A = 6+6+6 = 18. That is L = 8 If M= 7, then M+M+M= 7+7+7 = 21+10. That is EEL= 22

12 •

LOOKING BACK

Check the sum using the values you obtained to see if you have solved the problem correctly.

MA MA + MA EEL

76 76 +76 228