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Maths Assignment - Patterns and the 4 rule

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Introduction

Year 10 Maths Assignment

Task 1

  1. Compile an uninterrupted sequence of whole numbers up to 25 using R1

0 = 4-4+4-4

1 = 4-4+4/4

2 = (4x4)/(4+4)

3 = (4+4+4)/4

4 = 4+4-√4-√4

5 = 4/4+√4+√4

6 = 4+4-4+√4

7 = 4+√4+(4/4)

8 = (4x4)-4-4

9 = 4+4+(4/4)

10 = 4+4+4-√4

11 = 44/(√4+√4)

12 = (4(4+√4))/√4

13 = 44/4 + √4

14 = 4(√4+√4)-√4

15 = 4x4 - (4/4)

16 = 4x4+4-4

17 = 4x4 + 4/4

18 = 44/√4 -4

19 = 4x4+4-√√√√√√√√√√√√√√√√√√√4

20 = (44-4)/√4

21 = (44-√4)/√4

22 = 44/(4-√4)

23 = (44+√4)/√4

24 = (44+4)/√4

25 = (4+ 4/4)√4

  1. Using R2 to generate four different ways to create one number

                    .

40 = 4(4/0.4)+4  

40 = (√4+√4)(4/0.4)  

40 = (4+√4+4)x4

40 = (4!-4)x√4

Task 2

What happens when the process is followed for a different starting word? By using two (2) of your own examples, does it always happen?

image03.pngimage00.pngimage01.png

EXTERMINATED→TWELVE→SIX→THREE→FOURimage02.png

image03.pngimage00.pngimage01.pngimage02.png

DISPENSER→NINE→FOUR

Yes, it always happens. Four is the only number where it represents the number of letters its word has.

Task 3

Perform the Kaprekar process for 2 other four digit starting numbers. When the process continues indefinitely, what happens?

image00.pngimage03.pngimage01.pngimage02.png

1824→7173→6354→3087→8352→6174

image00.pngimage03.pngimage01.pngimage02.png

1002→2088→8514→7083→8352→6174

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Middle

image08.png

8→4→2→6→

image07.pngimage09.pngimage03.pngimage06.png

9→1→

The longest length bracelet for the above process is four numbers. There are four numbers that can make this length (2,3,7,8).

Task 5

What is the shortest length bracelet formed from the last digits of a generalised Fibonacci sequence?

image10.pngimage11.pngimage12.png

0→1→1→2→3→5→8→3→1→4→5→9→4→3→7→0→7→7→4→1→5→6→1→7→8→5→3→8→1→9→0→9→9→8→7→5→2→7→9→6→5→1→6→7→3→0→3→3→6→9→5→4→9→3→2→5→7→2→9→1→image13.png

image14.png

0→2→2→4→6→0→6→6→2→8→0→8→8→6→4→0→4→4→8→2

image10.pngimage12.pngimage11.png

0→3→3→6→9→5→4→9→3→2→5→7→2→9→1→0→1→1→2→3→5→8→3→1→4→5→9→4→3→7→0→7→7→4→1→5→6→1→7→8→5→3→8→1→9→0→9→9→8→7→5→2→7→9→6→5→1→6→7→3→image15.png

image14.png

0→4→4→8→2→0→2→2→4→6→0→6→6→2→8→0→8→8→6→4→

image07.pngimage06.pngimage16.pngimage03.png

0→5→5→

0→6→6→2→ Too Long

0→7→7→4

0→8→8→6

0→9→9→8

1→1→2→3

1→2→3→5

1→3→4→7

1→4→5→9

1→5→6→1→7

1→6→7→3

1→7→8→5

1→8→9→7

1→9→0→9

2→2→4→6

2→3→5→8

2→4→6→0

2→5→7→2→9

2→6→8→4

2→7→9→6

2→8→0→8

2→9→1→0

3→3→6→9

3→4→7→1

3→5→8→3

3→6→9→5

3→7→0→7

3→8→1→9

3→9→2→1

4→4→8→2

4→5→9→4→3

4→6→0→6

4→7→1→8

4→8→2→0

4→9→3→2

image07.pngimage06.pngimage03.pngimage17.png

5→5→0

5→6→1→7

5→7→2→9

5→8→3→1

5→9→4→3

6→6→2→8

6→7→3→0→3

6→8→4→2

6→9→5→4

7→7→4→1

7→8→5→3

7→9→6→5

8→8→6→4

8→9→7→6

9→9→8→7

The shortest length bracelet formed from the last digits of a generalised Fibonacci sequence is three numbers. Two bracelets can do this – one beginning with 0→5, one beginning with 5→5. But these are essentially two starting points of the same chain.

Task 6

        Abbreviations: Happy Number - HN

10,13,23,49,97 are happy numbers under 100 (given)

4,16,20,37,42,58,89 are not happy numbers under 100 (given)

99→162→41→17→50→25→29→85→89 89 is not a HN, therefore 99 is not

98→145→42 42 is not a HN, therefore 98 is not

97→Happy Number (Given) (1)

96→117→51→26→40→16→ 16 is not a HN, therefore 96 is not

95→106→37→58→89→145→42

...read more.

Conclusion

(See Task 6 for evidence)

image06.pngimage19.pngimage07.pngimage18.png

99→90→81→9            Does not fit in rule 1, therefore must be rule 2.

98→73→16→37→52→9→81→9 (2)

97→58→59→86→44→20→2→4→16… (2) 16 is in rule 2, therefore 97 must be

96→45→29→83→17→50→5→25→27→51→6→36→39→84→24→18→65→31→4→16… (2)

95→34→19→82→12→5… (2)

94→25… (2)

93→18→65… (2)

92→13→10→1 (1)

91→10→1(1)

90→9→81→65→31→4…(2)

89→89→ Does not fit in either R1 or R2. Therefore this must be the new rule. (3)

The three outcomes are:

  1. The numbers end up on the number one (1). They are happy numbers.

(See Task 6 for evidence)

  1. The numbers end in a loop of two (2) numbers (81,9)
  2. The number ends in a one number loop (89)

Task 8

99→80→88→60→66→20→22→40→44→80→ 1 is already given. It ends in a loop of 8 numbers (80,88,60,66,20,22,40,44).

image06.pngimage07.pngimage20.pngimage03.png

98→71→86→42→62→84→24→       2 ends in a loop of three numbers (62,84,24).

97→62→(2)

96→53→82→06→66→(1)

95→44→(1)

94→35→82→(1)

93→26→84→(2)

92→17→86→(2)

91→08→88→(1)

90→99→(1)

88→(1)

77→40→(1)

66→(1)

55→00→00  New Rule. It ends in a loop of 1 number (00).

44→80→(1)

33→60→(1)

22→40→(1)

11→20→(1)

The two other outcomes for this altered rule are:

  1. The numbers end in a loop of 3 numbers (62,84,24)
  2. The number ends in a loop of 1 number (00).

END OF ASSIGNMENT

...read more.

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