Mathematical Solutions To Problems IV And V

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Author: Adekola Taylor
January, 2015

(IV) Solution to Problem IV


Number of columns, (C) = 10, number of cards in each column = 10
Number of cards in the arrangement, n(E) = 100

Let represent each card with a number from 1 to 100 sequentially. Let d0 represents starting arrangement or distribution.

d0 d1
12345678910 9181716151413121111
11121314151617181920 9282726252423222122
21222324252627282930 9383736353433323133
31323334353637383940 9484746454443424144
41424344454647484950 9585756555453525155
51525354555657585960 9686766656463626166
61626364656667686970 9787776757473727177
71727374757677787980 9888786858483828188
81828384858687888990 9989796959493929199
919293949596979899100 100908070605040302010
d2 d3 = ds
100999897969594939291 102030405060708090100
90898887868584838281 9192939495969798999
80797877767574737271 8182838485868788598
70696867666564636261 7172737475767778797
60595857565554535251 6162636465666768696
50494847464544434241 5152535455565758595
40393837363534333231 4142434445464748494
30292827262524232221 3132333435363738393
20191817161514131211 2122232425262728292
10987654321 1112131415161718191
d4 = dt
12345678910
11121314151617181920
21222324252627282930
31323334353637383940
41424344454647484950
51525354555657585960
61626364656667686970
71727374757677787980
81828384858687888990
919293949596979899100

Figure 3: Regenerative Distribution of the Starting Arrangement of the Cards

From Figure 3, at d3 which is called ds (the Second to None Regenerative Distribution Number or the last transposed version distribution number of the starting arrangement), the position ranking values of cards at d0 are interchanged vertically and horizontally respectively at d3.

Therefore, at d3 all the cards' vertical position ranking values would be equal to the values of their horizontal position ranking values at the starting arrangement (d0).

(V) Solution to Problem V


(1) For the first mathematician
Number of columns, (C) = 10, number of cards in each column = 10, Number of cards in the arrangement, n(E) = 100

A. There are two methods through which the first mathematician could solve the problem.

a. Through the use of formula

n (E) = Cxe = C te/2
te denotes even regenerative distribution number, xe denotes even exponential number
100 = 102 = 10 te/2
2 = te/2
te = 4,
Therefore, it would take 4 distributions for the pupils to return to their starting arrangement.

b. By working it out

Number of columns, C = 10, number of pupils in each column = 10, and total number of pupils, n (E) = 100
Let represent each pupil with a number from 1 to 100 sequentially. Let d0 represents starting arrangement or distribution.

d0 d1
12345678910 9181716151413121111
11121314151617181920 9282726252423222122
21222324252627282930 9383736353433323133
31323334353637383940 9484746454443424144
41424344454647484950 9585756555453525155
51525354555657585960 9686766656463626166
61626364656667686970 9787776757473727177
71727374757677787980 9888786858483828188
81828384858687888990 9989796959493929199
919293949596979899100 100908070605040302010
d2 d3 = ds
100999897969594939291 102030405060708090100
90898887868584838281 9192939495969798999
80797877767574737271 8182838485868788598
70696867666564636261 7172737475767778797
60595857565554535251 6162636465666768696
50494847464544434241 5152535455565758595
40393837363534333231 4142434445464748494
30292827262524232221 3132333435363738393
20191817161514131211 2122232425262728292
10987654321 1112131415161718191
d4 = dt
12345678910
11121314151617181920
21222324252627282930
31323334353637383940
41424344454647484950
51525354555657585960
61626364656667686970
71727374757677787980
81828384858687888990
919293949596979899100

Figure 4: Regenerative Distribution of the Starting Arrangement of the Pupils

Considering the schematic distribution of Figure 4, the regenerative distribution (dt) of the starting arrangement = d4, while the regenerative distribution number (t) = 4
t is the number of distributions it would take for the pupils to return to their starting arrangement.

2. For the second mathematician

Number of columns, C = 5, number of pupils in each column = 20, and total number of pupils, n (E) = 100

Let represent each pupil with a number from 1 to 100 sequentially. Let d0 represents starting arrangement or distribution.

d0 d1 d2
12345 9691868176 25507510024
678910 7166615651 4974992348
1112131415 4641363126 7398224772
1617181920 21161161 9721467196
2122232425 9792878277 2045709519
2627282930 7267625752 4469941843
3132333435 4742373227 6893174267
3637383940 22171272 9216416691
4142434445 9893888378 1540659014
4647484950 7368635853 3964891338
5152535455 4843383328 6388123762
5657585960 23181383 8711366186
6162636465 9994898479 103560859
6667686970 7469645954 345984833
7172737475 4944394329 588373257
7677787980 24191494 826315681
8182838485 10095908580 53055804
8687888990 7570656055 295479328
9192939495 5045403530 537822752
96979899100 252015105 771265176
d3 d4 d5
775329582 1938577695 612182430
5834108763 1332517089 3642485460
3915926844 726456483 6672788490
2097734925 120395877 96171319
17854306 9614335271 2531374349
8359351188 908274665 5561677379
6440169369 842214059 85919728
4521987450 7897153453 1420263238
262795531 729192847 4450566268
784603612 668532241 748069298
8965411794 6079981635 39152127
7046229975 5473921029 3339455157
512738056 48676423 6369758187
328856137 4261809917 939941016
1390664218 3655749311 2228344046
95714723100 304968875 5258647076
765228481 24436281100 8288941005
573398662 1837567594 1117232935
3814916743 1231506988 4147535965
1996724824 625446382 7177838995
d6 to d52 d53 = ds d54 = dt
----- 20406080100 12345
----- 1939597999 678910
----- 1838587898 1112131415
----- 1737577797 1617181920
----- 1636567696 2122232425
----- 1535557595 2627282930
----- 1434547494 3132333435
----- 1333537393 3637383940
----- 1232527292 4142434445
----- 1131517191 4647484950
----- 1030507090 5152535455
----- 929496989 5657585960
----- 828486888 6162636465
----- 727476787 6667686970
----- 626466686 7172737475
----- 525456585 7677787980
----- 424446484 8182838485
----- 323436383 8687888990
----- 222426282 9192939495
----- 121416181 96979899100

Figure 5: Regenerative Distribution of the Starting Arrangement of the Pupils

Considering the schematic distribution of Figure 5, the regenerative distribution (dt) of the starting arrangement = d54, while the regenerative distribution number (t) = 54

t is the number of distributions it would take for the pupils to return to their starting arrangement.

Deductions

  • The design of the first mathematician would take 4 distributions for the pupils to return to their starting arrangement.
  • The design of the second mathematician would take 54 distributions for the pupils to return to their starting arrangement.

  • Therefore the design of the first mathematician would be more time saving than the design of the second mathematician.


    II. If a pupil was in the horizontal position ranking value 35th at the starting arrangement (d0) in the arrangement fashioned out by the second mathematician, what would be her vertical position ranking value (Pvd0) at that starting arrangement?

    n(E) = 20, C = 5, r = 20, Phd0 = 35, Pvd0 = ?

    Pvd0 can be calculated either (i) by assigning vertical position ranking value to the pupils by using the horizontal position ranking value given. (ii) or by using formula.

    (i) From the Figure 5, the pupil designated by number 35 was ranked to be in horizontal position 35th. The vertical position ranking was taken as the sequence of the pupils’ distribution. Therefore, a pupil who was in the horizontal position ranking value 35th would be in the vertical position ranking value 94th. Therefore, Pvd0 = 94.

    (ii) Similarly, by using formula

    Phdx =1/r Pvdx + [C + 1/r] khdx
    0 is greater than or equal to khdx, and khdx is less than or equal to r - 1

    n(E) = 20, C = 5, r = 20, Phd0 = 35, Pvd0 = ?

    Phd0 =1/r Pvd0 + [C + 1/r] khd0
    35 =1/20 Pvd0 + [5 + 1/20] khd0
    35 x 20 = Pvd0 + 101 khd0

    Calculation of khd0 (Horizontal position class interval rank)

    n(E)/r = 100/20 = 5
    Pupils in class interval (1 - 5) have their khdx = 0, pupils in class interval (6 - 10) have their khdx = 1, pupils in class interval (11 - 15) have their khdx = 2, etc. Since Phd0 = 35 fell into horizontal class interval (31 – 35),
    then its khd0 = 6
    35 x 20 = Pvd0 + 101 khd0
    700 = Pvd0 + 101 khd0 x 6
    Pvd0 = 700 – 606 = 94

    The vertical position ranking value (Pvd0) for the pupil at the starting position would be 94.

    Click here for solutions to Problems I, II and III

    Click here to go back to Problems I - IV

    References

    1. Taylor, Adekola A. (2016). Derivation of formulas for position change of entities in power inductive distribution. International Journal of Scientific and Research Publications 6(7):409-420.

    2. Taylor, Adekola A. (2015). Quadratic distribution patterns in Kola analysis. Mathematical Theory and Modeling 5: 60-66.

    3. Taylor, Adekola A. (2015). Derivation of formulas for position change in Kola analysis. International Journal of Scientific and Research Publications 5(11):101-109.

    4. Taylor, Adekola A. (2014). Logical-mathematical intelligence for teens. Mathsthoughtbook.com

    5. Taylor, Adekola A. (2013). Regenerative mathematics and dimurelo puzzles for children 8-12yrs. USA: Lulu Press Inc.

    6. Taylor, Adekola (2013). Card magic and my mathematical discoveries. USA:Lulu Publishing.

    7. Taylor, Adekola A. (2010). Kola Analysis: An inventive approach to logical-mathematical intelligence for secondary and advanced levels. Journal of Mathematical Sciences Education 1(1): 44-53.

        Recommended Readings:

        1. Regenerative Mathematics

        2. Card Mathematical Intelligence Games

        3. Dimurelo Puzzles

        4. Simple Dimension Positiomatics

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