Author: Adekola Taylor

January, 2015

**(1) Solution to Problem I**

Number of columns, (C) = 5, number of cards in each column = 48,828,125

Number of cards in the arrangement, n(E) = 5 x 48,828,125

= 244,140,625

Since n (E) could be expressed in term of C^{xe}

then n(E) = C^{xe} = C^{te/2}

t_{e} = even regenerative distribution number

x_{e} = even exponential number

This implies 244,140,625 = 5^{xe}

then 5^{xe} = 5^{te/2}

Therefore, 244,140,625 = 5^{12}

5^{12} = 5^{te/2}

12 =t_{e}/2

t_{e} = 24

**(2) Solution to Problem II**

Number of columns, (C) = 12, number of students in each column = 10

Number of students in the arrangement, n(E) = 120

Since n (E) could be expressed in term of C_{e}[C_{e} - 2]

then n(E) = C_{e}[C_{e} - 2] = 1/4[t_{e}]^{2}- 1

t_{e} = even regenerative distribution number

C_{e} = even column number

120 = 12[10] = 1/4[t_{e}]^{2}- 1

480 =[t_{e}]^{2}- 4

t_{e} = -22 or 22

Since we are dealing with concrete things

then t_{e} = 22

The regenerative distribution number (t) = 22

**(3a) Solution to Problem III**

Number of columns, (C) = 19, number of cards in each column = 21

Number of cards in the arrangement, n(E) = 399

Since n (E) could be expressed in term of C_{o}[C_{o} + 2]

then n(E) = C_{o}[C_{o} + 2] = [t_{e}]^{2}- 1

t_{e} = even regenerative distribution number

C_{o} = odd column number

Since t_{e} = s_{o} + 1

s_{o} = The last transposed version distribution number of the starting distribution

then, C_{o}[C_{o} + 2] = [s_{o}]^{2} + 2s_{o}

n(E) = 399 = 19[21] =[s_{o}]^{2} + 2s_{o}

[s_{o}]^{2} + 2s_{o} - 399 = 0

s_{o}[s_{o} + 21] - 19 [s_{o} + 21] = 0

[s_{o} + 21][s_{o} - 19] = 0

s_{o} = -21 or 19

Since we are dealing with concrete things

then s_{o} = 19

The last transposed version distribution number of the starting distribution = 19

**(3b) Solution to Problem III**

If the number of columns and rows were interchanged

then,number of columns, (C) = 21, number of cards in each column = 19

Number of cards in the arrangement, n(E) = 399

Since n (E) could be expressed in term of C_{o}[C_{o} - 2]

then n(E) = C_{o}[C_{o} - 2] = [t_{e}]^{2} - 1

t_{e} = even regenerative distribution number

C_{o} = odd column number

21 X 19 =[t_{e}]^{2} - 1

399 + 1 =[t_{e}]^{2}

t_{e} = -20 or 20

Since we are dealing with concrete things

then t_{e} = 20

The new regenerative distribution number (t) = 20

**Click here for solutions to Problems IV and V**

**Click here to go back to Problems I - IV**

**References**

- 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. - Taylor, Adekola A. (2015). Quadratic distribution patterns in Kola analysis.
*Mathematical Theory and Modeling*5: 60-66. - Taylor, Adekola A. (2015). Derivation of formulas for position change in Kola analysis.
*International Journal of Scientific and Research Publications*5(11):101-109. - Taylor, Adekola A. (2014).
*Logical-mathematical intelligence for teens.*Mathsthoughtbook.com - Taylor, Adekola A. (2013).
*Regenerative mathematics and dimurelo puzzles for children 8-12yrs.**USA: Lulu Press Inc.* - Taylor, Adekola (2013).
*Card magic and my mathematical discoveries.*USA:Lulu Publishing. - 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.<

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