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# Algebra

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A certain organism begins as two cells. Each cell splits and becomes two cells at the end of three days. At the end of another three days, every cell of the organism splits and becomes two cells. This process lasts for a total of 15 days, and no cells die during this time. How many cells are there at the end of the $15^\text{th}$ day?

Guest Oct 7, 2017
#1
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Let the number of cells after 15 days =C

15/3 =5 number of doublings of cells

C = F x  R^(N - 1), where F=First term, R=Common Ratio, N=Number of terms.

C =2  x  2^(5 - 1)

C =2  x  2^4

C =2  x  16

C = 32 cells after 15 days.

Guest Oct 7, 2017
#2
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Computer scan:  Result:  Blarney Banker formula detected.

Initiate subroutine for mathematical correction.

Minimize overt BB criticism (this instance).

Solve using table

$$\begin{array}{|rcll|} \hline \tiny \text { Time (Days) }| & \tiny \text {# of cells} \\ 0 |& 2 \\ 3 |& 4 \\ 6 |& 8 \\ 9 |& 16 \\ 12 |& 32 \\ 15 |& 64 \\ \hline \end{array}\\$$

Generalized formula for doubling per unit time.

$$\Large N = Z*2^{(t/d)}$$

Where (N) is the number of (cells) after time (t), (Z) is the starting number of (cells) at time zero (0), and (d) is the time required to double (cell) count.

$$\Large (2)*2^{(15/3)} = 64$$

-------------

Update Statistical counters:

Increment BB error count by 1: BB=(6.0244834478561245 E21) + 1

End subroutine.

Resume normal operation.

GingerAle  Oct 7, 2017