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Finding the population of 3-A at the end of the first "year"
In "year one", 2-A mutants come from 1-A. In each saturation cycle, 1-A has 10e29 divisions. One in 1.82*10e12 of these divisions will produce 2-A. 5.49450549*10e16 2-A will be produced per saturation cycle. One in 1.82*10e12 is also the number of 3-A generated by 2-A organisms
rem finds populations of 2-A and 3-A at the end of "year one" countit = 0 original = 5.49450549*10^16 rem 5.49450549*10^16 per saturation cycle (2-A) A3 = 0 population = 0 [start] population = population + original population = population * 1.01 A3 = A3 + (population/(1.82 * 10^12)) rem A3 means 3-A A3 = A3 * 1.02 countit = countit + 1 IF countit = 1000 then [end] GOTO [start] [end] print population rem meaning 2-A print A3 rem meaning 3-A end
answer: 1.16306248*10e23 2-A at end of "year one" 6.31648463*10e16 3-A at end of "year one" or about 6.3*10e16
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