math problem?
you are offered a job for the month of july. the employer tells you that he'll pay you $0.01 for the first day and double each days pay based on the pay of the previous day. How much should he pay you for you last day of work?
you are offered a job for the month of july. the employer tells you that he'll pay you $0.01 for the first day and double each days pay based on the pay of the previous day. How much should he pay you for you last day of work?
Comments
0.01 x 2^n where n =number of days so $21,474,836.48
All together, July has 31days.
There for :
Day 1 - $0.01
Day 2 - $0.01 x 2 = $0.02
Day 3 - $0.02 x 2 = $0.04
Day 4 - $0.04 x 2 = $0.08
Day 5 - $0.08 x 2 = $0.16
Day 6 - $0.16 x 2 = $0.32
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Day31- $5368709.12 x 2 = $10737418.24
The formula you want is [.01*2^(n-1)], where n is the number of days.
The first day is .01*(2^0) = .01
The next day is .01*(2^1) = .02
The next day is .01*(2^2), etc etc.
There are 31 days in July, so n = 31.
It's over ten million dollars on your last day. This is what they mean by "growing exponentially".
$10,737,418.24
just keep multiplying each days pay, day one is 1 cent day two would be 2 cents day 3 would be 4 cents day 4 would be 8 cents day 5 would be 16 cents and so on.
It would be .01*2*2*2*2...until you had 30 twos listed out or
.01 * 2^30
= .01 * 1073741824 which would equal $10,737,418.24
10,737,418,24