math problem?

Question

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?

Answer 1

Depends...Do you work week-ends.

If you do it's quite a bit more than if you don't

Roughly you'll get 2^22 pennies = 4 million pennies = 40k.
If you work weekends it'll be 2^31 pennies = 2 billion pennies = 20 million.
--------------
Sorry that was total.
Last day if you worked only weekdays 2^21 = 2 million pennies = 20k.
If you worked weekends 2^30 = 1 billion pennies = 10 million

Answer 2

You get a cent on the first day of 31 days.
Pay doubles each day and there are 31 days

Pay of day 1 = 1 cent
Pay of day 2 = 2 cents
Pay of day 3 = 4 cents
Pay of day 5 = 16 cents

and so on. You can notice that Pay of day 2 = 2^1 cents , day 3 = 2^2 cents, day 4 = 2^3 cents, day 5 = 2^4 cents.

Notice something?
Pay on day n following this pattern would be 2^(n - 1) cents

Pay on 31st day = 2^(31 - 1)
= 2^30
= 1073741824 cents
= $ 10737418.24

Answer 3

The nth day's pay will be 2^(n-1) cents.

Since July has 31 days,
and 2^30 equals 1073741824 that means
the last day you will earn $10737418.24,
or nearly $11 million dollars.

Not bad for a day's work, unless you're Bill Gates.

Answer 4

July has 31 days. on the last day he would pay you 2^30 ¢. for the month, you would get 2^31-1=2147483647 ¢ or
$21,474,836.47. Sounds like a great deal. Grab it!

Answer 5

working in cents, they pay you 2^(d-1) cents per day. On the last day of work, July 31 they will pay you 2^30 cents.... or $10,737,418.24. Not bad for one day of work!

Answer 6

Geometric progression
a=1, r=2, n=31

Tn = ar^(n-1)
T31
= 1(2)^(31-1)
= 2^30
= 1073741824 cents
= $10,737,418.24

Answer 7

.01 to the 30th power. In other words, the salary will times its self everyday for that month. BIG $$$$.

Answer 8

use geometry sequence
0.01 , 0.01(2) , 0.01(2)(2) ,.... 31 days
a,ar,ar^2,ar^3, ........
an= ar^(n-1)
a31 =0.01(2^30)
= 10,737,418.24$

if you want sum of all
you must use geometry series..

0.01 + 0.01(2) + 0.01(2)(2) + .... 31 days
Sn =(a-ar^n)/1-r
a = 0.01 and r = 2
S31 = (0.01 - 0.01*2^31)/(1-2)
= 0.01(2^31 -1)
= 21,474,836.47$

prove...Sn =(a-ar^n)/1-r
Sn = a +ar +ar^2 +ar^3 + .... +ar^n-1 -----#1
r x #1
rSn = +ar +ar^2 +ar^3 + .... +ar^n-1+ar^n -----#2
#1 - #2
Sn - rSn = a - ar^n
Sn(1-r)
Sn =(a-ar^n)/1-r

hope this help ^-^

Answer 9

well, first of all, I wouldn't want to work for any boss that'd be so cheap to pay me a penny... geez... but then it got me thinking... boss should pay me triple more on the last day of work. Don't ask me where I came up with this figure cuz I'm terriable in math ... lol.. duh

Answer 10

$5368709.12