what kind of system is 1->2->4->16 ? i can only figure out this:
4^0 = 1
4^1/2 = 2
4^1 = 4
4^2 = 16
cant do it like that. so i guess a workaround will do. with my poor math skills i come to this result:
3 + sum of: 4^(n-2)
3 is the amount of coins on day 1 and 2
"sum of" is where you put the sigma sign. below the sigma you put a 3 (as you start on day 3), on top you put "n" (the total amount of days).
and n is something around 3022 or so. i'll let someone else think about the leap years (did they have leap years at that time? 😀 ).
note that you didnt say explicitly, that you get 1 cent on the first day of your life, so it could be starting at any time, not necessarily on june 8, 1124.
Originally posted by crazyblueI think the intended sequence was 1,2,4,8,16 etc. The sequence number is represented by 2^n. The number of days given is 2950 not factoring in leap years, and assuming months have 30 days. I don't have that kind of time. The last amount you'll receive is 2^2950
what kind of system is 1->2->4->16 ? i can only figure out this:
4^0 = 1
4^1/2 = 2
4^1 = 4
4^2 = 16
cant do it like that. so i guess a workaround will do. with my poor math skills i come to this result:
3 + sum of: 4^(n-2)
3 is the amount of coins on day 1 and 2
"sum of" is where you put the sigma sign. below the sigma you put a 3 (as you s ...[text shortened]... rst day of your life, so it could be starting at any time, not necessarily on june 8, 1124.
The sum of a geometic sequence = (a(1)-a(n))/(1-r) = (1-2^2950)/(1-2) = (1-2^2950)/(-1) = 2^2950 + 1. I don't know how to reduce this though.
Originally posted by blindcheesecakeThe pattern seems to be 2^(number of cents the day before).
If on day 1 you got 1c then 2c on day 2 and 4c on day 3 and 16c on day 4 and so on, How much money would you have after 8 years 3 months and 12 days, if you were born on June 8 1124. Be shure to include leap years.
Day 1 is therefore 2^0 = 1
Day 2 is 2^1 = 2
Day 3 is 2^2 = 4
Day 4 is 2^4 = 16
I am not s(h)ure what the question is as the calendar is not specified, the planet is not specified nor is it specified whether the amount is a sum or an aggragate (ie on day 4 I have 16c not 1+2+4+16 = 23c... and what is a c?) If c is the speed of light, then if you have 16c then you really just have 1c due to relativity and the wave nature of light. So in that case, my anser would be c at the end of the unknown duration (which is presumably longer than nothing).
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Originally posted by blindcheesecakewell bill gates and excel never thought we'd try to figure this out cause my excel sheet stops at day 208 and the amount of coins being on that day to be [WORD TOO LONG] That 2,175,541,218,577,480 and 173 zeros following
If on day 1 you got 1c then 2c on day 2 and 4c on day 3 and 16c on day 4 and so on, How much money would you have after 8 years 3 months and 12 days, if you were born on June 8 1124. Be shure to include leap years.
i am not a smart person so thats all i can say, i would like to know how ever based on a 365 day year callander if my answer for this day is correct
thanks
Originally posted by blindcheesecakewhy would you ask it over such a long period of time. One can see after just 25 days that the amount of money you are getting at that point is phenomenal.
If on day 1 you got 1c then 2c on day 2 and 4c on day 3 and 16c on day 4 and so on, How much money would you have after 8 years 3 months and 12 days, if you were born on June 8 1124. Be shure to include leap years.
Originally posted by CoconutBecause it's much harder to do this way.
why would you ask it over such a long period of time. One can see after just 25 days that the amount of money you are getting at that point is phenomenal.
I am doing it manually, as in 1c + 2c + 4c + 16c all the way until I reach 2006. It may take until 2060....