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14 Sep 2025, 01:30
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Hi quiaitaque
When we start with 1 Water lilly, Then
Progression will be: 1(0th day), 2(2nd day), 4(4th day) , 8 (6th day).......2^30 (60th day)-----> Total number of lillies pond can hold is: 2^30
Now we are at 2 lillies which we have, so the number of lillies required to fill (\(2^(30)\))/2= \(2^(29) \)which means we need 29*2=58 Days
When we start with 1 Water lilly, Then
Progression will be: 1(0th day), 2(2nd day), 4(4th day) , 8 (6th day).......2^30 (60th day)-----> Total number of lillies pond can hold is: 2^30
Now we are at 2 lillies which we have, so the number of lillies required to fill (\(2^(30)\))/2= \(2^(29) \)which means we need 29*2=58 Days
quiaitaque
10 Dec 2025, 00:11
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Hey, I used the following algebraic method.
Scenario A: 1 x 2^x => 2^(x-1) in x days as day one we have only 1 lily.
Scenario B: 2 x 2^x => 2^(x+1) in x days.
We need them be be equal.
Scenario A: We get 2^59 lilies in 60 days.
Scenario B: To get 2^59 lilies we need to take x as 58 be cause 2 x 2^x => 2^(x+1) in x days.
So, we get same number of lilies in 58 days. I might have used numbers, but its still mostly logic, so bunnel's method is the most efficient.
Scenario A: 1 x 2^x => 2^(x-1) in x days as day one we have only 1 lily.
Scenario B: 2 x 2^x => 2^(x+1) in x days.
We need them be be equal.
Scenario A: We get 2^59 lilies in 60 days.
Scenario B: To get 2^59 lilies we need to take x as 58 be cause 2 x 2^x => 2^(x+1) in x days.
So, we get same number of lilies in 58 days. I might have used numbers, but its still mostly logic, so bunnel's method is the most efficient.
rainbooow
10 Dec 2025, 02:02
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process 1: - 1 + 2 + 4 + 8 + 16 ..... till 60 days(31 terms)
process 2:- 2+ 4+ 8+ 16+......
so we start with 2 directly so we dont need the 1st term here which will save us 2 days. But it also produced 1 flower. what do you think from where we add this 1? To cover this we need some extra days in the end bcz if all terms are same then from 2 to 2^31-1 is same for both process so we are just saving the first term which is in turn saving us 2 days but we should note something here. In 2 days we had 2+1 lily and not only 2 lily. and in the new process we have only 2 lily to start with so we are 1 less in the beginning which we need to cover later. so it will take more than 58 days
process 2:- 2+ 4+ 8+ 16+......
so we start with 2 directly so we dont need the 1st term here which will save us 2 days. But it also produced 1 flower. what do you think from where we add this 1? To cover this we need some extra days in the end bcz if all terms are same then from 2 to 2^31-1 is same for both process so we are just saving the first term which is in turn saving us 2 days but we should note something here. In 2 days we had 2+1 lily and not only 2 lily. and in the new process we have only 2 lily to start with so we are 1 less in the beginning which we need to cover later. so it will take more than 58 days
KarishmaB
