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A lily pad sits on a pond. It doubles in size every day. It takes 30 days for it to cover the pond. If you start with 8 lily pads instead, how many days does it take to cover the pond?

I think that the answer is $27$, but I don't really think that makes sense intuitively. I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.

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marked as duplicate by Carsten S, Dan, RRL, Micah, Namaste algebra-precalculus Jan 11 at 17:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ It'll take $27$ days only. Think of it like this that (for the first case) on the second day no of lily pads will become $2$, on the third day $4$ and on the fourth $8$. In your second case you start your first day at this point $\endgroup$ – Sauhard Sharma Jan 9 at 6:31
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    $\begingroup$ Just out of curiosity, where does 30/4 come from Joseph? $\endgroup$ – Peter Jan 9 at 10:11
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    $\begingroup$ Do not rely on your intuition; this problem is physically implausible and therefore your intuition, which has evolved in the real world, will be misleading. A lily pad that doubles in size every day will be larger than the sun very quickly. Ignore the nonsense window dressing about the lily pad and the pond and do math: We define two recurrences: f(0) = 1, f(n) = 2f(n-1), g(0) = 8, g(n) = 2g(n-1). What is the lowest integer k such that g(k) >= f(29)? No silly lily pads; just math. $\endgroup$ – Eric Lippert Jan 10 at 6:05
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    $\begingroup$ The question makes no sense. Lily pads are roughly round. You don't know how well 8 discs cover another disc unless you know their positions. If they are all on top of each other, the answer will be 30. $\endgroup$ – Jessica B Jan 10 at 6:25
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    $\begingroup$ In addition to @JessicaB we also don’t know if the assumption that the lake is round is a good one. But actually I like the question, it is suitable to derive some bounds taking the geometry of lake and pads into account and as already pointed out, one should impose some well-distribution assumptions. $\endgroup$ – Sebastian Bechtel Jan 10 at 8:18
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Starting with 8 pcs. is like 3 days have gone by. So 27 days remain.

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    $\begingroup$ How is this definite? Assuming the the lily pads grow from the middle out on all sides, then they will either overlap each other or have to grow fold over outside of the pool. Doesn't your answer require the lily pads to only grow in a certain direction, and start from particular spots in order to be true? $\endgroup$ – youngcouple10 Jan 10 at 11:45
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    $\begingroup$ @youngcouple10 It does not matter. The conditions are stated plainly: a lily-pad will double in size each day. Practically we know this 1) cannot happen and 2) does not scale with multiple lily-pads because the lily-pads will encroach on each other unless we change the shape of the pond. But — absent any additional information, and assuming there is a definite answer — we must assume this these lily-pads have been planted there by the Spherical Cow, and so we ignore the practicalities of the matter. $\endgroup$ – MichaelK Jan 10 at 13:13
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Hint $\#1$:

At the end of the $30$ days with one lilypad, the doubling means that the lilypad now encompasses the area of $2^{30}$ of the original lilypads.

In that light, starting with $2^3 = 8$ lilypads and each one doubling in size per day, how many doublings will it take for you to get to $2^{30}$?

(I know you've already solved this, I just think rewording/reframing the question might make it a bit easier to grasp on the intuitive level.)


Hint $\#2$:

If that doesn't help ease your intuition behind your answer (which to my understanding is correct), keep in mind that starting with $8$ lilypads is basically no different than your first scenario after $3$ days. Sure, you have more lilypads, but since each doubles in size, it's no different than one lilypad of the same size as those $8$ put together then doubling. The number of lilypads differs, but we're focused on the total area encompassed.

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Another way to look at it is to work backward.

First just consider the one lily pad. After $29$ days it covers half the pond. After $28$ days a quarter of the pond. After $27$ days an eighth of the pond.

So after $27$ days eight lily pads would cover the whole pond.

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I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.

Lily pad doubles in size every day, so it is increasing as a geometric progression. $$\begin{array}{c|c|c|c|c|c|c|c|c} \text{n-th day}&1&2&3&4&\cdots&26&27&28&29&30\\ \hline \text{size of $1$ lily pad}&1&2&2^2&2^3&\cdots&2^{25}&2^{26}&2^{27}&2^{28}&\color{red}{2^{29}} \end{array}$$

If you start with $8$ lily pads, each doubling on its own, then: $$\begin{array}{c|c|c|c|c|c|c|c|c} \text{n-th day}&1&2&3&4&\cdots&26&27&28&29&30\\ \hline \text{size of $8$ lily pads}&2^3&2^4&2^5&2^6&\cdots&2^{28}&\color{red}{2^{29}}&2^{30}&2^{31}&2^{32} \end{array}$$ Because when each of $8$ lily pads keeps doubling per day, the $8$ lily pads increase $8$ times faster in size altogether than that of one lily pad. So, you must multiply the size of one lily pad on any day by $8=2^3$ to find the total size of $8$ lily pads.

As this source informs, the Giant Water Lily may grow as large as $8$ to $9$ feet ($2.4-2.7$m) in diameter.

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  • $\begingroup$ Upvote for the visual listing. $\endgroup$ – Benxamin Jan 11 at 3:38
  • $\begingroup$ @Benxamin, thank you for finding this aspect useful and making me earn "Nice answer" badge. Wheat and chessboard problem is similar to the OP's problem, but it considers the total sum of the grains of wheat on the chessboard. Speaking in terms of the chessboard, the OP should start from the $4$-th cell and not divide $30$ by $4$, but subtract $30-4+1=27$ as mentioned by Arthur. $\endgroup$ – farruhota Jan 11 at 6:30
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$27$ is the right answer. Consider at which point the original lily pad is 8 times its original size (after three days), and how long it takes it to cover the lake from there.

Also, this relative "indifference" to a seemingly large disparity in starting point ("$x$ times more to start with means it takes $30-y$ days rather than $30/y$") is exactly what exponential growth means.

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  • $\begingroup$ +1 for your educated guess of the OP's $30/4$. $\endgroup$ – farruhota Jan 11 at 6:41
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Your answer is correct

If we denote the size of the lily pad by $x$, then after $1$ day, the coverage becomes $2x$ , . . . Also let's denote the size of pond by $y$. The assumptions imposes that:$$2^{30}x\ge y$$and $$2^{29}x<y$$Now starting with $8$ lily pads we obtain $$2^{27}\cdot 8x\ge y$$and$$2^{26}\cdot 8x< y$$which shows that the completion of the process takes long as much as $27$ days.

The intuition is that:

If we start with $1$ lily pad, after $3$ days we will have $8$ of them since the size of lily pads doubles each day. Starting with $8$ lily pads in first place and observing their growth is equivalent to observe the growth of $1$ lily pad during the days skipping the first $3$ days. If whole the process takes long $30$ days, then or modified process takes long $30-3=27$ days which is totaly intuitive.

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Answering the 'why is my intuition wrong?' aspect:

The key thing to remember here is not everything is linear. The unrealistic nature of the question isn't what makes it unintuitive. It's that we have a tendency to think things are linear, even when we know full well they are not.

The reasoning used seems to be 'if I start with twice as much, it must only take half as much time to get to the same end amount'. That's using linearity, and basic ideas of addition/multiplication. But exponential growth is very very much not like this.

There are other places this comes up in life: if you are speeding up then you cover most of the ground at the end; if you are saving for your pension, a large part of your savings comes from the last few years (when your salary is highest); twins newly separated are only half the size of a single baby the same age, but they'll only take minutes longer to reach full size, not twice as long.

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