This is a puzzle that has to be solved to find a geocache, and I'm stumped.
Flower Power
Geocache Description:
The cache is obviously not at the posted coordinates, but is within 1-2 miles of the pond. This is an easy puzzle if you think it through, and is based purely on math. It is really long though (so long there wasn't a FTF until May), and I dare you to do this without a calculator. Since the dandelions are popping up around this time, I figured it would be right to make a puzzle on dandelions, and since I came up with this problem and solved it about a year ago, I decided to adapt it to cache use.
Suppose there is one dandelion in the middle of a field. That dandelion will produce 200 seeds for that year. Suppose all of the seeds ever given off are certain to grow, all the offspring give off exactly 200 seeds the next years, and none of the dandelions ever die. How many dandelions will there be in field on the 100th year?
Once you have this number, take the 26 leftmost digits on the number. In order from left to right, they are A-Z. The coordinates for the cache are N41 CT.TMS W070 JG.HVQ. Good luck, and have fun!
Any help is appreciated!
Hugs, Devlyn
Each year, you have 201 times as many plants as the year before: the 200 new ones for each original one, plus the original ones that are still there. So, in year n, you have 201^(n-1) plants.
In year 100, you have 201^99 plants. That's a really big number. I can't even count the digits because, on my screen, it is wider than the full screen, but it's around 250 digits long. You need to calculate at least the first 26 digits to get the full alphabetic translation required for the coordinates. You don't actually use all 26 letters, but you have to get as far as the letter 'T', which means 20 digits. I tried the calculation in Excel, but it pooped out after 15 digits.
So, no, I can't get you the coordinates.
No calculator, my ***!
http://rabbit.eng.miami.edu/class/een218/calculator.html
I'd replace 200 with 3 and figure out the equation that way so you can validate. Then put the 200 back in there. Don't forget the one you started with.
only the offspring give seed ... but the others remain ...
1
200 +1 200^1 + 200^0
200 * 200 +200 +1 200^2 + 200^1 + 200^0
200*200*200 +200*200 +200 +1 200^3 + 200^2 + 200^1 + 200^0
There is a formula for that ...
hugs
The 200 is a trick, just use 2 since you ignore all but the first 26 digits. They only reproduce once. The rest looks straightforward.
Quote from: AnonyMs on May 04, 2016, 10:50:21 AMThey only reproduce once.
Hmm. I see that it could be interpreted that way. I hate ambiguous puzzles!
Quote from: KathyLauren on May 04, 2016, 11:05:43 AM
Hmm. I see that it could be interpreted that way. I hate ambiguous puzzles!
It's probably meant to be easy, once you understand it.
Quote from: KathyLauren on May 04, 2016, 10:08:46 AM
Each year, you have 201 times as many plants as the year before: the 200 new ones for each original one, plus the original ones that are still there. So, in year n, you have 201^(n-1) plants.
In year 100, you have 201^99 plants. That's a really big number. I can't even count the digits because, on my screen, it is wider than the full screen, but it's around 250 digits long. You need to calculate at least the first 26 digits to get the full alphabetic translation required for the coordinates. You don't actually use all 26 letters, but you have to get as far as the letter 'T', which means 20 digits. I tried the calculation in Excel, but it pooped out after 15 digits.
So, no, I can't get you the coordinates.
No calculator, my ***!
This looks really good and gives coordinates right on the edge of a bike path in the target area. Will let you know when I get verification or find my way down there.
Thanks for all the help, everyone! :)
Hugs, Devlyn
I'm not sure the 201 is accurate. Wouldn't it be the algorithm using 200, then adding a +1 after the calculation? Each of the 200 will reproduce, but the original one has already done it.
I also just tried it with 200 and 2. With scientific notation, the displayed digits are identical between the two with the only difference being the total number of digits, 228 and 30 respectively. Using 2 has enough digits to get the full 26. Whoever said that... genius!
Quote from: KathyLauren on May 04, 2016, 10:08:46 AM
Each year, you have 201 times as many plants as the year before: the 200 new ones for each original one, plus the original ones that are still there. So, in year n, you have 201^(n-1) plants.
In this formula its 201 squared.
But only 200 are reproduced, the old ones do not reproduce.
It needs to be a sum imo.
The old ones plus the new ones.
Imo its a geometrical line, and there is a formula for that.
Unless there is some kind of hidden formulation which makes it all a diiferent thing.
"all the offspring give off exactly 200 seeds the next years"
meaning there is 200 offspring always every year ?
hugs
Quote from: Emileeeee on May 05, 2016, 08:30:02 AM
I'm not sure the 201 is accurate. Wouldn't it be the algorithm using 200, then adding a +1 after the calculation? Each of the 200 will reproduce, but the original one has already done it.
As noted above, it depends how you interpret the question. It doesn't actually state that the plants only reproduce once. I know that my dandelions reproduce every year!!
If the plants only reproduce once each, then the answer is 200^99 +1 (or 2^99 +1 since you only need the first 26 digits). If the plants reproduce every year, then it is 201^99. The bit about doing it without a calculator might be a hint about which way to interpret it, since 2^99 is feasible (though tedious) without a calculator. 201^99 is not.
Quote from: KathyLauren on May 05, 2016, 08:58:29 AM
It doesn't actually state that the plants only reproduce once.
"That dandelion will produce 200 seeds for that year."
Yes. meaning could be : only for that year or each year.
With only for that year its a geometrical line.
If the formula is used then results are over 60 degrees ,and even if recacculated for degrees the result is in the sea.
So each year might be a possibility.
But without calculator ?
hugs
Quote from: Laura_7 on May 05, 2016, 09:07:11 AM
But without calculator ?
For 2^99, sure. Computer nerds know that 2^16 = 65536. So 2^99 = 8 * 65536^6. That would be a pain in the butt to do longhand, but only involves 6 multiplications.
Quote from: KathyLauren on May 05, 2016, 08:58:29 AM
As noted above, it depends how you interpret the question. It doesn't actually state that the plants only reproduce once. I know that my dandelions reproduce every year!!
If the plants only reproduce once each, then the answer is 200^99 +1 (or 2^99 +1 since you only need the first 26 digits). If the plants reproduce every year, then it is 201^99. The bit about doing it without a calculator might be a hint about which way to interpret it, since 2^99 is feasible (though tedious) without a calculator. 201^99 is not.
Ah yes it is ambiguous. I hadn't read it that way, but I do see that it could mean that each one produces 200 offspring each year. What a nightmare. I'd skip that one :)
I have the one set of promising coordinates, do any other math whizzes want to give this a shot before I commit to a hundred mile ride? ;D
Hugs, Devlyn
I think you should go with two sets of coordinates, one derived from 2^99, and the other from 201^99. The two sets of 26 digits are:
63382530011411470074835160
10385076131385329514228303
Only one set is on land, though! ;D
Hugs, Devlyn
Quote from: Devlyn Marie on May 07, 2016, 09:30:24 AM
Only one set is on land, though! ;D
Hugs, Devlyn
LOL! I guess that would make a difference! :)
Quote from: Devlyn Marie on May 07, 2016, 08:31:14 AM
I have the one set of promising coordinates, do any other math whizzes want to give this a shot before I commit to a hundred mile ride? ;D
Hugs, Devlyn
I agree with the solutions posted so far, and my calculator agrees with Kathy's.
In [1]: print `201**99`[:26]
Out[1]: 10385076131385329514228303
Enjoy your ride!
I guess they figure the mathematical incantations were camouflage enough! Truthfully, I think the construction crews in the area relocated it here.
(https://www.susans.org/proxy.php?request=http%3A%2F%2Fi1279.photobucket.com%2Falbums%2Fy537%2FDevlynMarie%2FMobile%2520Uploads%2F20160507_125852_zpsqpfevvb2.jpg&hash=d6cfbd19e6a95d34af97c06d0d02a579e913870c)
Next math problem. If Devlyn uses a $600 GPS to find soggy paper in empty pill bottles, how many rocks does she have in her head?
(https://www.susans.org/proxy.php?request=http%3A%2F%2Fi1279.photobucket.com%2Falbums%2Fy537%2FDevlynMarie%2FMobile%2520Uploads%2F20160507_130121_zpsa3vz2ycr.jpg&hash=90c9c0e847ab1d4ef55651e29650a0bb3c67abf7)
It has begun!
(https://www.susans.org/proxy.php?request=http%3A%2F%2Fi1279.photobucket.com%2Falbums%2Fy537%2FDevlynMarie%2FMobile%2520Uploads%2F20160507_130259_zpscvnf75zm.jpg&hash=fbe13b0999f6df9f51eff7f80f8a3545a3311f0b)
Mucho thanks for the help with this! ;D The Geohound and I had fun finding it and acting nonchalant while people zipped by on their bikes. I filled the jar with toys and trinkets for the next finders.
Hugs, Devlyn
Kewl! Glad to help. Being a math nerd is good for something after all. :)
So what did you use? Did the dandelions only reproduce once? If so, I think the formula should have been:
N=sum (200^t) for t={0,1,2, ... n}
If not, it should have been :
N=1 for t=0
N=(1+200)×200^(t-1) for t>0
A more elegant and commonly used version of the second equation is:
N=201*e^(ln(200)*(t-t0)) so that it follows the form
b*e^r*(t-t0)
Quote from: Newfie on May 15, 2016, 05:12:21 PM
So what did you use? Did the dandelions only reproduce once? If so, I think the formula should have been:
N=sum (200^t) for t={0,1,2, ... n}
If not, it should have been :
N=1 for t=0
N=(1+200)×200^(t-1) for t>0
I tried the formula for reproducing once, coordinates made no sense.
The author of the puzze used this solution :
https://www.susans.org/forums/index.php/topic,208870.msg1851080.html#msg1851080
(and suggested to do it without calculator :) )
hugs
Haha ok, that's not actually the answer but I'm glad you got the goods in the end! That equation would have each plant producing 201 seeds every year instead of 200.
The reason I'm invested in this particular question is that I do population dynamics. This is a well- known equation in my field for exponential growth.
Quote from: Newfie on May 15, 2016, 05:12:21 PM
So what did you use? Did the dandelions only reproduce once? If so, I think the formula should have been:
N=sum (200^t) for t={0,1,2, ... n}
If not, it should have been :
N=1 for t=0
N=(1+200)×200^(t-1) for t>0
A more elegant and commonly used version of the second equation is:
N=201*e^(ln(200)*(t-t0)) so that it follows the form
b*e^r*(t-t0)
Kathy's solution brought me right to it. Some of the other geocachers said it turned out to be easier than they thought, they just needed to find big calculators to do it.
Quote from: KathyLauren on May 04, 2016, 10:08:46 AM
Each year, you have 201 times as many plants as the year before: the 200 new ones for each original one, plus the original ones that are still there. So, in year n, you have 201^(n-1) plants.
In year 100, you have 201^99 plants. That's a really big number. I can't even count the digits because, on my screen, it is wider than the full screen, but it's around 250 digits long. You need to calculate at least the first 26 digits to get the full alphabetic translation required for the coordinates. You don't actually use all 26 letters, but you have to get as far as the letter 'T', which means 20 digits. I tried the calculation in Excel, but it pooped out after 15 digits.
So, no, I can't get you the coordinates.
No calculator, my ***!
Quote from: Newfie on May 15, 2016, 05:33:09 PM
That equation would have each plant producing 201 seeds every year instead of 200.
The formula was 201^(n-1). The 201 comes from the 200 child plants plus the parent plant.
In the first year, you have 1 plant.
In the second year, you have the 200 offspring plus the parent ( = 200+1 = 201)
In the third year, each of the 201 plants from the second year has produced 200 more plants ( = 201 * 200) plus the 201 plants from the second year are still there. So the total in the third year is 201 * 200 + 201 = 201*201.
In the fourth year, each of the 201*201 plants from the third year has produced 200 new plants ( = 201*201*200), plus the 201*201 plants from the third year are still there. The total in the fourth year is therefore 201*201*200 + 201*201 = 201*201*201.
The general formula is therefore 201^(n-1).
Oops, you're right. And I made an additional mistake.
Let me amend my first case to be:
N = e^ln(200)(t-t_0)
Proof:
N_t=e^ln(200)(t-t_0)=(e^ln(200)t)*(e^-ln(200)t_0)
e^-ln(200)t_0 = 1 because t_0 = 0, so in this case the t_0 is really quite redundant. I'll eliminate it for now to make the math more legible. Unfortunately the sub and sup options don't seem to be working for me.
N_(t+1) = e^(ln(200)(t+1)) = (e^ln(200)t)*(e^ln(200)) = 200*e^ln(200) = 200*N_t
This follows the be^rt with b = 1 and r = ln(200). r is the instantaneous rate of change, given by (dN/dt)/y(t)
For the other case, you're absolutely right that I failed to take into account that it was cumulative past the first year. As you showed, the effect of it being cumulative is to increase r by 1, because at any time t the number which we multiply by 200 remains, and therefore the cumulative at time t+1 is Nt * 201.
The end result would be:
Nt = e^ln(201)(t-t_0)
So my boss proposed a fun math problem. Using the information below,
The total land surface area of Earth is about 57,308,738 square miles, of which about 33% is desert and about 24% is mountainous. Subtracting this uninhabitable 57% (32,665,981 mi2) from the total land area leaves 24,642,757 square miles or 15.77 billion acres of habitable land.
and the same dandelion rules as the original equation, with each dandelion covering one square inch, how long until the dandelions populate all the habitable land on earth?
Go! Tick, tock, tick, tock..........
Do you want us to do this without a calculator as well? :P
Let's call this population maximum N_inf, and the time at which we reach N_inf to be t_a.
Then, N_inf = e^(ln(201)t_a
ln(N_inf) = ln(201)t_a
ln(N_inf) / ln(201) = t_a
N_inf = 24642757 sq mi * (5280^2 sq ft / 1 sq mi) * (144 sq in / 1 sq ft) is approximately 9.8928092 * 10^16
=> ln(N_inf) / ln(201) is approximately 7.379. Assuming that the dandelions spawn only once a year, that would mean you need 8 years to cover the total land surface of the earth.
Enjoy!
To be consistent with the first problem, we have to count the year with the single parent dandelion as year 1. The dendelions would cover every square inch in year 9.
Newfie is correct that the elapsed time would be 8 years.