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Math Riddles

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Post math riddles.

Here's a challenging one:

>Suppose you’re seeding a lawn with grass sprouts. One grass sprout can cover the entire lawn your working on in 30 days. How long would it take two grass sprouts to do the same?
>>
are you talking about stem tillering rate?
because theoretically a long lived perennial wit strong tillering tendencies could cover an entire lawn within a few years from one single seedling, 30 days seems a bit excessive
please clarify further
>>
>>7628786
15 days
>>
>>7642420
Clover is some bullshit
>>
>>7642420
I don't get it, where's the riddle supposed to be? Is there something about grass sprouts that I'm missing here?
>>
Depends on the shape of the lawn and grass growth.

If both are perfectly square then it would still take 30 days, you would need 4 seeds to get 15 days.
>>
>>7642443
>>7642420
>>
>>7642443
How do you need two more to shorten it to 15 days?
>>
>>7642461
Green is sprout.
Each shade of blue is 10 days worth of growth.
With two seeds you can speed up either vertical or horizontal growth but not both, with three you can speed up both but only for half of the lawn, you need 4 to speed up evenly.
>>
>>7642475
Forgot pic
>>
How can you add eight 8s to make 1000?
>>
>>7642507
888+112
>>
>>7642580
>eight 8s
>>
>>7642507

888+88+8+8+8
>>
>>7642438
the riddle is that he's wrong and that intuitive thinking about exponential growth is often wrong,
consider:
1 * 2^0 on the first day for one seed
So after 30 days there is
1* 2^30 = 1 lawn
so with two seeds you're looking for
2^x + 2^x = 1 lawn, solve for x, that is the number of days needed
>>
>>7642929
this poster again, in case you can't follow the logic
the answer is 29 days for 2 seeds to cover the lawn.
this seems almost obvious, right?
>>
>>7642443
>>7642475
confirmed for still in grade 10, just put more seeds directly in the middle and see what happens.
with 1 seed in the middle, the second day there will be 2, the third day there will be 4,
so if you put 4 seeds in on the first day, it's like being in day 3, so it will take 28 days to cover the whole lawn if you put 4 seeds in the first day

source: i'm in grade 11
>>
You win a lottery contest where the prize is $1 the first day, $2 the second day, $3 the third day, etc, so you get $1 more than the previous day forever. However, when you check your bank account, you find your balance is -.08333... cents.

What is the formula used to determine your prize?
>>
>>
1. Is that one sprout for the entire lawn? That's a big fucking grass sprout.

2. It takes 30 days. It's grass, do you think they were going to compete with each other?
>>
>>7643050
It's easy to confuse yourself with this shit but it's quite simple.

All of the ramanujan summation stuff, cutoff and zeta regularization do is look at the smoothed curve at x = 0. What sums usually do is to look at the value as x->inf.

It's just a unique value you can assign to a sum, really they have many such values.

https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#/media/File:Sum1234Summary.svg
>>
>>7642929
>>7642933
>>7642966
That's stupid. What if the seed sprouts 4 more seeds? or 100 seeds? Or 1 seed? Then you get a different amount of days. Not to mention that coverage is ill defined and the way you are solving it ignores density and simply says if you have x amount of sprouts then you have covered the lawn. But clearly coverage implies a certain area of uniform density rather than a certain average density.
>>
>>7643076
Doubling time is a useful concept when dealing with exponential growth.
>>7642933 is assuming that the doubling time is one day.
However the riddle at >>7642391 doesn't absolutely support that the growth is exponential i.e. a certain % per day.
>>
>>7643052
8
>>
If thou art diligent and wise, O stranger, compute the number of cattle of the Sun, who once upon a time grazed on the fields of the Thrinacian isle of Sicily, divided into four herds of different colours, one milk white, another a glossy black, a third yellow and the last dappled. In each herd were bulls, mighty in number according to these proportions: Understand, stranger, that the white bulls were equal to a half and a third of the black together with the whole of the yellow, while the black were equal to the fourth part of the dappled and a fifth, together with, once more, the whole of the yellow. Observe further that the remaining bulls, the dappled, were equal to a sixth part of the white and a seventh, together with all of the yellow. These were the proportions of the cows: The white were precisely equal to the third part and a fourth of the whole herd of the black; while the black were equal to the fourth part once more of the dappled and with it a fifth part, when all, including the bulls, went to pasture together. Now the dappled in four parts were equal in number to a fifth part and a sixth of the yellow herd. Finally the yellow were in number equal to a sixth part and a seventh of the white herd. If thou canst accurately tell, O stranger, the number of cattle of the Sun, giving separately the number of well-fed bulls and again the number of females according to each colour, thou wouldst not be called unskilled or ignorant of numbers, but not yet shalt thou be numbered among the wise.
>>
>>7643434
But come, understand also all these conditions regarding the cattle of the Sun. When the white bulls mingled their number with the black, they stood firm, equal in depth and breadth, and the plains of Thrinacia, stretching far in all ways, were filled with their multitude. Again, when the yellow and the dappled bulls were gathered into one herd they stood in such a manner that their number, beginning from one, grew slowly greater till it completed a triangular figure, there being no bulls of other colours in their midst nor none of them lacking. If thou art able, O stranger, to find out all these things and gather them together in your mind, giving all the relations, thou shalt depart crowned with glory and knowing that thou hast been adjudged perfect in this species of wisdom.
>>
>>7642391
29 days
>>
>>7642391
not enough information. next shit-tier riddle plz
>>
>>7643434
>>7643435
story problems used to be so good...
>>
100 prisoners are given an ultimatum: if they succeed they will all go free and if they fail they will all be executed.

Each prisoner's name is placed in a box so that there is one name per box and the boxes are placed on a table in a room.

The warden leads each prisoner into the room, where they may open exactly 50 of the 100 boxes. The boxes are then closed so that the boxes and room are left exactly in their original condition, and the next prisoner is led into the room. The prisoners may discuss a strategy beforehand, but once the game starts there can be no communication between prisoners.

If every prisoner finds his name, every prisoner is freed. In any prisoner does not find his name, every prisoner is executed.

What strategy can the prisoners follow to maximize their chance of survival?
>>
>>7644483

Will the game end as soon as one prisoner fails?
>>
>>7644542
It doesn't matter.
>>
>>7644483
As there are 100 prisoners and 100 boxes, for any chance for the prisoners to be saved, every box must be opened at some point.

If the prisoners cannot communicate with each other to confirm which boxes contain names that are "found" and thus do not need to be re-opened, I would suggest that each prisoner's openings be staggered such that priz 1 opens boxes 1-50, priz 2 opens 2-51, priz 3 opens 3-52... priz 51 opens 51-100, priz 52 opens 52-100 and 1, priz 53 opens 53-100 and 1-2... until priz 100 opens boxes 100 and 1-49. This means that each box would be opened an optimal 50 times.
>>
>>7644637
That's retarded. Box 1 would only be opened by 1 person. So that's a 1/100 chance. Also, why would #53 only open 53-100? That's not 50 boxes. You're a fucking idiot.

The best way would just be for the first 50 to open the first 50 and the second 50, the second 50. They're fucked anyway because the odds of that are (1/2)^100
>>
>>7644637
It's like optimizing area given a perimeter. Of course the answer is a square.
>>
>>7646231
With the best strategy the chance that all survive is
[math]1 - \sum_{k=51}^{100} \frac{1}{k} \approx 31\%[/math]
Thread posts: 36
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