# Complicated Math Problem. Can Anyone Figure This Out?

**URL:** <https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727>\
**Category:** Factual Questions\
**Created:** [March 28, 2014, 1:18am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727 "2014-03-28T01:18:34Z")\
**Posts on this page:** 20\
**Page:** 1

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**Author:** ![Einstein\_was\_a\_Hippie](https://avatars.discourse-cdn.com/v4/letter/e/e99b99/32.png) [@Einstein\_was\_a\_Hippie](https://boards.straightdope.com/u/Einstein_was_a_Hippie)\
**Post date:** [March 28, 2014, 1:18am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/1 "2014-03-28T01:18:34Z")

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I have every confidence that someone here can figure this out…

Let’s say a cricket farmer has 1 million crickets. They are active, healthy crickets that reproduce at a rate of doubling their population every 8 weeks. These crickets have a lifespan of 12 weeks. 2/3 or 667,000 are juveniles and 1/3 or 333,000 are adults (this is the average ratio of adults to juveniles at any given time). Only the adults lay eggs. Adults are crickets that are 6 weeks old and they lay eggs until they die at 12 weeks. Juveniles under 6 weeks old do not lay eggs. Bear in mind the total population doubles every 8 weeks. How many crickets can this cricket farmer sell each week, and still maintain a 1 million cricket population?

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**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\
**Post date:** [March 28, 2014, 1:44am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/2 "2014-03-28T01:44:40Z")

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Just a guess for fun. 93,781

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**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\
**Post date:** [March 28, 2014, 1:57am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/3 "2014-03-28T01:57:28Z")

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> [@Einstein\_was\_a\_Hippie](#):
>
> Let’s say a cricket farmer has 1 million crickets. They are active, healthy crickets that reproduce at a rate of doubling their population every 8 weeks. These crickets have a lifespan of 12 weeks. 2/3 or 667,000 are juveniles and 1/3 or 333,000 are adults (this is the average ratio of adults to juveniles at any given time). Only the adults lay eggs. Adults are crickets that are 6 weeks old and they lay eggs until they die at 12 weeks. Juveniles under 6 weeks old do not lay eggs. Bear in mind the total population doubles every 8 weeks. How many crickets can this cricket farmer sell each week, and still maintain a 1 million cricket population?

If a cricket spend the first half of its life as a juvenile and the second half of its life as an adult, why do juveniles outnumber adults two to one? Do half the juveniles die off at six weeks?

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**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\
**Post date:** [March 28, 2014, 1:58am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/4 "2014-03-28T01:58:48Z")

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Change my answer

333,000 adults=1,000,000 crickets in 8 weeks  
333,000 adults = 125,000 crickets in 1 week.  
25% adults die=82,500  
surplus crickets = 125,000-82,500=42,500

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**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\
**Post date:** [March 28, 2014, 2:10am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/5 "2014-03-28T02:10:13Z")

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Screwed up again, 1/6 die every week =55,500  
125,000-55,500=69,500 for sale  
I give up if I find any more screw ups.

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**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\
**Post date:** [March 28, 2014, 2:19am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/6 "2014-03-28T02:19:41Z")

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The 2/3 juvenile number doesn’t seem stable.

If the population doubles every 8 weeks, then on a daily basis there should be a 1.26% increase in population (x = 2^(1/56) = 1.0126). So, that means that after a day we have 12,600 new juveniles.

But if 666,667 of the total are juveniles, and they have an even age distribution, and they turn adult after 6 weeks, then after a day we’ll lose 15,900 juveniles to adulthood. It seems like there should really be 42\*12,600 total juveniles, or 52.9%.

At any rate, it doesn’t change the answer. If the crickets are collected randomly (so as not to change the ratio), then after a week there are 90,507 new crickets to collect.

If they’re collected daily, then you only get 88,200 (and slightly less if you collect “continuously”).

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**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\
**Post date:** [March 28, 2014, 2:29am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/7 "2014-03-28T02:29:25Z")

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Oops–I ignored death, though I think it’s still unstable (in the other direction). After a day, 7940 out of those 333,333 adults will die. So, to maintain 1.25% daily growth, you actually need 20,400 new juveniles, but again you are losing 15,900 to adulthood.

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**Author:** ![Jragon](https://avatars.discourse-cdn.com/v4/letter/j/e19b73/32.png) [@Jragon](https://boards.straightdope.com/u/Jragon)\
**Post date:** [March 28, 2014, 2:37am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/8 "2014-03-28T02:37:23Z")

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The problem is that we don’t know how many crickets are poised to die and how many are poised to mature at the beginning.

Either way, this problem is going to be really unstable for a while until you get into an equilibrium of crickets relatively uniformly growing up and dying. At this point you should hit an equilibrium which is contradictory to the given information (barring additional info about juvenile mortality rate).

At this point the “doubles every 8 weeks” number can no longer be trusted, since now we have a different number of adults and we’re **given** that only adults may lay eggs. It all comes back to the initial conditions. The only thing you can do is make assumptions about the initial conditions, and calculate the average eggs laid per week per adult and then project that onto the time when the population becomes stable.

After that it’s an easy problem, but at this point it’s severely underconstrained.

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**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\
**Post date:** [March 28, 2014, 2:42am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/9 "2014-03-28T02:42:56Z")

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I think the correct ratio is 60.4% juveniles.

At the start of the day, we have:  
604,000 juveniles  
396,000 adults

During the course of a day:  
604,000 / 42 = 14,381 juveniles turn to adults  
396,000 / 42 = 9,429 adults die  
21,883 new juveniles are born as replacements

So at the end of the day, we have:  
604,000 + 21,883 - 14,381 = 611,502 juveniles  
396,000 + 14,381 - 9,429 = 400,952 adults  
611,502 + 400,952 = 1,012,454 total

That’s the right amount for an 8-week doubling, and the 60.4% ratio is maintained.

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**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\
**Post date:** [March 28, 2014, 2:46am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/10 "2014-03-28T02:46:56Z")

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> [@Jragon](#):
>
> After that it’s an easy problem, but at this point it’s severely underconstrained.

Or overconstrained, depending on how you look at it. If you assume a stable population distribution, and that the population doubles every 8 weeks (i.e., adults are _defined_ to reproduce at the rate needed to double that often), then the number of juveniles is wrong–but one can fix that and get a reasonable answer.

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**Author:** ![Einstein\_was\_a\_Hippie](https://avatars.discourse-cdn.com/v4/letter/e/e99b99/32.png) [@Einstein\_was\_a\_Hippie](https://boards.straightdope.com/u/Einstein_was_a_Hippie)\
**Post date:** [March 28, 2014, 3:07am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/11 "2014-03-28T03:07:11Z")

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> [@Dr.Strangelove](#):
>
> The 2/3 juvenile number doesn’t seem stable.

> [@Dr.Strangelove](#):
>
> I think the correct ratio is 60.4% juveniles.

Consider then that the number of juveniles versus the number of adults is simply an estimate, and that your 60.4% number is probably good. Seems about right.

> [@Jragon](#):
>
> The problem is that we don’t know how many crickets are poised to die and how many are poised to mature at the beginning.

Yes, I understand, you need to know how many 11 week olds there are and then how many 10 week olds…this is a problem. But if you go backwards using Dr Strangelove’s 60.4% figure on adults, and consider all other factors involved in the problem, shouldn’t you be able to get a rough determination of how many are in each age category (ie 11 weeks, 10 weeks, 9 weeks old etc)?

You can see what I mean by “complicated”…

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**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\
**Post date:** [March 28, 2014, 1:09pm UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/12 "2014-03-28T13:09:29Z")

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How old are the one you sell? I think you can sell more if you sell them the day before they die and fewer if you sell only new adults. More detail is needed.

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**Author:** ![puddleglum](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/puddleglum/32/137_2.png) [@puddleglum](https://boards.straightdope.com/u/puddleglum)\
**Post date:** [March 28, 2014, 1:19pm UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/13 "2014-03-28T13:19:42Z")

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What does the first sentence mean? Does the population double every 8 weeks or does each cricket produce 2.5 offspring is his lifetime? If the population does double every 8 weeks then the number of dying crickets makes no difference and the answer is 125,000. The second scenario is more complex and takes more math than I am willing to undertake.

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**Author:** ![Darth\_Panda](https://avatars.discourse-cdn.com/v4/letter/d/ee7513/32.png) [@Darth\_Panda](https://boards.straightdope.com/u/Darth_Panda)\
**Post date:** [March 28, 2014, 1:25pm UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/14 "2014-03-28T13:25:43Z")

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> [@Hari\_Seldon](#):
>
> I think you can sell more if you sell them the day before they die

Did you used to work at PetSmart?

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**Author:** ![74westy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/74westy/32/3950_2.png) [@74westy](https://boards.straightdope.com/u/74westy)\
**Post date:** [March 28, 2014, 3:25pm UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/15 "2014-03-28T15:25:46Z")

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> [@Hari\_Seldon](#):
>
> How old are the one you sell? I think you can sell more if you sell them the day before they die and fewer if you sell only new adults. More detail is needed.

I wish to register a complaint!

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**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\
**Post date:** [March 28, 2014, 3:33pm UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/16 "2014-03-28T15:33:48Z")

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> [@Little\_Nemo](#):
>
> If a cricket spend the first half of its life as a juvenile and the second half of its life as an adult, why do juveniles outnumber adults two to one? Do half the juveniles die off at six weeks?

You’re thinking that the juveniles should be equal in number to the adults. But that applies only if the overall population is stable - if there’s a permanent population boom then the younger generation will always outnumber the older. (If you have one mated pair of crickets then as soon as you have four offspring your juvies are outnumbering your adults…)

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**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\
**Post date:** [March 28, 2014, 5:12pm UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/17 "2014-03-28T17:12:52Z")

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> [@Einstein\_was\_a\_Hippie](#):
>
> Bear in mind the total population doubles every 8 weeks. How many crickets can this cricket farmer sell each week, and still maintain a 1 million cricket population?

Does any of the other stuff matter? You haven’t specified whether the farmer only sells adults or just sells a random selection from the entire population. I’ll assume the latter.

If the population doubles every 8 weeks, then that means it grows by 9.0508% per week. It doesn’t matter how many are juveniles or adults, or how many die. The population, if left unfettered, grows by 9.0508% per week.

Because:  
(1+x/100)^8 = 2

Where X = the increase in cricket population per week, expressed as a percent of the starting population

That means if you start with 1M living crickets, and wait one week, you’ll have 90,508 living crickets you can sell, and then you’re back to 1M living crickets. One week later, you again have 90,508 living crickets you can spare to sell.

So my answer is that if he started with 1M live crickets, he can sell 90,508 live crickets every week while maintaining a population of 1M.

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**Author:** ![Senegoid](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/senegoid/32/6606_2.png) [@Senegoid](https://boards.straightdope.com/u/Senegoid)\
**Post date:** [March 29, 2014, 8:36am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/18 "2014-03-29T08:36:06Z")

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Is this a Common Core problem?

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**Author:** ![brickbacon](https://avatars.discourse-cdn.com/v4/letter/b/898d66/32.png) [@brickbacon](https://boards.straightdope.com/u/brickbacon)\
**Post date:** [March 29, 2014, 8:41am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/19 "2014-03-29T08:41:51Z")

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Is this a real life problem? If so, I was always under the impression that things like this were done by weight. Since most people aren’t gonna bother to count thousands of crickets, they would just take the average weight, then multiple. Using that logic, after separating out the dead crickets, you would just sell the weight overage on a daily basis, not having to worry about the math not accounting for real world variance.

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**Author:** ![Einstein\_was\_a\_Hippie](https://avatars.discourse-cdn.com/v4/letter/e/e99b99/32.png) [@Einstein\_was\_a\_Hippie](https://boards.straightdope.com/u/Einstein_was_a_Hippie)\
**Post date:** [March 30, 2014, 1:30am UTC](https://boards.straightdope.com/t/complicated-math-problem-can-anyone-figure-this-out/684727/20 "2014-03-30T01:30:45Z")

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> [@brickbacon](#):
>
> Is this a real life problem? If so, I was always under the impression that things like this were done by weight. Since most people aren’t gonna bother to count thousands of crickets, they would just take the average weight, then multiple. Using that logic, after separating out the dead crickets, you would just sell the weight overage on a daily basis, not having to worry about the math not accounting for real world variance.

No, they sell by the each, not by weight.

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