Old Guy Needs Help With Math Problem! Non Cube Related

[kbrune]

I'm looking for some help from anyone who possesses math skills!

I have a scenario that I need an answer to. What I'm looking for is an equation that I can use to give me the answer to a question that
I would be able to change a few of the variables to if I needed to.

The scenario:

One student starts a club. Well say a cubing club! He challenges himself to find one person per month to join the club for 12 months.
To grow the club to a larger number of people, he tells each person that he has found that they must find 1 person every 4 months to also join the club.

So what I would like to know is how to calculate how many people are in the club after 12 months and after 24 months. I'm hoping to get an equation that would enable to change the frequency in which the organizer finds new recruits, the frequency in which the recruits find new people, and the amount of people in the club after a certain number of months.

Please forgive me if any of this seems off. I haven't done any formal math since highschool and I am now 33 yrs old!!

Thank you to anyone who can help!

 

[TheNextFeliks]

Well let's see.

Where m is the number of months and S is the number of total students, I think this would work.

Spoiler: Reverse Engineering

After 12 months, there would 33
After 8, 17
After 4, 9

Hmmm.

\( S=2^((m/4)+2)+1 \)

That works.

 

[CheesecakeCuber]

This question seems a bit flawed. Can we assume that he only finds one member every month for 12 months? And what if a new member joins in December, would his 4 months overflow into the next year?

Anyway, here's my stab at it. Assuming that the leader finds one member every month for 12 months, the leader starts in January, each new member finds another recruit in the exact time allotted (e.g. 4 months) then I believe this werks, or doesn't:

Spoiler

Let t = answer
m = goal number of people leader must find every month
x = number of months new members have to find someone new
a = number of months leader must find someone to join

t = a(m) + a(m)/x

Okso, in your problem you said the leader has 12 months and finds 1 person every month: a = 12, m = 1
Also, you state that every new member must find another person in exactly 4 months: x = 4

Solution: t = 12(1) + 12(1)/4
t = 12 + 12/4
t = 12 + 3
t = 15 people in a months (12 months in this case)
To find members in 24 months you can make a = 24 or just multiply t, where a = 12, by 2
t = 15(2)
t = 30 people in 24 months

Geez, we need ben to come back here and render us all stupid...

 

[Kelsparov]

Solution: t = 12(1) + 12(1)/4
t = 12 + 12/4
t = 12 + 3
t = 15 people in a months (12 months in this case)

Are you sure? When the leader recruits someone, the first 7/8 can definitely recruit someone (possibly the 9th can as well because of the ambiguity in the question), so ~8 of those recruited by the leader can recruit at least one new member, so the answer must be over 12 + 8 = 20.

Am I missing something? I don't really like solving questions about months and years because of ambiguities :S

 

[CheesecakeCuber]

Am I missing something? I don't really like solving questions about months and years because of ambiguities :S

Same here, looking back at my solution, I can see flaws, but it's so difficult to answer this question when there are barely any details given. Like I mentioned above, what happens if a member joins in December, would his allotted time overflow into the next year? But then that wouldn't be the amount of people in 12 months. And also, lol, how do we know that each person takes exactly 4 months to find a new recruit? Right?

 

[Kelsparov]

This might be closer to the solution the OP is looking for:

Leader recruits 12 members (minions)

Every minion can recruit at least one new member because we must assume that the minions recruit a new member as soon as the leader recruits them.

Some minions can recruit 2 members (noobs), others will recruit one; the minion recruited in September can only recruit one noob (September, October (1), November (2), December (3), January -- 4), so the means the September-October-November-December minions can only recruit one. The rest can recruit 2, (not 3 -> Jan + 5 = May, May + 5 = Sept, Sept + 5 = no solution).

When a minion recruits two noobs, then that is a group of 3, (minion + 2 noobs), and when a minion recruits 1 noob, that is a group of 2.

There are 4 groups of 2, 8 groups of 3, and the leader.

(4 x 2) + (8 x 3) + 1 = 33 (which is the same figure as TheNextFeliks calculated, so his solution & formula are probably correct

 

[CheesecakeCuber]

Ah, I see my mistake. I incorrectly read that each new member could only recruit one more person. (Basically after four months, they were done recruiting)

 

[blokpoi]

k i asked ben and his results:

0months=1
4months=1
8months=1
12months=2
16months=3
20months=5
24months=10
28months=19
32months=37
36months=74
etc

he solved it into a recurrence relation
f(x)=2f(x-4)-1+floor(|cos(xpi/12)|), f(0)=1

 

[Kelsparov]

k i asked ben and his results:

0months=1
4months=1
8months=1
12months=2
16months=3
20months=5
24months=10
28months=19
32months=37
36months=74
etc

he solved it into a recurrence relation
f(x)=2f(x-4)-1+floor(|cos(xpi/12)|), f(0)=1

12months=2 WHAT?