catmg

Your answer looks very cool, but I wanted to try to see if it can be solved with calculus too because I haven't learned it yet. :))

m = maximum of ab(a-b)

x = a - b (like the hint said)

9 = a + b

a = (9+x)/2

b = (9-x)/2

m = maximum of (9+x)/2*(9-x)/2*x

Note: x has to be be nonegative since if b > a, and a and b are positive integers, ab(a-b) will be negative.

This is a cubic with roots at -9, 0, and 9.

By plugging in points and drawing a sketch, we can get an idea of what the cubic looks like. 

The graph approaches 0 from the positive side from as x approaches -9 from the negative side. 

Then, the graph crosses the x axis at -9, and goes down, but then goes back up to the x axis at 0. 

After that, the graph dips up but returns back to 0 at x = 9, and just continues going down.

So our maximum is at the turning point between 0 and 9, this is the distance m. 

Then we can use the method written here:

https://www.themathdoctors.org/max-and-min-of-a-cubic-without-calculus/

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Thank you Anthrax, I am able to understand the problem now, it took a quite a while to process. :))

Dear Melody,

I'm not sure if this is helpful, but it's always nice to think about it in small cases.

Let's say I have a pair of jokers (j, j) and a pair of ones (1, 1). I randomly mix it up and draw cards until I get a pair, by symmetry, I will draw the pair of ones first 50% of the time, and will draw the pair of jokers first 50% of the time. Notice how this is the same as the probability of having two cards (1, j), and seeing which card I draw first. I will draw the one first 50% of the time, and will draw the joker first 50% of the time.This is because since both cards are pairs(1, 1 and j, j), the probability of which pair you draw first is still the same as is if you were to draw two non-paired cards (1, j).

Hope it makes a bit more sense.

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I'm going to assume that the pairs of jokers are the same, and since the probability of drawing a pair is the same for all pairs, we can simply ignore that fact and simplify it to ordering 7 N’s and 2 J’s randomly. 

I am not sure if this is the correct logic, but it's starting to make a bit more sense to me.

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Aww, thank you.

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Thank you GA.

It seems quite stupid to be mean on a math forum to me. 

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I don't really understand why this question is useless. 

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I think that there may be a small mistake. 

The sum of the first six terms is 70, so

a + ar + ar^2 + ar^3 + ar^4 + ar^5  =  70

54 + ar^3 + ar^4 + ar^5  =  70

ar^3 + ar^4 + ar^5  = 16

I think your method is really smart since by keeping the amount of terms in both expressions constant, you can just divide. :))

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It's nice to see you too. :))

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I don't this so.

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