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Wizard
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Wizard
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May 2nd, 2020 at 4:18:00 PM permalink
Good problem. I had seen it before, so stayed out of it. I wouldn't have filed it as an "easy math problem" (the title of the thread).
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
ThatDonGuy
ThatDonGuy
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May 2nd, 2020 at 4:54:32 PM permalink
Quote: Wizard

Good problem. I had seen it before, so stayed out of it. I wouldn't have filed it as an "easy math problem" (the title of the thread).


Well, here's an easy one:

For what positive value p does 1 + 2 p + 3 p2 + 4 p3 + 5 p4 + ... = 25?
Show your work - none of this "I approximated it on Excel" stuff.
rsactuary
rsactuary
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May 2nd, 2020 at 5:19:20 PM permalink
Quote: ThatDonGuy

Well, here's an easy one:

For what positive value p does 1 + 2 p + 3 p2 + 4 p3 + 5 p4 + ... = 25?
Show your work - none of this "I approximated it on Excel" stuff.



Let equation A be 25=1 +2p +3p^2 etc... (the equation as stated)
Let equation B be 25p = p + 2p^2 + 3p^3 etc.

Let C equal A minus B to get 25(1-p) = 1 + p + p^2 + p^3 + p^4 etc

Let D be p*C = p + p^2 + p^3 etc

Then C minus D is 25(1-p)^2 = 1

solve to get p=4/5
Last edited by: rsactuary on May 2, 2020
ThatDonGuy
ThatDonGuy
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May 2nd, 2020 at 5:53:43 PM permalink
Quote: rsactuary

Let equation A be 25=1 +2p +3p^2 etc... (the equation as stated)
Let equation B be 25p = p + 2p^2 + 3p^3 etc.

Let C equal A minus B to get 25(1-p) = 1 + p + p^2 + p^3 + p^4 etc

Let D be p*C = p + p^2 + p^3 etc

Then C minus D = 25(1-p)^2 = 1

solve to get p=4/5


Correct.

Another way to do this is to note that:
1 + 2 p + 3 p2 + 4 p3 + ... = (1 + p + p2 + p3 + ...)2.
rsactuary
rsactuary
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May 2nd, 2020 at 6:00:05 PM permalink
Quote: ThatDonGuy

Correct.

Another way to do this is to note that:
1 + 2 p + 3 p2 + 4 p3 + ... = (1 + p + p2 + p3 + ...)2.



Yeah, it's been too long since I did stuff like this on a regular basis or I would have recognized that. I kind of got there by first principles.
Gialmere
Gialmere
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May 2nd, 2020 at 7:11:00 PM permalink
This one made a splash in India a few years back.

Have you tried 22 tonight? I said 22.
unJon
unJon
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Thanks for this post from:
Gialmere
May 3rd, 2020 at 5:20:49 AM permalink
Quote: Gialmere

This one made a splash in India a few years back.



I count
18
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Gialmere
Gialmere
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May 3rd, 2020 at 11:30:10 AM permalink
Quote: unJon

I count

18


Correct!

Your geometry teacher would be proud.
Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
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May 3rd, 2020 at 12:24:52 PM permalink

You're playing a track and field board game with your family during the lockdown. If each of the runner pawns travels the indicated number of spaces every turn, at which numbered spot will all of the runners end up next to one another?
Have you tried 22 tonight? I said 22.
rsactuary
rsactuary
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Thanks for this post from:
Gialmere
May 3rd, 2020 at 2:42:26 PM permalink
At t=6, they will all be at 19. Again at t=36 etc

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