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Thread: fun math problem
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2014-09-10, 10:01 AM (ISO 8601)
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Re: fun math problem
Ignoring the division by n for now, we can see that the two last terms of the formula are always canceling the first and last items of the expansion: - 2nxn /n and - (-1)nyn.
So, each of the two "side ones" in the Pascal triangle disappears in the end. (The right one would have been 1/n, so always a fraction, capping n at 1 if it stayed there.)
So, the triangle:
First line (2)1 (-1)1
Second line 221 21(-1)12 (-1)21
Third line 231 22(-1)13 21(-1)23 (-1)21
We can see we can easily start ignoring the minus ones regardless of the power as they have no bearing on the fractionality of the coefficient.
Fourth line 241 234 226 214 201
(Bringing back the division by n that we left off the screen for clarity, we can easily verify at this stage that the terms that matter on the fourth line are 32/4 (integer), 24/4 (integer), 8/4 (integer). So, all integers.)
On the nth line of the triangle, the rth coefficient is 2n-r times the rth number of the nth line of the Pascal triangle which is n!/(r!(n-r)!). Divided by n, of course.
i.e.
2n-r(n-1)!
----------
r!(n-r)!
So, the problem becomes, this, above, has to be an integer for all values of r from 1 to n-1 (we can ignore the first and last terms as they're cancelled in the formula).
2n-r is always an integer so it doesn't bring anything more to the table, so we can ignore it.
I am pretty sure there's no upper limit to n satisfying the condition of this quotient being an integer:
(n-1)!
-------
r!(n-r)!
So, at first sight, no upper limit to n.
Take everything I've written with a huge grain of salt, it's been well over a decade since I last did such things and found out I was extremely rusty.
Edit: It's obvious that Jay R knows what he's talking about!! You're right, let's keep (n)!/(r!(n-r)!) and the division by n separate.
Problem reduces itself to, for the n you pick, for all possible values of r from 1 to n-1, (n)!/(r!(n-r)!) has to be divisible by n. Correct?Last edited by lio45; 2014-09-10 at 10:12 AM.
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2014-09-10, 10:08 AM (ISO 8601)
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2014-09-10, 10:14 AM (ISO 8601)
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Re: fun math problem
Offer good while supplies last. Two to a customer. Each item sold separately. Batteries not included. Mileage may vary. All sales are final. Allow six weeks for delivery. Some items not available. Some assembly required. Some restrictions may apply. All entries become our property. Employees not eligible. Entry fees not refundable. Local restrictions apply. Void where prohibited. Except in Indiana.
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2014-09-10, 10:17 AM (ISO 8601)
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2014-09-10, 10:23 AM (ISO 8601)
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Re: fun math problem
I reserve the right to be wrong and will use that right whenever it happens
78% of DM's started their first campaign in a tavern. If you're one of the 22% that didn't, copy and paste this into your signature.
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2014-09-12, 01:04 AM (ISO 8601)
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Re: fun math problem
All right. This whole thread is bugging me.
I hold two BSci degrees in Maths and Physics. I'm the math nerd among all of my friends who are math nerds. I've learned to trust my gut when it comes to math and math-people, and my gut's telling me, right now, that this is either somebody's homework or a random solved problem pulled off the internet.
But if it's not, and this is the part I keep coming back to, then why post it here? If anything, we have a forum specifically devoted to science and technology, a forum where the problem might be better presented as an object for serious mutual discourse and not a cryptic, unsolicited challenge to weed out those who "get it" and those who don't.To see the World in a Grain of Sand
And a Heaven in a Wild Flower
Hold Infinity in the palm of your hand
And Eternity in an hour
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2014-09-12, 09:07 AM (ISO 8601)
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Re: fun math problem
I reserve the right to be wrong and will use that right whenever it happens
78% of DM's started their first campaign in a tavern. If you're one of the 22% that didn't, copy and paste this into your signature.
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2014-09-12, 01:29 PM (ISO 8601)
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2014-09-12, 03:30 PM (ISO 8601)
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