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JEBZ Komics
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« Reply #1680 on: September 30, 2020, 09:29:29 PM »

Yoooo teach me your secrets
Brother E: Lol! Grin

Actually I have four legs. Just bend over and touch the ground. Vola! Four legs! Smiley
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« Reply #1681 on: September 30, 2020, 09:39:48 PM »

Brother E: Lol! Grin

Actually I have four legs. Just bend over and touch the ground. Vola! Four legs! Smiley

Power balls have 0 legs. Pensive
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JEBZ Komics
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« Reply #1682 on: September 30, 2020, 10:51:49 PM »

Power balls have 0 legs. Pensive
Brother E: True  Hmm
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Kismala
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« Reply #1683 on: September 30, 2020, 11:41:15 PM »

There's in total 6 users active that joined in 2012: Rize, Nickito, Theo, Plankton5165, Sag and AskJoe.
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Weird Quotes:
perri should bomb Koilee lake
Matt and clover became dead after playing Sushiria with SoundCloud on the car
Kookie [Jul 12 12:28 AM]:   Next time I will fill your mouth with oatmeal and raisins. What do you think about that?
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« Reply #1684 on: September 30, 2020, 11:50:54 PM »

Papa's Cheeseria (PC) came out the day after my birthday!
Cars and Animal Crossing: New Leaf came out ON my birthday!!!
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♥ ~ My Important Topics ~ ♥

♥ ~ My Social Media ~ ♥
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My Twitter: https://twitter.com/Shawna92272306
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My Favorite Customers:
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JEBZ Komics
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« Reply #1685 on: October 10, 2020, 06:53:38 PM »

Brother Z: Today is 10/10/20, and 10+10=20.
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-Phil-
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« Reply #1686 on: October 29, 2020, 07:02:46 PM »

Customers that have lost in the Customer Elimination Contest with double digit votes against them:

- Connor (10, Placed 124th.)
- Wylan B (10, Placed 121st.)
- Janana (10, Placed 120th.)
- Bertha (13, Placed 119th.)
- Emmlette (11, Placed 107th.)
- Quinn (10, Placed 98th.)
- Iggy (11, Placed 97th.)
- Sienna (11, Placed 93th. Allan also reached 10 this round.)
- Foodini (10, Placed 92nd.)
- Steven (13, Placed 82nd.)
- Wally (10, Placed 79th.)
« Last Edit: October 30, 2020, 02:51:54 PM by -Phil- » Logged

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« Reply #1687 on: October 29, 2020, 08:30:02 PM »

Customers that have lost in the Customer Elimination Contest with double digit votes against them:

- Connor (10, Placed 124th.)
- Wylan B (10, Placed 121st.)
- Janana (10, Placed 120th.)
- Bertha (13, Placed 119th.)
- Emmlette (11, Placed 107th.)
- Quinn (10, Placed 98th.)
- Iggy (11, Placed 97th.)
- Sienna (11, Placed 93th. Allan also reached 10 this round.)
- Foodini (10, Placed 92nd.)
- Steven (Currently 12, Placed 82nd.)
This is just wrong
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AstroSmokey
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« Reply #1688 on: October 30, 2020, 01:59:51 PM »

Borat's middle name is Margaret
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-Phil-
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« Reply #1689 on: October 30, 2020, 02:52:56 PM »

Forgot to mention I'll update this each time someone hits 10+ votes there. Also rip Wally who has now been added. Pensive

Customers that have lost in the Customer Elimination Contest with double digit votes against them:

- Connor (10, Placed 124th.)
- Wylan B (10, Placed 121st.)
- Janana (10, Placed 120th.)
- Bertha (13, Placed 119th.)
- Emmlette (11, Placed 107th.)
- Quinn (10, Placed 98th.)
- Iggy (11, Placed 97th.)
- Sienna (11, Placed 93th. Allan also reached 10 this round.)
- Foodini (10, Placed 92nd.)
- Steven (13, Placed 82nd.)
- Wally (10, Placed 79th.)
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AskJoe
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« Reply #1690 on: Today at 03:01:03 PM »

The product of 4 consecutive positive numbers is always 1 less than a perfect square. Examples:
1*2*3*4 = 5^2 -1
2*3*4*5 = 11^2 - 1
3*4*5*6 = 19^2 - 1
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« Reply #1691 on: Today at 03:33:27 PM »

 
The product of 4 consecutive positive numbers is always 1 less than a perfect square. Examples:
1*2*3*4 = 5^2 -1
2*3*4*5 = 11^2 - 1
3*4*5*6 = 19^2 - 1
Actually, in trying to prove it directly, I found that it works for all integers, not just positive numbers.

Well, I found that the general form is:
x^4 + 6x^3 + 11x^2 + 6x = y^2 -1, where x is any integer and y is any whole number.

Doing the steps:
sqrt(x^4 + 6x^3 + 11x^2 + 6x + 1) is what we're trying to find.

Assuming that equation has an correct polynomial solution, we're looking for a polynomial in the following fashion:
(ax^2 + bx +c)^2, because of a degree four polynomial.

Because of the coefficients of the first and last terms, we have verified that our square root polynomial must be in the form:

x^2 + ax + 1

Squaring that then solving, we find that our square root polynomial is:

x^2 + 3x + 1.

Any polynomial is an integer for all integers.

Therefore, we can definitively say that for all integers x and whole numbers y,

x(x + 1)(x + 2)(x + 3) = y^2 - 1, or:

The product of 4 consecutive integers is always 1 less than a perfect square.
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« Reply #1692 on: Today at 04:17:17 PM »

Here is an alternative method which doesn't require factoring a fourth degree polynomial. We start with n(n+1)(n+2)(n+3), and since the order of the factors don't matter we can say that
n(n+1)(n+2)(n+3) = n(n+3)(n+1)(n+2).
Focusing on the right hand side we multiply the first and second factors as well as the third and fourth to get
n(n+1)(n+2)(n+3) = (n^2+3n)(n^2+3n+2).
Which can be rewritten as
n(n+1)(n+2)(n+3) = [(n^2+3n+1)-1][(n^2+3n+1)+1].
This is a difference of squares, hence
n(n+1)(n+2)(n+3) = (n^2+3n+1)^2-1.
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