Make a note of the last digit of your original number
See the last digit of the new number
They're the same
OOOOOHHHHHHHH
--
senorhousemouse
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WhoopsyDaisy 04/18/12 4:09:00 AM #1: |
Make a note of the last digit of your original number
See the last digit of the new number They're the same OOOOOHHHHHHHH -- senorhousemouse ... Copied to Clipboard!
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Psycho_Kenshin 04/18/12 4:10:00 AM #2: |
I just came
-- One Piece: Pirates with style! -= Metal Gear Solid: Tactical Espionage Action =- ... Copied to Clipboard!
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pjbasis 04/18/12 4:11:00 AM #3: |
HOW IS THIS POSSIBLE
-- http://i498.photobucket.com/albums/rr345/Rakaputra/pjbasis.png SuperNiceDog - 1, pjbasis - 0 ... Copied to Clipboard!
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JDTAY 04/18/12 6:19:00 AM #4: |
Eoin posted...
Works with ninth powers as well. WHOAMG mind kasploded -- PSN: JDTAY87 ... Copied to Clipboard!
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Eoin 04/18/12 6:19:00 AM #5: |
Works with ninth powers as well (and every 4 powers, such as 13th power, 17th power, etc.).
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azuarc 04/18/12 7:31:00 AM #6: |
Considering that a lot of digits work on this principle on any power or on any odd-numbered power, this isn't really that amazing.
xxx1^anything => xxx1 xxx2^5 = xxx2 xxx3^5 = xxx3 xxx4^3 = xxx4 xxx5^anything => xxx5 xxx6^anything => xxx6 xxx7^anything => xxx7 xxx8^5 => xxx8 xxx9^3 => xxx9 xxx0^anything => xxx0 There's a very simple reason for this, too. Even + Even = Even. Odd * Odd = Odd. Therefore you only have five choices for last digit. And numbers ending in 5 or 0 are strictly multiples of 5, so any number not already a multiple of 5 can't become one by being raised to a power. So really, you only have 4 choices. So you get a cycle of 1, 2, or 4 of these end choices. (You don't get 3 because 3 isn't a factor of 4. Explaining why that's important is a little trickier.) ... Copied to Clipboard!
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