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03-23-2015, 04:06 PM | #16 |
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03-23-2015, 04:06 PM | #17 |
Most Valuable Villain
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03-23-2015, 04:06 PM | #18 |
Kind of a mod
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03-23-2015, 04:07 PM | #19 |
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03-23-2015, 04:09 PM | #20 |
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It should be twelve. Order of operations makes (1x0) = 0 and there are twelve other ones.
edit: oh I see - 1+1+1+1+11+1+1+1+11+(1x0)+1 bleh. |
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03-23-2015, 04:09 PM | #21 |
Stroking to the SB Champs!
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What did I win?
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03-23-2015, 04:09 PM | #22 |
GBM 8-12-15
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03-23-2015, 04:10 PM | #23 | |
It's just a ride.
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Quote:
1+1+1+1+11+1+1+1+11+1x0+1 = ? or 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 x 0 + 1 = ?
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12·12·11 回 RIP Turd Haley |
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03-23-2015, 04:10 PM | #24 |
Kind of a mod
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03-23-2015, 04:22 PM | #25 |
error 404
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A really shitty written PEMDAS problem is the correct answer.
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It's Easier to Fool People Than It Is to Convince Them That They Have Been Fooled. Mark Twain. |
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03-23-2015, 04:31 PM | #26 |
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This is all about a PEDMAS problem, just from a computer's perspective. That's why I put the disclaimer in the OP.
As far as I know (the languages I use) all non-arithmetic characters are ignored (or puked) in an arithmetic equation. None of the characters in this test would be puked. |
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03-23-2015, 04:34 PM | #27 |
Ain't no relax!
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40.
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03-23-2015, 04:36 PM | #28 | |
Kind of a mod
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Quote:
x = 1 + 1 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1 1 + 1 x 0 + 1 The first line would result in x =5 and the rest would fail syntax check. I know it'd work in PHP and Javascript, though (which I pretend to muddle my way through from time to time). |
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03-23-2015, 04:38 PM | #29 |
Gimme My Berries Back!
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13
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03-23-2015, 04:38 PM |
beach tribe |
This message has been deleted by beach tribe.
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03-23-2015, 04:41 PM | #30 |
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1x0 goes first and yield 0. Then add the rest and you should get 12.
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