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Cost of War


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61 replies to this topic

#41 I.C.U.P (i see you pee)

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Posted 04 March 2008 - 01:53 AM

100-70=30
100-75=25
100-85=15
100-80=20 100-90=10 (100-30,25,15,20= 10)
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#42 Naruto Uzumaki

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Posted 05 March 2008 - 05:41 PM

Nevermind. I'm retarded. I'm calculating 75% as 25% and I don't know why. PLEASE DISREGARD.

I'd edit it but the edit feature isn't working properly for some reason.

its okay i still believe your a smart person :)
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#43 IThinkImMe

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Posted 16 March 2008 - 01:11 AM

Wouldn't the minimum number of soldiers that lost all 4 limbs be 1?
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#44 JohnSantaFe

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Posted 26 March 2008 - 01:28 AM

To me an interesting problem is to create a generic equation to solve any problem like this.

I think this is the answer but would be interested for someone to confirm.

Let:

T = total number of soldiers
N = number of different types of injuries
X[i] = number of soldiers injured for each injury (index i from 1 to N)
Z = minimum number of soldiers with all N injuries


Z = (T - (N)(T)) + summation i=1 to N (X[i])


for this problem

Z = (100 - (4)(100)) + 70 + 75 + 80 + 85 = 10
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#45 hopstop

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Posted 27 March 2008 - 09:53 AM

Indeed, however, gangreen could claim more body parts.


this is only a puzzle, just stick with what it gave you, and dont think about all the other possibilities of gangreen etc
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#46 Dissident420

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Posted 05 April 2008 - 04:44 AM

If you want to get into semantics,

I think that the answer is 70 because the question is not well stated enough for the answer to be logically 10.
Maybe if it was 100 'out of' the answer would be 10.
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#47 jitHU

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Posted 08 April 2008 - 11:26 AM

its 10 no duh!!
:blush: :angry:
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#48 DrakenFrost

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Posted 22 April 2008 - 05:16 PM

if you name all 100 solders first one being named 1 and so on this is easy
Spoiler for my salution

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#49 iSlowmotion

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Posted 25 April 2008 - 03:22 PM

This is super easy. If the least common injury totals 70, it cant exceed that number because if only 70 had injury A and B,C, and D are more than that, the average of those becomes 77.5 but bearing in mind only 70 had A and 75 had B, it can only be the least.
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#50 Naruto Uzumaki

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Posted 29 April 2008 - 03:57 PM

This is super easy. If the least common injury totals 70, it cant exceed that number because if only 70 had injury A and B,C, and D are more than that, the average of those becomes 77.5 but bearing in mind only 70 had A and 75 had B, it can only be the least.

?????????????? :huh: :huh: :huh: :huh: :huh:
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