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Cost of War


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#21 newton

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Posted 27 August 2007 - 02:36 PM

If 100 soldiers had 3 injuries, and 10 soldiers had 4 injuries, wouldn't that make
3*100 + 4*10 = 340 injuries?
This does not seem right.

Steve



There are 100 soldiers total, meaning that 90 have 3 injuries and 10 have 4 injuries.

You are double counting. Your math suggests that you have 100 soldiers with 3 injuries (and, yes, a soldier with 4 injuries does have 3), BUT then you say that there are an additional 10 soldiers, who each have 4 injuries.

You can either say there are 90 soldiers with exactly 10 injuries, and 10 with exactly 4, or say there are 100 soldiers with at least 1, 2, and 3 injuries, and an additional 10 soldiers with an least 4 injuries. Thus:

3*90 + 4*10 = 310 injuries
100 + 100 + 100 + 10 = 310 injuries
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#22 RedSox2233

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Posted 01 September 2007 - 06:15 AM

not really sure why everyone is trying to use math for the answer....but..the obvious answer is 70!!! 70 is the smallest number of anything lost everything else had a higher number...!!
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#23 unreality

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Posted 01 September 2007 - 11:53 PM

You are mistaken. If you don't understand, look at my proof posted earlier;

70 soldiers lost an eye... so 30 DID NOT lose an eye

75 lost an ear. 25 DID NOT lose an ear

85 lost a leg. 15 DID NOT

80 lost an arm. 20 DID NOT

To lose all 4, you can't "DID NOT" any of those...

so 30+25+15+20 = 90... that means 90 people out of a 100 AT LEAST lost 1 body part.

leaving 10 out of 100 to have lost ALL FOUR
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#24 oranfry

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Posted 11 September 2007 - 11:13 PM

Here's how I got the answer. Hopefully this will convince some people who aren't willing to get bogged down in math!

Start with an empty space of size 100, one place for each soldier.
Add each injury group to the space one by one, all the while trying to avoid the current intersection of all sets added so far:

The empty space:
|----------------------------------------------------------------------------------------------------|

Add the eye-losers (70):
|[----------------------------------------------------------------------]------------------------------|

Add the ear-losers (75), overlapping with eye-losers as little as possible:
|[-------------------------[---------------------------------------------]------------------------------]|

Add the leg-losers (85), overlapping with ear-and-eye-losers as little as possible. This time it is necesarry to 'split up' the group to minimise the overlap with the current intersection.
|[[-------------------------[------------------------------]---------------][------------------------------]]|


Add the arm-losers (80), overlapping with ear-eye-and-leg-losers as little as possible. Again it is necesarry to 'split up' the group.
|[[[-------------------------[----------]--------------------][---------------][------------------------------]]]|
0

The final intersection (marked in pink) of all four injury groups is, as you can see, 10 soldiers in size. Since we have been keeping the intersection as small as possible the whole way, we can be sure this is the true minumum number of soldiers who fall into all four groups.

There are many other slightly different ways I could have filled in the space and kept the intersection minimal, but they are all logically equivalent and we would still have gotten the same answer.
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#25 BoilingOil

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Posted 29 September 2007 - 12:55 AM

This one is so nicely simple that people might believe it's too simple and start looking for hidden traps

The solution: 30 did not lose an eye, 25 lost none of their ears, 15 lost no leg, and 20 still have both arms.
If you add those together, you'll have the maximum number of people who did NOT suffer 4 injuries. This number is 90.
So a minimum of 10 people must have lost an eye, an ear, a leg AND an arm.

Interesting to note is, that in this case the other 90 MUST ALL have lost 3 items... This division is rendered completely useless, from a warfare point of view.
Although it's a morbid way of looking at it, the more people would suffer all four injuries (max 70), the less would be too handicapped to ever fight again. That would at least leave you a few man to continue the fight with.

There was once an army general who uttered: my men should die in battle, that's what they're trained for.
Someone hearing this comment then replied: then this must be not their lucky day!


Enough of this morbid talk.

BoilingOil
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#26 credels

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Posted 16 October 2007 - 10:15 PM

But no-one has mentioned the over time factor. Thats when an amputation leads to another loss of limb, which happened alot in pre-mid 20th century wars. They all died, except for, maybe 1 or 2.
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#27 rookie1ja

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Posted 16 October 2007 - 11:31 PM

But no-one has mentioned the over time factor. Thats when an amputation leads to another loss of limb, which happened alot in pre-mid 20th century wars. They all died, except for, maybe 1 or 2.


well, the question was not asking who died ... it was as follows:
What is the minimum number of soldiers who must have lost all 4?
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#28 credels

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Posted 16 October 2007 - 11:35 PM

Indeed, however, gangreen could claim more body parts.
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#29 rookie1ja

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Posted 16 October 2007 - 11:37 PM

Indeed, however, gangreen could claim more body parts.


which has no immediate impact on the minimum number of soldiers who must have lost all 4 body parts, does it?
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#30 Chaos

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Posted 02 December 2007 - 06:39 AM

You know, my favorite part about questions like this is that all the correct solutions people come up with are exactly the same mathematically. Adding up the uninjured and finding the intersection are the same mathematically.

Note: I shortened the mathematics heavily because I don't want to take a page to explain.

Adding the uninjured:
100 - 70 = 30
100 - 75 = 25
100 - 80 = 20
100 - 85 = 15

30 + 25 + 20 + 15 = 90

100 - 90 = 10

Lumping all these calculations together and simplifying:
100 - (100 - 70) - (100 - 75) - (100 - 80) - (100 - 85) = 10
100 - 100 + 70 -100 + 75 - 100 + 80 - 100 + 85 = 10
70 + 75 + 80 + 85 - 100 - 100 - 100 = 10

Intersection (shortened heavily):
70 + 75 - 100 = 45
45 + 80 - 100 = 25
25 + 85 - 100 = 10

Lumping all these equations together and simplifying:
70 + 75 - 100 = 45

(70 + 75 - 100) + 80 - 100 = 25
70 + 75 + 80 - 100 - 100 = 25

(70 + 75 + 80 - 100 - 100) + 85 - 100 = 10
70 + 75 + 80 + 85 - 100 - 100 -100 = 10

As you see, the last line is the same in both methods. The only difference between the two methods is the order of operations. This also leads to the solution I got, add them all together (70 + 75 + 80 + 85) and subtract 300.
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