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Cost of War


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#1 rookie1ja

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Posted 30 March 2007 - 03:17 PM

Cost of War - Back to the Logic Puzzles
Here's a variation on a famous puzzle by Lewis Carroll, who wrote Alice's Adventures in Wonderland.
A group of 100 soldiers suffered the following injuries in a battle: 70 soldiers lost an eye, 75 lost an ear, 85 lost a leg, and 80 lost an arm.
What is the minimum number of soldiers who must have lost all 4?

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#2 ptditz

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Posted 22 May 2007 - 02:17 PM

Alternate solution or bad math and a coincidence? 100 soldiers, max uninjured = 15 x 4 different injuries = 60 subtracted from minimum # injured 70 = 10
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#3 robxmccarthy

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Posted 25 May 2007 - 04:22 PM

I have this xml on my google homepage.

The way I did it was to simply add up the uninjured people from each injury and subtract it from the total. A lot easier in my mind.

So: 25 + 30 + 15 + 20 = 90

Total soldiers - ones lacking at least one injury = 100 - 90 = 10

~Rob

(And yeah, the second solution above was just lucky )
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#4 dishkyaaoonn

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Posted 30 May 2007 - 01:56 PM

The number of soldiers who lost three body parts cannot be hundred. In fact, the maximum number of soldiers who lost three body parts can be only 80. Hence, as it has been reported in the first post, the statement that at a minimum, 100 soldiers lost 3 body parts is wrong.
According to me, the minimum number of people who lost four body parts will be 15.
Let w = number of people who lost exactly one body part
x = number of people who lost exactly two body parts
y = number of people who lost exactly three body parts
z = number of people who lost exactly four body parts

Since, the total number of soldiers is 100,
w + x + y + z = 100
And, when we add all the injuries (70 + 75 + 85 + 80), we get 310. In this 310, w is counted once, x is counted twice, y is counted thrice and z is counted four times.
Hence, w + 2x + 3y + 4z = 310 .... subtracting this equation from the first equation, we get...
x + 2y + 3z = 210

Now, in the above equation, we have to minimise z, which means we have to maximize y and x.
The set of injuries are 70, 75, 85, and 80. If we try to maximize y, then the maximum number of people with three injuries can be 75 (from the second, third and fouth group) and after this the standing will be 70, 0, 10 and 5 (subtracting 75 from second, third and fouth groups). Now, we can get another five people who have three injuries from the first, third and fourth groups which changes the group standings to 65, 0, 5 and 0. So, the maximum value of y can be 75 + 5 = 80. And with the new group standings the maximum value of x can be 5 (after which the group standings become 60, 0, 0 and 0). So, our last equation now is...
x + 2y + 3z = 210 where x = 5 and y = 80
3z = 45 -> z = 15
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#5 chubaca

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Posted 31 May 2007 - 11:08 PM

I also believe that the answer is 10. Let's go injury by injury:

1.-First, 70 soldiers lost an eye, so 30 soldiers have their two eyes healthy.

2.-75 soldiers lost an ear. So, to get the MINIMUM number of soldiers who lost his ear AND eye you have to supose that every one of the 30 soldiers with two healthy eyes lost his ear. That gives us a total of 75-30=45 people who lost one eye AND one ear, and 55 people who lost only one of those two, or none.

3.-85 soldiers lost a leg. Following the same logic, we had 55 people who didn't lost his eye AND ear, and suposing that every one of these people loses one of their legs, then we have a minimum of 85-55=30 people who have lost one eye AND one ear AND one leg, and the rest of the soldiers are 70.

4.-Finally, 80 soldiers lost an arm. Again, we have a minimum of 80-70=10 soldiers who lost everything.

At first glance, I believe this is kind of equivalent to the answer of robxmccarthy
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#6 larryhl

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Posted 05 June 2007 - 08:46 PM

the answer of 10 is correct. the solution presented by rookie1ja is using the method of filling pigeonholes and figuring out minimums. all this method does is average out the injuries over the whole group and figuring out, at minimum, how many will be over that average.

the more easily understood explanation has been shown by various others: figuring out the maximum number of people who can't have all 4 injuries by adding up the amount of people who don't have a specific injury.
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#7 dishkyaaoonn

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Posted 08 June 2007 - 04:45 AM

Oh yeah... thanks for the explanation chubaca. My grouping was incorrect.
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#8 jdba

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Posted 12 June 2007 - 04:52 PM

Hi,
I am wondering why the following reasoning would not be correct.

Consider, 70% have a first type of injury
75% have a second one.
Then, if we manually pick the maximum non-intersecting set, there are only 20 persons that have both of those injuries.

Now consider the third type of injury, affects 80% of the soldiers. Now manually pick those 80% to not include those 20% having both injury 1 and 2, then there are zero persons with injuries 1,2,3.

It's not even necessary to consider the 4th injury type that affects 85%.

Everyone is in agreement on 10%, but where did the above logic fail?
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#9 larryhl

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Posted 13 June 2007 - 01:43 PM

Hi,
I am wondering why the following reasoning would not be correct.

Consider, 70% have a first type of injury
75% have a second one.
Then, if we manually pick the maximum non-intersecting set, there are only 20 persons that have both of those injuries.

Now consider the third type of injury, affects 80% of the soldiers. Now manually pick those 80% to not include those 20% having both injury 1 and 2, then there are zero persons with injuries 1,2,3.

It's not even necessary to consider the 4th injury type that affects 85%.

Everyone is in agreement on 10%, but where did the above logic fail?



The logic fails in that you're assuming you can manually pick a non-intersecting set. Remember, this is only a pool of 100 soldiers in total. Your logic assumes 100 soldiers per set of injuries (making a total of 400), in which case, yeah, there's no one who would suffer all four injuries. Follow chubaca's explanation, it's the most well-explained one for those who haven't taken discrete logic.
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#10 wrzesinski

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Posted 15 July 2007 - 12:14 AM

the answer would be 70 because at least 70 people lost an eye and if you had to have lost all four at lest 70 had to be injured with them all
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