rookie1ja 13 Report post Posted March 30, 2007 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? This old topic is locked since it was answered many times. You can check solution in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. Cost of War - solution Add up all the injuries, and you find that 100 soldiers suffered a total of 310 injuries. That total means that, at a minimum, 100 soldiers (Edit: it is, of course, not 100 soldiers, but 100 as calculation from 310 person-injuries out of 400 possible) lost 3 body parts, and 10 (the remainder when dividing 310 by 100) must have lost all 4 body parts. (In reality, as many as 70 may have lost all 4 body parts.) Edit: another way to solve it is to draw a line of 100 parts and compare injuries from opposite ends of the line, finding the intersection part of all 4 injuries. If left side of line (LS), right side (RS) and intersection (I), then: 70 (LS) and 75 (RS), then 45 (I) 45 (LS) and 85 (RS), then 30 (I) 30 (LS) and 80 (RS), then 10 (I) soldiers must have lost all 4 parts Share this post Link to post Share on other sites

Guest Report post Posted May 22, 2007 Alternate solution or bad math and a coincidence? 100 soldiers, max uninjured = 15 x 4 different injuries = 60 subtracted from minimum # injured 70 = 10 Share this post Link to post Share on other sites

Guest Report post Posted May 25, 2007 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 ) Share this post Link to post Share on other sites

Guest Report post Posted May 30, 2007 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 Share this post Link to post Share on other sites

Guest Report post Posted May 31, 2007 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 Share this post Link to post Share on other sites

Guest Report post Posted June 5, 2007 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. Share this post Link to post Share on other sites

Guest Report post Posted June 8, 2007 Oh yeah... thanks for the explanation chubaca. My grouping was incorrect. Share this post Link to post Share on other sites

Guest Report post Posted June 12, 2007 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? Share this post Link to post Share on other sites

Guest Report post Posted June 13, 2007 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. Share this post Link to post Share on other sites

Guest Report post Posted July 14, 2007 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 Share this post Link to post Share on other sites

unreality 1 Report post Posted July 14, 2007 i see why you think that, but... just because you lose an eye doesnt mean you lose all the other stuff. there are much less soldiers that lost ALL FOUR than those that lost JUST ONE. just look at it this way... 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 Share this post Link to post Share on other sites

Guest Report post Posted July 20, 2007 total number of injured soldiers 70+75+80+85 = 310 Even if we assume each has 3 injuries even then 10 injuries remain so at least 10 soldiers have all 4 kind of injuries Thnx, Adarsh Share this post Link to post Share on other sites

Guest Report post Posted July 20, 2007 For those who are getting answers other than 10... "70 soldiers lost an eye" That means 30 have not lost any eye, or have good eyes. "75 lost an ear" Of those 75, max. 30 can have good eyes, other 75-30 or 45 will not. Therefore, in the best case scenario, min. 45 have lost both an eye and an ear, and 55 will have lost single organ only. "85 lost a leg" Of these 85, max. 55 can have either-eye-or-ear-loss (and not both-eye-ear), 85-55 or 30 will have eye-ear-leg loss. So, in the best case scenario, min. 30 (=85-55) have lost all 3 organs, and 70 will have lost 2 organs. "80 lost an arm" Of these 80, max. 70 will have lost 2 organs (of eye, ear & leg). So, in the best case scenario, min. 10 (=80-70) have lost all 4 organs. Algebraic solutions are neater, shorter but more difficult to explain... Share this post Link to post Share on other sites

unreality 1 Report post Posted July 21, 2007 umm quoting myself... just look at it this way... 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 Share this post Link to post Share on other sites

Guest Report post Posted July 29, 2007 ok so 85 lost a leg. that leaves 15 who didnt so subtract that from 80 who lost an ear so 65 lost both subtract from 100 and u have 35. do the same for the 70 who lost an ear and you have 35. 35 from one hundred is 65. 65 minus those that lost an arm is 5. 5 is the minimum Share this post Link to post Share on other sites

unreality 1 Report post Posted July 30, 2007 ur logic is very mistaken. You should write that math out to see that you are wrong... Share this post Link to post Share on other sites

Guest Report post Posted August 9, 2007 and I wonder why didn't anyone take into account a simple fact that 70 soldiers who lost an eye might be doubled, so 35 soldiers lost 2 eyes, and got counted twice same with arms, ears, legs it's all about the way to count right way or army way =) so all your math is wrong Share this post Link to post Share on other sites

Guest Report post Posted August 9, 2007 ok 10 is fine but how can u b so sure that all the ten hv lost all 4 parts cos the value asked is minimum and it can b "0" taking that 10 are not injured... Am i proving u as an a** hole AdMin ??? Share this post Link to post Share on other sites

Guest Report post Posted August 19, 2007 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. the problem with this is the fact that averages can be altered by one extremity. f/e. when a job advertisement says the average pay is, lets say, $55 an hour, one would think the job pays a good amount. however, the minimum pay, the pay the person would actually get, could be only $10 an hour, while the owner gets paid $100 an hour. Share this post Link to post Share on other sites

Guest Report post Posted August 24, 2007 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 Share this post Link to post Share on other sites

Guest Report post Posted August 27, 2007 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 Share this post Link to post Share on other sites

Guest Report post Posted September 1, 2007 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...!! Share this post Link to post Share on other sites

unreality 1 Report post Posted September 1, 2007 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 Share this post Link to post Share on other sites

Guest Report post Posted September 11, 2007 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. Share this post Link to post Share on other sites

Guest Report post Posted September 28, 2007 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 Share this post Link to post Share on other sites