## Math Brain Teaser, The Wounded

While visiting a military hospital overseas, I was told that exactly 2/3 of soldiers had lost an eye, 3/4 had lost an arm, and 4/5 had lost a leg. I remarked to my friend, “Then it must be that at least 26 of the men must have lost all three- an eye, an arm, and a leg”.

This being the case, can you say exactly how many men were in the hospital?

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suineg| PUZZLE MASTER | Profile August 11th, 2010 - 11:06 pmOk, I think the answer could be like this:

The order from the smallest to the biggest value is: 2/3, 3/4, 4/5 so if 26 is the minimal number possible of the men that lost all three: the minimal intersection, you what mix the biggest complement with the biggest remaining parts:

way 1: way 2:

3/4 – 1/3 = 5/12 4/5 – 1/3 = 7/15

5/12 – 1/5 = 13/60 7/15 – 1/4 = 13/60

Either way you get: 13/60*X= 26

X=120, so there were 120 men in the hospital, cool man.

Hex| PUZZLE MASTER | Profile August 12th, 2010 - 7:55 amThe number of soldiers N must be a multiple of 3, 4 and 5. Hence, N = 3x4x5xK = 60 x K where k is a positive integer.

The given is ambiguous here: I remarked that the wounded with all 3 injuries are at least 26.

Taken at face value, the following scenario can apply:

N=60 soldiers

E=eye injury=40

A=arm injury=45

L=leg injury=48

26 soldiers have all 3 injuries: EAL=26

19 soldiers have only A and L: !EAL=19

3 soldiers have only E and L: E!AL=3

11 soldiers have only E: E!A!L=11

As all the above are mutually exclusive, the number of soldiers in the hospital is: 26+19+3+11=59

engjs1960| Profile August 12th, 2010 - 8:28 am120.

Hex| PUZZLE MASTER | Profile August 12th, 2010 - 1:38 pmFollow-up:

Not being convinced of the puzzle’s given, I dug furthermore:

First, it is not clear if all the soldiers are in the hospital or a percentage of them as I had assumed previously. I did the following exercise:

Assume 60 soldiers all in hospital.

The minimum number of soldiers with 3 wounds is 13.

It occurred to me immediately that 26 is the double of 13, and hence the number of soldiers must be 120.

Bobo The Bear| PUZZLE MASTER | Profile August 12th, 2010 - 5:12 pmIf 1/3 had both eyes, 1/4 had both arms, 1/5 had both legs; and if there is no overlap between these three group; then this would be (1/3 + 1/4 + 1/5) = 47/60 of the soldiers with at least one of these facilities intact. This implies that a minimum of 13/60 have lost all three. If 26 represents 13/60 of the men, then there are at least 120 men in the hospital.

(Posted by text while on the road from San Diego to Phoenix. No, I’m not the one driving.)

Shawn| PUZZLE GRANDMASTER | Profile August 16th, 2010 - 10:19 am39 men

Shawn| PUZZLE GRANDMASTER | Profile August 16th, 2010 - 10:46 amYes, I see. I solved for a specific case instead of an “at-least” case, will have to learn to read more carefully.

RK| Founder | Profile August 19th, 2010 - 6:47 amThe official answer I have is 120

deepside0058| Profile August 21st, 2010 - 3:02 am120???