Mathematical Fallacy - The $17$ camels Problem.

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So the Problem goes like this :-

  • An old man had $17$ camels . He had $3$ sons and the man had decided to give each son a property with his camels.
  • Unfortunately however, the man dies, and in his last will he says that his $1$st son will get $1/2$ of the total camels, his $2$nd son will get $1/3$ on the total camels, and his $3$rd son will get $1/9$ of the total camels.
  • This looks like a major problem , because $17$ is not divisible by neither of $2,3,9$ ; and it looks impossible to divide $17$ camels like that .

The Solution however, is very interesting and goes like this :

  • First bring another camel ( from a camel owner ) and add it to this group of $17$ camels. The total no. of camels now becomes $18$ .
  • Also $18$ is divisible by each of $2,3,9$; so the $1$st son gets $9$ camels, the $2$nd son gets $6$ camels, and the third son gets $2$ camels.
  • Miraculously , we get $9 + 6 + 2 = 17$ camels , hence the extra camel that was brought before can be returned back to the owner.

After doing this problem , I suppose that there is some mathematical fallacy involved here. When we had $17$ camels , it was impossible to divide them. However after adding $1$ extra camel , everything divides very nicely among the sons as well as the extra camel could be returned. How is this working ?.

I am posting this Problem so that I can get some opinions or discussions about this Problem.

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The way the puzzle is told there is supposedly no fallacy and this is supposed to be the solution. The division doesn't add up to $100\%$ so there's nothing wrong with there being a camel left over. And the sons get $\frac 12, \frac 13, \frac 19$ of $18$ camels so the division worked out evenly. And every version of the puzzle I've ever heard claims this is the legitimate solution.

Which drives me nuts because this obviously isn't a solution. The first son got $\frac 12$ of $18$ camels but that wasn't what the will speculated. The will speculates that he should get $\frac 12$ of $17$ camels. He ended up with half a camel more! Likewise the second son ended up with $\frac 13$ of a camel more and the third son with $\frac 19$ of a camel more.

Which makes sense as the will only speculated what to do with $16 \frac 1{18}$ of the camels and not what to do with the remaining $\frac {17}{18}$ of a camel. But they didn't need the "wise man" for that. They could have simply said: Son 1 gets $8\frac 12$ camel. Son 2 gets $5 \frac 23$ of a camel. Son 3 gets $1\frac 89$ of a camel. There is $\frac {17}{18}$ of a camel left over. Let's make an agreement among ourself that Son 1, Son 2,and Son 3 get an extra $\frac 12, \frac 13, \frac 19$ camel to make it even. As we each get more than the will speculated we should all be happy.

But I don't think that "solves" any problem.

And it the end the sons didn't end up with $16\frac 1{18}$ camels divide in $\frac 12, \frac 13$ and $\frac 19$ as the will stipulated. They ended up with $17$ camels dived $\frac 9{17},\frac {6}{17}, \frac 2{17}$ as the will did not stipulated.

So in MY opinion.... I hate this stupid puzzle and think it is false and its fallacy is....

.......

The mathematical fallacy is the are returning the camel in disproportion to their inheritance.

The 1st son who inherits $\frac 12$ the camels should be borrowing and returning $\frac 12$ a camel. He is borrowing and returning $\frac {9}{17}$ of a camel. The 2nd son who inherits $\frac 13$ is returning $\frac {6}{17}$s of a camel and not $\frac 13$ and the 3rd son who inherits $\frac 19$ is return $\frac 2{17}$s of a camel and not $\frac 19$.

The father's will divided his camels. Not his camels plus some other number of camels and then return the borrowed camels back.

Suppose the extra man instead of having $1$ camel had $37$ camels. So that makes $37+17= 54$ camels. The first son gets $27$ and the second gets $18$ and the third gets $6$. The wise man gets back $3$ camels. Then the older son give him $21$ back and keeps $6$ and the second son gives him $12$ back and keeps $6$ and the third son gives him $1$ back and keeps $5$. so the wise man gets back his $37$ camels and we end up with the sons having $5,6,6$ camels each.

Problem solved?

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So because the ratios don't add to 100 percent of the camels and the will remains silent on the last 1/18 of the camels.

LCD of 1/2 and 1/3 and 1/9 is 18.

  • son 1 gets 9/18 of camels
  • son 2 gets 6/18 of camels
  • son 3 gets 2/18 of camels
  • nobody gets 1/18 of camels (not assigned by the will!)

Pretend son 4 was supposed to get the 1/18 not assigned by the ratio, bringing us to 100% of camels assigned.

So, what really happens is that

  • son 1 "steals" 1/2 camels from the unassigned 1/18
  • son 2 "steals" 1/3 camels from the unassigned 1/18
  • son 3 "steals" 1/9 camels from the unassigned 1/18
  • son 4 looses out on his 1/18 camels!

The solution distributes an unassigned fractional camel.