# More Solutions To The “17 Elephants/Cows/Horses/Camels” Type Puzzle Previously, I posted a famous puzzle about an owner of seventeen elephants with three sons who died. In his will he wrote, “My eldest son gets half of my elephants, my middle son gets one-third, and my youngest son one-ninth of the seventeen elephants.” The man owned 17 elephants so it’s impossible to follow the will without killing a couple of elephants. Then an old man suggested that he’d add one of his elephants so now, the three sons have eighteen elephants.

Now, it’s easy to divide the elephants! The eldest son got 9, the second son got 6, and the youngest son got 2. There’s one elephant left which was returned to the old man.

Of course what they have done was illegal and also mathematically incorrect. Note that $\frac{1}{2}+\frac{1}{3}+\frac{1}{9}=\frac{17}{18}$

instead of 1. This makes it impossible to make a proper division since a portion of one-eighteenth will be left unaccounted for.

Jack Shalom brought up an interesting question in the comments section of the original post. He wrote:

I always wondered whether it was possible to create a solution to a similar problem as above, but with different numbers. In other words, find X such that the sum of X+1 divided by A, B, and C would equal X. Now in the classic problem above, A=2, B=3, and C=9. Is it possible to find X given *any* A, B, and C? Anyone want to take a stab at generalizing the solution in this manner? Would there have to be restrictions placed on A, B, and C so that X would turn out to be a whole number?

This means that he wanted to know if there are other solutions for the equation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{x}{x+1}.$

As I looked at the Diophantine equation¹, I realized how similar it was to the Erdős–Straus conjecture².

Using a brute-force algorithm, I was able to find seven solutions (including the solution in the classic puzzle) while taking into account the restrictions:

7: 2, 4, 8
11: 2, 4, 6
11: 2, 3, 12
17: 2, 3, 9
19: 2, 4, 5
23: 2, 3, 8,
41: 2, 3, 7

Note that there are two solutions for 11.

I’m certain that these are the only solutions since as x increases, $\frac{x}{x+1}$ slowly approaches 1. However, the values of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ are not large enough to produce a value larger than $\frac{41}{42}$ for $\frac{x}{x+1}$. While I am too lazy to write a formal proof, I’m sure that this can be proven using induction.

## Notes

1. A Diophantine equation is an equation, normally with two or three unknowns, such that only integer solutions are studied and required. The most basic Diophantine equation is the linear Diophantine equation of the form ax by = c, where a, b and are given integers, and and are the unknown variables. For instance, 3x – 4y = 7. The Diophantine equation is named after Diophantus of Alexandria who first studied this type of equation.

2. The  Erdős–Straus conjecture is a form of a Diophantine equation. In 1948, Paul Erdős and Ernst G. Straus conjectured that for all integers n ≥ 2, 4/n can be expressed as the sum of three unit fractions $\frac{1}{a},\frac{1}{b}$ and $\frac{1}{c}$. To put it more simply, there exist positive integers a, b and c for $\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

as long as n is equal or greater than 2.

For example, for n = 7, there are a couple of solutions: $\frac{4}{7}=\frac{1}{4}+\frac{1}{4}+\frac{1}{14}=\frac{1}{3}+\frac{1}{6}+\frac{1}{14}$.

Anyway, a friend of mine gave an interesting perspective to the 17 Elephants puzzle. She theorized that the man must have worked for the tax department. If you do your own taxes, you may know that a fractional part of a dollar is disregarded unless if it amounts to 50 cents or more, in which case, you round up to the next whole number (Note: I’m not sure if this is the case in your country). Now, to solve the puzzle:

For the eldest son: One-half of 17 is 8.5 then rounded up to 9 elephants.

For the second son: One-third of 17 5.67 then rounded up 6 elephants.

For the youngest son: One-ninth of 17 is 1.89 then rounded up to 2 elephants.

So, 9 + 6 + 2 = 17. Problem solved!

Now, we can say that the old man who added one elephant to help the brothers figure out how to divide their father’s estate was not of any help. Then he took one of the brothers’ elephants too (Note: In many versions of this puzzle, the wording is that the old man “gave” one elephant to the brother. I just used the word “added” to make it less ambiguous)! It’s like hiring a tax accountant to prepare your taxes, which you could have prepared yourself. ### Posted by Edmark M. Law

My name Edmark M. Law. I work as a freelance writer, mainly writing about science and mathematics. I am an ardent hobbyist. I like to read, solve puzzles, play chess, make origami and play basketball. In addition, I dabble in magic, particularly card magic and other sleight-of-hand type magic. I live in Hong Kong. You can find me on Twitter` and Facebook. My email is edmarklaw@learnfunfacts.com

## 15 thoughts on “More Solutions To The “17 Elephants/Cows/Horses/Camels” Type Puzzle”

1. natuurfreak says:

I love elephants

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2. You are so smart! I used to love math a lot, but I only use my knowledge of geometry recently. It helps. I regret a lot of people who want to learn drawing have no clue about geometry. It helps when doing perspectives for instance.

Liked by 1 person

1. Edmark M. Law says:

I personally know an artist who told me that his knowledge of projective geometry helped him a lot.

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3. pick1solution says:

This proves that it is best if you try to be the firstborn.

Liked by 1 person

1. Edmark M. Law says:

Hm, that must have been the point of tee puzzle after all :D

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4. New Bloggy Cat says:

Trying to understand this mammoth-matically is giving me a herd-ache. ʅฺ(・ω・。)ʃฺ？？

Liked by 1 person

1. Edmark M. Law says:

I’ll make sure to never forget those puns…

Liked by 1 person

1. New Bloggy Cat says:

Yes, such mammoth-rable puns!
ﾍ(^･･^=)~

Liked by 1 person

2. Edmark M. Law says:

After all, elepuns never forget…

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3. New Bloggy Cat says:

Ivory-member this for a very long time.
(*•̀ᴗ•́*)و ̑̑

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5. emmitsomo says:

… so, there are still three sons. And now what? :-p ´-´ What does this change??

Liked by 2 people

1. Edmark M. Law says:

The number of elephants. In the classic puzzle, the man had 17 elephants. Now, there can be 6 more possible variations.

For instance, the puzzle can be as follows: A man who has 3 sons died. His will stated that the first son gets 1/2, the second son gets 1/4 and the third son gets 1/5 of the elephants. The man owned 19 elephants in total.

To help the sons divide the elephants, the old man added one elephant (which made it 20).

Now for the division: The first son got 1/2 of 20 or 10; the second son got 1/4 of 20 or 5 and the third son got 1/5 of 20 or 4.

10 +5+4=19

So, there’s one elephant left whch the old man took back.

Liked by 1 person

1. emmitsomo says:

okay… thank you for the answer… only, know there keeps to be the question, who is the elephant and who is the old man…? Or “what” is the elephant? But I know I guess already. Only the old man keeps to be the mystery.

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6. B. says:

I wish i were much more gifted in maths to follow the calculations…😉😜🤗

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