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
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
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, slowly approaches 1. However, the values of are not large enough to produce a value larger than for . While I am too lazy to write a formal proof, I’m sure that this can be proven using induction.
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 c are given integers, and x and y 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 and . To put it more simply, there exist positive integers a, b and c for
as long as n is equal or greater than 2.
For example, for n = 7, there are a couple of solutions:
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.