Problems & Puzzles: Problems

Problem 35. More wrong turns...

"I once asked Andre Weil, the legendary mathematician, and also a superb historian,
 why the Egyptians did this"
[to represent fractions as a sum of unit fractions],
[Ronald] Graham, "and he said 'They took a wrong turn..."
The Man who loved only numbers, by Paul Hoffman, p. 170


Any Unitary fractions 1/a, a = integer is also called an Egyptian fraction.

One particular interesting equation involving distinct Unitary fractions and widely studied is the following one:

S(1/ai ) = 1, for i =1 to k>1such that if ai = aj then i = j  ...................... [eq. 1]

The sum of every set of ai integers that satisfy [1] is called an "Egyptian number"

The least particular example of [eq. 1] is:

1 = 1/2+1/3+1/6, that provides the Egyptian number 11 = 2 + 3 + 6.

Is well known that there is not any solution to [eq. 1] if all the ai integers are asked to be:

* Consecutive numbers (Taeisinger, Kurschak)
* Numbers in Arithmetic Progression  (Nagell)
* Prime numbers

Other peculiar numerical results are:

1. John Leech says that if all the ai integers are distinct odd numbers, then you need at least 9 of them to satisfy [eq. 1], being the largest denominator at least 105. For some unknown reason, in the D11 section of the well known  R. K. Guy's book (UPiNT2) is offered only an example composed by 11 odd numbers: 3, 5, 7, 9, 15, 21, 27, 35, 63, 105, 135

2. Barbeau found a solution of [eq. 1] using 101 distinct positive integers no one dividing each other.

3. Allan Johnson found a solution of [eq. 1] using 48 distinct positive integers all of them having two prime factors (*)


Regarding the Leech problem & predictions, Jaime Ayala and myself got some weeks ago not one but four solutions to [eq. 1] using only 9 odd values (**) , confirming both of the Leech's predictions.

Here is our 'minimal' solution using 9 distinct odd numbers and having the least Egyptian number:

1/3+1/5+1/7+1/9+1/11+1/15+1/35+1/45+1/231 = 1


1) Can you find the other 3 solutions using 9 distinct odd numbers? Is there other of these solutions unknown to us?

2) We think - according to our method - that our solution, shown above, is the least solution to [eq. 1] using only distinct odd numbers, in the double sense that it has only 9 odd numbers and also has the minimal Egyptian number. Is this true?

3) Can you find a solution that improves the Barbeau solution?

4) Can you find a solution that improves the Allan Johnston solution?

Other interesting references:

a) Eric Weisstein World of Mathematics
b) Kevin S. Brown

* In the Barbeau and in the Johnson respective problems, the condition of the Leech problem (of being all the numbers odd numbers) does not applies.

** As a matter of fact the August 6, 2000 we informed of these solutions to R. K. Guy who answered the 22th of the same month: "... I am filing ready for UPINT3, should it ever appear. Best wishes, RKG"


Jud McCranie found (October 5, 2000) the other 3 solutions that we have found and a fifth one more!...

3 5 7 9 11 15 21 135 10395  <--- this is the unknown for us
3 5 7 9 11 15 21 165 693
3 5 7 9 11 15 21 231 315
3 5 7 9 11 15 33 45 385

(Of course that he also found the minimal solution with 9 members and 231 the largest one)

Jud thinks that there are only these 5 solutions with 9 membres and that there is not any smaller one.


Jean-Charles Meyrignac sent the following lines the 21/5/2001:

Digging in my books to search for some documentation, I found Martin Gardner's book of Mathematical Diversions. The problem was to decompose 1 as the sum of odd inverses. This problem has been solved by S. Yamashita in 1976 (and then Jean-Charles send the 5 solutions discovered between Jud, Jaime and myself 24 years later, partially due to the lack of this info in the R.K. Guy's book).

So Yamashita is the first solver of this question 1.



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