Problems & Puzzles: Problems Problem 35. More wrong turns...
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/a_{i }) = 1, for i =1 to k>1such that if a_{i }= a_{j} then i = j ...................... [eq. 1] The sum of every set of a_{i }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 a_{i }integers are asked to be:
Other peculiar numerical results are:
*** 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 Questions: 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 _____________________ ** As a matter of fact the August 6, 2000 we informed of these solutions to R. K. Guy who answered the 22^{th} of the same month: "... I am filing ready for UPINT3, should it ever appear. Best wishes, RKG" Solution 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 (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. *** JeanCharles 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 JeanCharles 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. ***






