All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Extra Egyptians (Posted on 2013-02-14) Difficulty: 3 of 5

(1) Consider Egyptian Number.

(2) Conjecture 1. There is no number greater than 125 that cannot be expressed as a series of fractional terms (1/a+1/b+1/c...=1) so that at least one partial sum of those terms is either 1/2 or 1/3.

(3) Conjecture 2. There is at least one finite set {p1,p2,p3,...} of prime factors of the denominators of the fractional terms so that all integers greater than 125 can be so expressed (it seems that this would have to follow, but can it be proved?)

(4) Can the limit 125 in (2) or (3) be further sharpened?

(5) Is there a further limit beyond which both (2) and (3) can be made to hold simultaneously for all integers? If so, what is it?

No Solution Yet Submitted by broll    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Most of a solution; I think this is what is wanted Comment 1 of 1

I assume a partial sum can be just one term. Let's start with assuming that a 3 is present (a 1/3 in the reciprocal total).


In the table below, the reciprocals of the numbers on the right add up to 2/3, rather than 1, so each can be supplemented by a 3 to cause their reciprocals to add up to 1. So the list on the left, incremented by 3 are all reachable under the terms of part 1, with the partial sum 1/3.

This list can be extended by doubling each number and adding a 3 to the list. That will give you the next batch of odd numbers for this list; for even numbers, add a 4 and a 12. The doubling changes the 2/3 to 1/3, then the 1/4+1/12 gives you another 1/3. Again, at the end,to get an Egyptian number, the requisite 3 is added as that's what we're seeking: a set that includes a 3.

For example 127: We want reciprocals for the numbers that add to 124 to themselves add up to 2/3. As 124 is even we subtract 4+12=16 making 108, half of which is 54, so use twice the numbers from 54 and add in those 3, 4 and 12: {6,18,24,24,36} union {3,4,12} = {3,4,6,12,18,24,24,36}. The 3 involves a partial sum of 1/3, as does the portion representing 1/3: 1/6+1/18+1/24+1/24+1/36 or the 1/4+1/6.

              addends whose reciprocals
              add up to 2/3
             
38            3  5  15  15
39            6  6  9  9  9
40            3  5  12  20
41            3  6  8  24
42            6  6  6  12  12
43            4  9  9  9  12
44            4  8  8  12  12
45            4  8  8  10  15
46            4  6  12  12  12
47            4  6  10  12  15
48            3  5  10  30
49            4  6  9  12  18
50            4  6  8  16  16
51            2  7  42
52            3  10  12  12  15
53            3  10  10  15  15
54            3  9  12  12  18
55            3  4  24  24
56            2  18  18  18
57            2  15  20  20
58            2  14  21  21
59            3  7  14  14  21
60            3  8  10  15  24
61            3  4  18  36
62            2  12  24  24
63            2  12  21  28
64            2  12  20  30
65            3  6  14  21  21
66            2  14  15  35
67            3  8  8  24  24
68            2  11  22  33
69            3  6  12  24  24
70            3  6  12  21  28
71            3  4  16  48
72            2  10  30  30
73            3  6  14  15  35
74            4  4  11  22  33
75            3  6  11  22  33
76            2  10  24  40
77            3  5  20  21  28
78            2  12  16  48
79            3  6  10  30  30
80            3  12  15  15  15  20
81            3  12  12  18  18  18
82            3  4  15  60
83            2  9  36  36
84            3  7  8  24  42
85            3  6  12  16  48
86            2  9  30  45
87            3  5  14  30  35
88            3  10  12  15  24  24
89            2  12  15  60
90            3  6  9  36  36
91            3  8  14  21  21  24
92            2  9  27  54
93            3  6  9  30  45
94            3  7  21  21  21  21
95            3  7  8  21  56
96            3  6  12  15  60
97            3  8  12  20  24  30
98            2  24  24  24  24
99            2  21  24  24  28
100           2  20  24  24  30

In fact, as the list starts with 38, all numbers starting with 41 can be expressed in Egyptian form with at least one 3, assuring a term (and therefore partial sum) equal to 1/3.

The problem with 40 is that the only Egyptian ways of expressing it are:

4+4+8+8+8+8
4+5+5+8+8+10
4+6+6+6+6+12
5+5+5+5+10+10

But 1/6 + 1/6 is in fact a partial sum of 1/3, so we can go lower than 41 as the first such expressible number.

33 through 39 are expressible as:

3 + 3 + 9 + 9 + 9
3 + 3 + 8 + 8 + 12
3 + 4 + 4 + 12 + 12
3 + 3 + 6 + 12 + 12
3 + 3 + 6 + 10 + 15
3 + 4 + 4 + 9 + 18
3 + 3 + 6 + 9 + 18

and 29 through 32 as:

2 + 3 + 12 + 12
2 + 3 + 10 + 15
2 + 4 + 5 + 20  (the first need for 1/2 in this)
2 + 3 + 9 + 18

but 26 through 28 can be expressed:

4 + 4 + 6 + 6 + 6 = 26
3 + 6 + 6 + 6 + 6 = 27
4 + 4 + 4 + 8 + 8 = 28 and
4 + 4 + 5 + 5 + 10 = 28 also

with subsets whose reciprocals add to 1/2 or 1/3.

25 stops us from going lower as its only Egyptian representation is

5 + 5 + 5 + 5 + 5

Conjecture 2:

So far as I can see, the only prime factors used in any of the numbers above, including the multiples and the added numbers used to extend the list, are 2, 3, 5 and 7. That would seem to be the finite set sought.

 


  Posted by Charlie on 2013-02-14 17:03:00
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information