- 10 May '11 05:01Actually, there is at least one solution:

1/3 + 1/4 + 1/5 + 1/6 + 1/20 = 1

This must be the greatest E (= 20) when A = 3, since 1/3 + 1/4 + 1/5 + 1/6 is the highest possible 4-term sum we can get starting at 1/3, making the last term (1/20) the smallest possible last term (i.e., the greatest value for the denominator).

There are no solutions for A > 3 since 1/4 + 1/5 + 1/6 + 1/7 + 1/8 < 1, and so it's impossible to reach a sum of 1 if the first term is 1/4 or smaller.

The only question remaining is whether it's possible to start with A=2 and find a combination that yields a higher value for E. I'll let someone else take it from here, as I'm very tired at the moment... ;-) - 10 May '11 09:33 / 1 edit

I found 1/2+1/3+1/7+1/43+1/1806.*Originally posted by mtthw***1/2 + 1/4 + 1/7 + 1/14 + 1/28 = 1**

(Sum of reciprocals of factors, excluding one, of a perfect number is 1, so I looked for one with the right number of factors)

No idea if it's the biggest E or not, though,

My idea was to start try to get the largest sum possible such that what remains is the lowest possible 1/x at each step but still > 0. This means I get the remainder and simply add 1 to its denominator.

First 1/2 => 1- 1/2 = 1/2

Second => 1/2-1/3 = 1/6

Third => 1/6-1/7 = 1/42

Fourth => 1/42-1/43 = 1/1806

So E is 1806. I don't think this can be bettered but I'm not sure I've proven it. - 10 May '11 22:23

An exhaustive search shows you are right Palyanka, nice one! It would be interesting to prove it without a search though...*Originally posted by Palynka***I found 1/2+1/3+1/7+1/43+1/1806.**

My idea was to start try to get the largest sum possible such that what remains is the lowest possible 1/x at each step but still > 0. This means I get the remainder and simply add 1 to its denominator.

First 1/2 => 1- 1/2 = 1/2

Second => 1/2-1/3 = 1/6

Third => 1/6-1/7 = 1/42

Fourth => 1/42-1/43 = 1/1806

So E is 1806. I don't think this can be bettered but I'm not sure I've proven it. - 11 May '11 00:09

Interesting would be if you can rewrite it to a formula.*Originally posted by iamatiger***An exhaustive search shows you are right Palyanka, nice one! It would be interesting to prove it without a search though...**

What number would 'n' be when 1/A + 1/B + ... 1/n = 1?

Where A, B, ... n are all positive integers.

First Lets make a difference between an 'ending integer' and an 'integer in line'

'ending integer' = 'integer in line' - 1

Below I will only work with 'integers in line'

A=2

B=3

C=7

D=43

This seems like a function line!

B= A+1

C= B*A+1

D= C*B*A+1

Lets see if this works for E

E= 2*3*7*43+1 = 1807

'ending integer' = 'integer in line' - 1

1807 - 1 = 1806

Continuing

F= 2*3*7*43*1807+1 = 3263443

and the ending number 3263442 is indeed the right answer to F

G would be 2*3*7*43*1807*326443+1 and so on

This is where my math ends. I learned how this can be made into a formula. Unfortunately my memory fails. Also it is 2 AM here and I CBA to solve this. - 11 May '11 07:41

Cool, now we just need to prove that a line designed by that formula is optimal. It must be something like showing that this line grows the denominators at the maximum rate possible.*Originally posted by yashin***Interesting would be if you can rewrite it to a formula.**

What number would 'n' be when 1/A + 1/B + ... 1/n = 1?

Where A, B, ... n are all positive integers.

First Lets make a difference between an 'ending integer' and an 'integer in line'

'ending integer' = 'integer in line' - 1

Below I will only work with 'integers in line'

A=2

B=3

C=7

D=43

Th ...[text shortened]... la. Unfortunately my memory fails. Also it is 2 AM here and I CBA to solve this. - 11 May '11 08:34 / 1 edit

Err... Isn't that what I said? The remainder is always 1/(A*B*...*(m-1)) at each step m.*Originally posted by yashin***Interesting would be if you can rewrite it to a formula.**

What number would 'n' be when 1/A + 1/B + ... 1/n = 1?

Where A, B, ... n are all positive integers.

First Lets make a difference between an 'ending integer' and an 'integer in line'

'ending integer' = 'integer in line' - 1

Below I will only work with 'integers in line'

A=2

B=3

C=7

D=43

Th la. Unfortunately my memory fails. Also it is 2 AM here and I CBA to solve this. - 11 May '11 12:30

Yes you did more or less, but you can see by iamatiger's reaction he didn't understand the method you used. It seemed like you found the number where ... + 1/m =1 and then added 1 to m. It wasn't clear to me how you got the answer, but since you were first the credit is all yours .*Originally posted by Palynka***Err... Isn't that what I said? The remainder is always 1/(A*B*...*(m-1)) at each step m.** - 11 May '11 12:51 / 1 edit

iamatiger is a pretty smart guy, I'd give him more credit. But it's true that you explained it much more clearly.*Originally posted by yashin***Yes you did more or less, but you can see by iamatiger's reaction he didn't understand the method you used. It seemed like you found the number where ... + 1/m =1 and then added 1 to m. It wasn't clear to me how you got the answer, but since you were first the credit is all yours .**

My idea was first to start with the largest 4 element sum possible for integers larger than one (1/2+1/3+1/4+1/5) and work backwards. The idea is that E will be the largest the closer I get to 1. But as I was doing it I realized that at every step of the sum the remainder was of the form 1/x and so if I subtract the largest possible number _ 1/(x+1) _ that still leaves the sum < 1 then the remainder is also of the same form (namely _ 1/[x*(x+1)] _ ). This is how I got there, with a bit of luck and intuition. I'm usually terrible with puzzles that deal with integers only.