10 May 11
Actually, 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
1 edit
Originally posted by mtthwI found 1/2+1/3+1/7+1/43+1/1806.
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.
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10 May 11
Originally posted by PalynkaAn exhaustive search shows you are right Palyanka, nice one! It would be interesting to prove it without a search though...
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
Originally posted by iamatigerInteresting would be if you can rewrite it to a formula.
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.
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11 May 11
Originally posted by yashinCool, 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.
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
1 edit
Originally posted by yashinErr... Isn't that what I said? The remainder is always 1/(A*B*...*(m-1)) at each step m.
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
Originally posted by PalynkaYes 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 🙂.
Err... Isn't that what I said? The remainder is always 1/(A*B*...*(m-1)) at each step m.
11 May 11
1 edit
Originally posted by yashiniamatiger is a pretty smart guy, I'd give him more credit. But it's true that you explained it much more clearly. 🙂
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.