09 May '11 11:19>
What is the largest integer E, such that 1/A+1/B+1/C+1/D+1/E=1 ??
A<B<C<D<E are all positive integers
A<B<C<D<E are all positive integers
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,
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.
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...
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.
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.
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.
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 🙂.