1. Joined
    19 Apr '11
    Moves
    59
    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
  2. SubscriberAThousandYoung
    All My Soldiers...
    tinyurl.com/y9ls7wbl
    Joined
    23 Aug '04
    Moves
    24791
    09 May '11 15:24
    There is no such integer. As E approaches infinity, 1/E approaches zero.
  3. Joined
    07 Sep '05
    Moves
    35068
    09 May '11 17:31
    Why does that prove there's no solution?
  4. Joined
    15 Mar '11
    Moves
    925
    10 May '11 05:01
    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... ;-)
  5. Joined
    07 Sep '05
    Moves
    35068
    10 May '11 09:07
    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,
  6. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    10 May '11 09:331 edit
    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,
    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.
  7. Joined
    26 Apr '03
    Moves
    25823
    10 May '11 22:23
    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.
    An exhaustive search shows you are right Palyanka, nice one! It would be interesting to prove it without a search though...
  8. Delft, Netherlands
    Joined
    17 Oct '03
    Moves
    64193
    11 May '11 00:09
    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...
    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
    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.
  9. Joined
    26 Apr '03
    Moves
    25823
    11 May '11 07:41
    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.
    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.
  10. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    11 May '11 08:341 edit
    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.
    Err... Isn't that what I said? The remainder is always 1/(A*B*...*(m-1)) at each step m.
  11. Delft, Netherlands
    Joined
    17 Oct '03
    Moves
    64193
    11 May '11 12:30
    Originally posted by Palynka
    Err... Isn't that what I said? The remainder is always 1/(A*B*...*(m-1)) at each step m.
    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 🙂.
  12. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    11 May '11 12:511 edit
    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 🙂.
    iamatiger is a pretty smart guy, I'd give him more credit. But it's true that you explained it much more clearly. 🙂

    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.
  13. Joined
    24 Jan '09
    Moves
    5514
    11 May '11 21:55
    no response from thudnblunder, but it seems you are right.
  14. Joined
    19 Apr '11
    Moves
    59
    12 May '11 13:27
    Yes, Palynka's solution is correct. Well done!
  15. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    12 May '11 13:41
    Originally posted by ThudnBlunder
    Yes, Palynka's solution is correct. Well done!
    Thanks!

    Do you have a definite proof?
Back to Top