1. Joined
    11 Nov '05
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    04 Jan '08 10:55
    The sum of 5 integers is 25.
    The product of the same 5 integers is 2520.
    What are the numbers?
  2. Standard membercelticcountry
    Copyright ©2001-2006
    Eastbourne
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    20 Sep '04
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    04 Jan '08 11:012 edits
    Originally posted by FabianFnas
    The sum of 5 integers is 25.
    The product of the same 5 integers is 2520.
    What are the numbers?
    *EDITED ANSWER*

    *EDITED METHOD*
  3. Joined
    11 Nov '05
    Moves
    43938
    04 Jan '08 11:42
    Originally posted by celticcountry
    *EDITED ANSWER*

    *EDITED METHOD*
    It took six minutes for you to figure this one out!
    (He mailed the answer to me)
  4. Joined
    31 May '07
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    696
    04 Jan '08 12:391 edit
    2520 = 2x2x2x3x3x5x7
    2+2+2+3+3+5+7 = 24 and is 7 numbers, so we need to get to 25 by combining two of the numbers.
    2+2+6+3+5+7 = 25 and is 6 numbers, so combine the two which doesn't change their sum.
    4+6+3+5+7 = 25 and 4x6x3x5x7 = 2520.

    as this is the only answer, a nicer question might have been "5 numbers have a sum of 25 and product of 2520. Are the number's consecutive?"
  5. Standard memberwolfgang59
    howling mad
    In the den
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    04 Jan '08 12:46
    to start with factorise 2520

    the answer pops out when you realise 5 numbers adding to 25 are gonna be averaging 5....

    anyone clever enough to prove answer is unique? I'm not!
  6. Standard memberMitsurugi
    Not the dead
    Porto, Portugal
    Joined
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    04 Jan '08 15:241 edit
    def notInSolutions(sols, i1, i2, i3, i4, i5):
    for sol in sols:
    if i1 in sol and i2 in sol and i3 in sol and i4 in sol and i5 in sol:
    return False
    return True

    solutions = []
    for i1 in xrange(25):
    for i2 in xrange(25):
    for i3 in xrange(25):
    for i4 in xrange(25):
    for i5 in xrange(25):
    if i1+i2+i3+i4+i5 == 25 and i1*i2*i3*i4*i5 == 2520:
    if notInSolutions(solutions,i1,i2,i3,i4,i5):
    print i1,i2,i3,i4,i5
    solutions.append([i1,i2,i3,i4,i5])

    Run this in Python and you'll find out that the only solution is any combination of the one posted above 🙂

    EDIT: The editor here messed up my tabs so you'll have to actually tab it correctly for it to work in Python. Nevertheless notice that this is not an optimal implementation of a solver for this problem :p
  7. Joined
    11 Nov '05
    Moves
    43938
    04 Jan '08 16:41
    Thank you all of you for the effort.
    Yes, the numbers are 3, 4, 5, 6, and 7

    Trial and error is on good method, but I liked the faktorisation!

    If the integers don't have to be positive, is there any more solutions then? (I don't know myself.)
  8. Joined
    26 Nov '07
    Moves
    1085
    04 Jan '08 16:57
    Originally posted by wolfgang59
    to start with factorise 2520

    the answer pops out when you realise 5 numbers adding to 25 are gonna be averaging 5....

    anyone clever enough to prove answer is unique? I'm not!
    No, it's not unique. We're working over the integers, so {21, 5, 4, -3, -2} works.
  9. Standard memberwolfgang59
    howling mad
    In the den
    Joined
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    04 Jan '08 21:10
    Nice one. I didnt think the solution was unique!
  10. Joined
    26 Nov '07
    Moves
    1085
    04 Jan '08 21:18
    Originally posted by wolfgang59
    Nice one. I didnt think the solution was unique!
    I actually assumed it was a trick question and was looking for positive and negative numbers from the start. It was only after I had my solution that I read the rest of the thread, and that actually I was being overly complicated...
  11. Standard memberMitsurugi
    Not the dead
    Porto, Portugal
    Joined
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    11360
    06 Jan '08 03:37
    Originally posted by Swlabr
    No, it's not unique. We're working over the integers, so {21, 5, 4, -3, -2} works.
    Ah negative numbers.... didn't think of that :p
  12. Joined
    26 Nov '07
    Moves
    1085
    06 Jan '08 09:201 edit
    Originally posted by FabianFnas
    Thank you all of you for the effort.
    Yes, the numbers are 3, 4, 5, 6, and 7

    Trial and error is on good method, but I liked the faktorisation!

    If the integers don't have to be positive, is there any more solutions then? (I don't know myself.)
    Edited while I think about it some more...
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