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Posers and Puzzles

Posers and Puzzles

  1. Standard member wolfgang59
    Infidel
    25 Oct '07 17:14 / 1 edit
    I have 3 numbers, lets call them a, b and c. I am surprised to see that one of the pairs is such that the difference between the two numbers is exactly equal to the difference between their squares.

    ie a-b = a^2-b^2

    when I look at the other pairs I find that in one pair the difference between their squares is twice their difference and in the final pair the difference between their squares is thrice their difference.

    What is the sum of a, b and c?

    (The problem is easy ... your solution must be concise!)
  2. 25 Oct '07 17:58 / 1 edit
    I think you need one further condition...unless you state that a, b and c are different numbers then there is no unique solution.
  3. 25 Oct '07 18:20
    if a!=b!=c, then the answer is 3.
  4. Standard member PBE6
    Bananarama
    25 Oct '07 18:26
    Originally posted by wolfgang59
    I have 3 numbers, lets call them a, b and c. I am surprised to see that one of the pairs is such that the difference between the two numbers is exactly equal to the difference between their squares.

    ie a-b = a^2-b^2

    when I look at the other pairs I find that in one pair the difference between their squares is twice their difference and in the final ...[text shortened]... e.

    What is the sum of a, b and c?

    (The problem is easy ... your solution must be concise!)
    At first I thought there were no solutions if a =/= b =/= c, but after a little algebra I believe I have an answer, and that a+b+c = 3.
  5. Standard member PBE6
    Bananarama
    25 Oct '07 18:27 / 2 edits
    Dang! Didn't see your post there zzyw.
  6. Standard member wolfgang59
    Infidel
    26 Oct '07 08:37
    Originally posted by PBE6
    At first I thought there were no solutions if a =/= b =/= c, but after a little algebra I believe I have an answer, and that a+b+c = 3.
    Yes. The solution is unique and a+b+c=3

    But how did you get there? Its the technique I was after!!!

    What if the ratios in question are; 1, 49.5 & 149.5 ?????


    Show your working for the marks!!

  7. Standard member wolfgang59
    Infidel
    26 Oct '07 08:41
    Originally posted by zzyw
    if a!=b!=c, then the answer is 3.
    This is NOT true.
  8. 26 Oct '07 09:05 / 2 edits
    Originally posted by wolfgang59
    But how did you get there? Its the technique I was after!!
    a^2 - b^2 = a - b
    b^2 - c^2 = 2(b - c)
    c^2 - a^2 = 3(c - a)

    Note that a^2 - b^2 = (a - b)(a + b)

    If a, b and c are all different, we can divide each line by (a - b), (b - c) and (c - a) respectively.

    a + b = 1
    b + c = 2
    c + a = 3

    Add them up (and divide by 2): a + b + c = 3.

    If the ratios are p, q and r, then the same process shows that:
    a + b + c = (p + q + r)/2


    If you allow a, b and c to be equal, then you always have (0, 0, 0) as a possible solution.
  9. Standard member wolfgang59
    Infidel
    26 Oct '07 09:18
    Originally posted by mtthw
    a^2 - b^2 = a - b
    b^2 - c^2 = 2(b - c)
    c^2 - a^2 = 3(c - a)

    Note that a^2 - b^2 = (a - b)(a + b)

    If a, b and c are all different, we can divide each line by (a - b), (b - c) and (c - a) respectively.

    a + b = 1
    b + c = 2
    c + a = 3

    Add them up (and divide by 2): a + b + c = 3.

    If the ratios are p, q and r, then the same process shows that:
    a + b + c = (p + q + r)/2
    Note that a^2 - b^2 = (a - b)(a + b)

    Yes!

    If you dont spot this and instead solve the simultaneous equations you can get in a mess .. particularly for the second scenario!!