Results 1 to 5 of 5

Math Help - "sum of numbers = product of that numbers" problem

  1. #1
    Newbie
    Joined
    Jul 2011
    Posts
    10

    "sum of numbers = product of that numbers" problem

    Hi guys,

    find all solutions of equation a + b + c = a . b . c, where a \leq b \leq c are positive integers.

    Well, by trial and error method I found out that 1+2+3 = 1.2.3
    Probably there is no other solution, but how can I proove that? I tried case analysis, but unsuccessfully. Any idea? Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: "sum of numbers = product of that numbers" problem

    Quote Originally Posted by jozou View Post
    Hi guys,

    find all solutions of equation a + b + c = a . b . c, where a \leq b \leq c are positive integers.

    Well, by trial and error method I found out that 1+2+3 = 1.2.3
    Probably there is no other solution, but how can I proove that? I tried case analysis, but unsuccessfully. Any idea? Thanks!


    Hint(I believe):

    \frac{a+b+c}{3}\geq \sqrt[3]{abc}

    a+b+c\geq 3\sqrt[3]{abc} <abc

    The last inequality holds for: abc\geq 6
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2010
    From
    Israel
    Posts
    148

    Re: "sum of numbers = product of that numbers" problem

    Quote Originally Posted by jozou View Post
    Hi guys,

    find all solutions of equation a + b + c = a . b . c, where a \leq b \leq c are positive integers.

    Well, by trial and error method I found out that 1+2+3 = 1.2.3
    Probably there is no other solution, but how can I proove that? I tried case analysis, but unsuccessfully. Any idea? Thanks!
    I try a straighforward approach.

    The integers a,b,c cannot be equal; otherwise 3a=a+b+c=abc=a^3 and then a^2=3, but 3 is not a perfect square.

    So the following inequality holds: a+b+c<c+c+c=3c. Now if ab\geq3, then 3c\leq abc implying a+b+c<abc.
    Therefore, ab<3, i.e. a=1 and b=2. The equation a+b+c=abc gives 3+c=2c and hence c=3.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2011
    Posts
    10

    Re: "sum of numbers = product of that numbers" problem

    Quote Originally Posted by melese View Post
    I try a straighforward approach.

    The integers a,b,c cannot be equal; otherwise 3a=a+b+c=abc=a^3 and then a^2=3, but 3 is not a perfect square.

    So the following inequality holds: a+b+c<c+c+c=3c. Now if ab\geq3, then 3c\leq abc implying a+b+c<abc.
    Therefore, ab<3, i.e. a=1 and b=2. The equation a+b+c=abc gives 3+c=2c and hence c=3.
    Thank you. This proof is great!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2011
    Posts
    10

    Re: "sum of numbers = product of that numbers" problem

    another nice proof (I'm not an author):

    Let a ≤ b ≤ c be positive integers, s.t. equation a+b+c = a.b.c holds.

    If a > 1, then 4c ≤ a.b.c = a+b+c. So 3c ≤ a+b, what contradicts a ≤ b ≤ c.
    So a= 1. Now we have to solve 1+b+c = b.c. Then b.(c-1) = 1+b+c - b = 1 + c, (b-1).(c-1) = 1 + c - (c-1) = 2.
    So (b-1).(c-1) = 2. Hence the only solution satistying assumptions is b = 2, c = 3.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 6th 2011, 03:00 PM
  2. Replies: 0
    Last Post: June 26th 2011, 01:56 AM
  3. Euclid's "perfect numbers theorem"
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: October 1st 2010, 01:48 AM
  4. Replies: 1
    Last Post: July 16th 2009, 01:06 PM
  5. The old "Four numbers equal 24" problem
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: February 16th 2009, 05:44 AM

/mathhelpforum @mathhelpforum