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

  1. #1
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    "sum of numbers = product of that numbers" problem

    Hi guys,

    find all solutions of equation a + b + c = a . b . c, where $\displaystyle a \leq b \leq c$ 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!
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  2. #2
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    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 $\displaystyle a \leq b \leq c$ 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):

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

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

    The last inequality holds for: $\displaystyle abc\geq 6$
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  3. #3
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    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 $\displaystyle a \leq b \leq c$ 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 $\displaystyle a,b,c$ cannot be equal; otherwise $\displaystyle 3a=a+b+c=abc=a^3$ and then $\displaystyle a^2=3$, but $\displaystyle 3$ is not a perfect square.

    So the following inequality holds: $\displaystyle a+b+c<c+c+c=3c$. Now if $\displaystyle ab\geq3$, then $\displaystyle 3c\leq abc$ implying $\displaystyle a+b+c<abc$.
    Therefore, $\displaystyle ab<3$, i.e. $\displaystyle a=1$ and $\displaystyle b=2$. The equation $\displaystyle a+b+c=abc$ gives $\displaystyle 3+c=2c$ and hence $\displaystyle c=3$.
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  4. #4
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    Re: "sum of numbers = product of that numbers" problem

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

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

    So the following inequality holds: $\displaystyle a+b+c<c+c+c=3c$. Now if $\displaystyle ab\geq3$, then $\displaystyle 3c\leq abc$ implying $\displaystyle a+b+c<abc$.
    Therefore, $\displaystyle ab<3$, i.e. $\displaystyle a=1$ and $\displaystyle b=2$. The equation $\displaystyle a+b+c=abc$ gives $\displaystyle 3+c=2c$ and hence $\displaystyle c=3$.
    Thank you. This proof is great!
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  5. #5
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    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.
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