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

1. "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!

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

Originally Posted by jozou
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: $\displaystyle abc\geq 6$

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

Originally Posted by jozou
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 $\displaystyle c=3$.

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

Originally Posted by melese
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!

5. 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.