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Hi guys,
find all solutions of equation a + b + c = a . b . c, where a ≤ b ≤ 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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Hi jozou;
You are correct that is the only solution but I do not have a proof to that.
Welcome to the forum.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi jozou
The restrictions on a, b and c make it very likely that's the only solution and a proof by exhaustion (trying all cases) should do it.
If a = b = c = 3 then the product is way too big compared with the sum.
Make any one variable bigger than three and the product gets even larger than before.
So you only need to consider cases less than or equal to three. That about wraps it up for one solution only.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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a + b + c = abc
abc - a = b + c
a(bc - 1) = b + c
Last edited by reconsideryouranswer (2011-08-03 02:03:49)
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hi reconsideryouranswer
Where did that line come from?
If c = 3 then b le 1.6666 which would force b to be 1 ??
I think the end of your proof needs a re-think.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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One guy found really nice proof:
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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hi jouzo
If a > 1, then 4c ≤ a.b.c = a+b+c. So 3c ≤ a+b, what contradicts a ≤ b ≤ c.
Don't understand where the 4c bit came from. Can you expand this please?
Bob
If a > 1, then a ≥ 2. Next b ≥ a, so b ≥ 2. Hence a.b ≥ 4.
So a+b+c = a.b.c ≥ 4c.
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hi
Sorry. I was being stupid. (too early in the morning for me) I worked it out, so I deleted my post. Thought you had gone off-line.
Nice proof. Thanks.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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hi reconsideryouranswer
Where did that line come from?
If c = 3 then b le 1.6666 which would force b to be 1 ??
I think the end of your proof needs a re-think.
Bob
My post has been reedited to show the
correct denominator of (c - 1).
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I wish a had a more interesting signature line.
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hi reconsideryouranswer,
Thanks for the edit. That makes good sense to me now.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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