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"Find positive integer solutions to the equation:
."I played around with it a bit and got it into this form:
But I don't think this helps. I've seen one solution as a = 3, b = 3, c = 4 (by observation). And I've been able to rule out the case a = b = c (has no integer solutions). Could I have a hint on this one, please? (No solution if possible!)
I should clarify, I've been asked to find all the positive integer solutions.
I've noticed that the LHS is even so the RHS must also be even (as a! = 1, b! = 1 is not a possible pairing).
All factorials above 1! are even, so that doesn't tell you much.
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Hi;
That is the only solution in the interval:
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
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Hi;
That is the only solution in the interval:
Did you reverse \ge and \le, because there are no numbers in that interval.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
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hi zetafunc
a!b! = a! + b! + c! will re-arrange to
a! = (b! + c!)/(b! -1)
So you might try to prove that's the only solution.
[The olympiad question setters are looking for good maths ... that would include proving the negative case]
Maybe something like this:
if b and c ≥ 3 then the top of this fraction is divisible by 3
But b! -1 will not be ............
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 anonimnystefy;
Yes, thanks. The latex bug got me. Fixed it all up.
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 bobbym
You're welcome.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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hi zetafunc
a!b! = a! + b! + c! will re-arrange to
a! = (b! + c!)/(b! -1)
So you might try to prove that's the only solution.
[The olympiad question setters are looking for good maths ... that would include proving the negative case]
Maybe something like this:
if b and c ≥ 3 then the top of this fraction is divisible by 3
But b! -1 will not be ............
Bob
Thanks for the reply...
I think the negative case proof is unnecessary though as they said a, b and c are positive integers?
I'm not sure what to do with the re-arrangement you got... and have never really got anything solid. Just for example, if you consider
a!b! = a! + b! + c!
Then, we can assume without loss of generality that a ≤ b (from the symmetry). If we consider the case a = b, then dividing both sides by a! gives:
b! = (c! + 2) mod a!... which implies that b ≤ c? But not sure where to go from here. Maybe if I consider the different cases separately, e.g. c > a, a < c?
Hang on, maybe I mis-understood your re-arrangement... what do you do about the b! + 1 term?
(Sorry, I meant the b! - 1 term)
Hi zetafunc.
When both b and c are greater than 2, b! and c! will be divisible by 3, but b!-1 won't be divisible by 3. So you have the other few cases to check now.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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That doesn't really get you anywhere though. b! + c! is the numerator, b! - 1 is the denominator. You don't learn much by saying that the numerator has factors that aren't in the denominator. Consider the example we already know - 3! + 4! = 30, 3! - 1 = 5. The numerator is divisible by 3 and the denominator is not, but you can still divide them and get another factorial as the quotient.
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Hi;
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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Those solutions have some good ideas, but there are flaws in both of them.
Edit: Rather, the first solution is flawed, and while the second looks correct he skips a lot of steps and makes it hard to follow.
Last edited by TheDude (2012-07-11 02:59:20)
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Hi;
I can not check that. The whole page uses Mathjax to display its latex and I can hardly see it in my browser.
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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