- #1

jobyts

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where a, b, c are positive integers.

I can think of only one solution to this. {1, 2, 3}.

Is there any other solution to it?

Can you prove or disprove?

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- Thread starter jobyts
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- #1

jobyts

- 218

- 58

where a, b, c are positive integers.

I can think of only one solution to this. {1, 2, 3}.

Is there any other solution to it?

Can you prove or disprove?

- #2

Office_Shredder

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[itex] a*b*c \geq 4a [/itex]

Which is necessarily larger than [itex] a+b+c \leq 3a[/itex].

So c=1 necessarily. Then we have

[itex] a+b+1 = a*b[/itex]

Now assume that b>2. The right hand side is at least 3a, and the left hand side is smaller than 2a+1, and we know that a is larger than 1 so these two cannot be equal. Therefore b=1 or b=2

If b=1 and c=1 there is obviously no solution (we get a+2 = a). If c=1 and b=2 we get a+3 = 2a which is solved by a=3. So {1,2,3} is the only positive integer solution.

- #3

jobyts

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Thank you.

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