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## #1 2005-08-03 01:03:43

eleusis
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### Easy problem

Find three positive whole numbers such that for any two of them, the number one less than their product is divisible by the third number.

This is one of my favorite integer problems.  The wording is a little off.  I think it should read, "their product is evenly divisible", no fractions.  I think there is also a proof that shows there is exactly one solution to this problem.  When posting solutions, please post your method as well.

(x × y – 1)  ÷ z = (int)
(y × z – 1)  ÷ x = (int)
(z × x – 1)  ÷ y = (int)

## #2 2005-08-03 01:40:27

mathsyperson
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### Re: Easy problem

Does 1, 1 and 1 count?

Why did the vector cross the road?
It wanted to be normal.

## #3 2005-08-03 01:59:58

eleusis
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### Re: Easy problem

Ahh, good question.  For this solution, x ≠y ≠z.  For all values of x, y, z such that they are all integers, positive, and unique.

## #4 2005-08-03 04:10:48

mathsyperson
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### Re: Easy problem

Here's my reasoning, hidden for anyone who wants to do it themselves:

Sorry to anyone uptight about punctuation, but the hide tag has a problem with apostrophes.
Anyway, if you read that, you can see that it was mostly guesswork that got it and we're still nowhere near proving that that's the only combination. I've got as far as showing that there needs to be 1 even and 2 odd, but beyond there I'm stuck.

Last edited by mathsyperson (2005-08-03 04:12:21)

Why did the vector cross the road?
It wanted to be normal.

## #5 2005-08-03 05:53:42

kylekatarn
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### Re: Easy problem

I coded a small program that searches for solutions.
Below is a copy of the output file.

#### Code:

```Bounded integer solutions list
Bounds: x:[1;500] y:[1;500] z:[1;500]
--------------------------
x           y           z
2           3           5
2           5           3
3           2           5
3           5           2
5           2           3
5           3           2
--------------------------
6 solutions found```

And with more and more attempts I always get this result. 6 Solutions! Could it be?
I know that brute-forcing the equation WILL NEVER PROVE that *ONLY* 6 solutions exist...
Even if I asked the program to search within bounds [1 to 50000000000]...we don't know if 50000000001 has a solution...
Its not the maths way. But it can give us great clues!
Just for curiosity...even from 1 to 5000 there are still only these 6 solutions : )

## #6 2005-08-03 05:58:47

eleusis
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### Re: Easy problem

mathsyperson got it.  He used the same method that I did, "plug-and-chug".  Sometimes that's the best way to solve a problem.  I also wrote a program to cycle all combinations of all integers from 1-100,000 and only found one unique solution.  Later, I found on the same website that I originally found the problem a proof showing there was a single unique solution.

I said it was easy.

## #7 2005-08-03 06:01:53

eleusis
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### Re: Easy problem

kylekatarn is also correct.  Technically, there are six solutions but they are all permutations of the same three numbers.

I know I am new here but I'm quite impressed with the speed that you found an answer.

## #8 2005-08-03 16:09:14

MathsIsFun
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### Re: Easy problem

#### mathsyperson wrote:

... but the hide tag has a problem with apostrophes.

I think I got the hide tag straightened out. Quotes should work now

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #9 2005-08-03 17:11:24

ganesh
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### Re: Easy problem

#### eleusis wrote:

Later, I found on the same website that I originally found the problem a proof showing there was a single unique solution.

Can you post the proof? I got as far as Mathsy did, that two of the numbers should be odd and one even. Maybe, the proof has got something to do with 2 being the only even prime

Character is who you are when no one is looking.

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