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- #1

- Mar 10, 2012

- 834

- Thread starter caffeinemachine
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- Thread starter
- #1

- Mar 10, 2012

- 834

- Jan 26, 2012

- 268

Total number of pairwise differences $\displaystyle =\binom{20}{2}=190$. Each difference must belong to the set $\{1,\ 2,\ \ldots,\ 68\}$: there are only 68 distinct values for each pairwise difference.

I don't know how to proceed...

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- #3

- Mar 10, 2012

- 834

How does "Challenging Puzzles.." forum work?Total number of pairwise differences $\displaystyle =\binom{20}{2}=190$. Each difference must belong to the set $\{1,\ 2,\ \ldots,\ 68\}$: there are only 68 distinct values for each pairwise difference.

I don't know how to proceed...

Am I supposed to post the solution(in case no one is able to solve it) within a day, a week or what?

- Jan 26, 2012

- 268

If no one replied, you could do that. But since I've posted an attempt, you could just give me a hint.How does "Challenging Puzzles.." forum work?

Am I supposed to post the solution(in case no one is able to solve it) within a day, a week or what?

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- #5

- Mar 10, 2012

- 834

Let the $20$ numbers be ordered as $a_1 < a_2 < \ldots < a_{20}$.If no one replied, you could do that. But since I've posted an attempt, you could just give me a hint.

Consider the differences:

$a_{20}-a_{19}, a_{19}-a_{18}, \ldots, a_2-a_1$

- Jan 26, 2012

- 268

Assume, for the sake of argument, that among the pairwise differences there are at most 3 equal numbers.Let the $20$ numbers be ordered as $a_1 < a_2 < \ldots < a_{20}$.

Consider the differences:

$a_{20}-a_{19}, a_{19}-a_{18}, \ldots, a_2-a_1$

$(a_{20}-a_{19})+(a_{19}-a_{18})+\ldots+(a_2-a_1)\ge 1+1+1+2+2+2+\ldots+6+6+6+7=3*6*7/2+7=70$

$a_{20}-a_1\ge 70$

$a_{20}\ge 70+a_1>70$ (contradiction)

So, among the pairwise differences there are at least 4 equal numbers.

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- #7

- Mar 10, 2012

- 834

Great!Assume, for the sake of argument, that among the pairwise differences there are at most 3 equal numbers.

$(a_{20}-a_{19})+(a_{19}-a_{18})+\ldots+(a_2-a_1)\ge 1+1+1+2+2+2+\ldots+6+6+6+7=3*6*7/2+7=70$

$a_{20}-a_1\ge 70$

$a_{20}\ge 70+a_1>70$ (contradiction)

So, among the pairwise differences there are at least 4 equal numbers.