Existence of 2006 distinct natural numbers
$begingroup$
Does there exists $2006$ distinct natural numbers such that the sum of any two divides the sum of all the given numbers?
combinatorics elementary-number-theory recreational-mathematics
asked Jan 15 at 7:11
Jerry TaoJerry Tao
906
$endgroup$
add a comment |
$begingroup$
Does there exists $2006$ distinct natural numbers such that the sum of any two divides the sum of all the given numbers?
combinatorics elementary-number-theory recreational-mathematics
asked Jan 15 at 7:11
Jerry TaoJerry Tao
906
$endgroup$
1
$begingroup$
Why $2006{}{}$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 7:14
$begingroup$
It's from a 2006 contest I guess. But I was told that it works for all $nge 3$.
$endgroup$
– Jerry Tao
Jan 15 at 7:18
2
$begingroup$
Perhaps this might help: cut-the-knot.org/arithmetic/NumbersDivideSums.shtml
$endgroup$
– crskhr
Jan 15 at 7:18
$begingroup$
Has to be a contest problem, indeed. Nevertheless, it is a good one.
$endgroup$
– Aaron
Jan 15 at 15:03
add a comment |
$begingroup$
Does there exists $2006$ distinct natural numbers such that the sum of any two divides the sum of all the given numbers?
combinatorics elementary-number-theory recreational-mathematics
asked Jan 15 at 7:11
Jerry TaoJerry Tao
906
$endgroup$
Does there exists $2006$ distinct natural numbers such that the sum of any two divides the sum of all the given numbers?
combinatorics elementary-number-theory recreational-mathematics
combinatorics elementary-number-theory recreational-mathematics
asked Jan 15 at 7:11
Jerry TaoJerry Tao
906
asked Jan 15 at 7:11
Jerry TaoJerry Tao
906
edited Jan 15 at 7:32
Jerry Tao
asked Jan 15 at 7:11
Jerry TaoJerry Tao
906
asked Jan 15 at 7:11
Jerry TaoJerry Tao
906
asked Jan 15 at 7:11
Jerry TaoJerry Tao
906
906
1
$begingroup$
Why $2006{}{}$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 7:14
$begingroup$
It's from a 2006 contest I guess. But I was told that it works for all $nge 3$.
$endgroup$
– Jerry Tao
Jan 15 at 7:18
2
$begingroup$
Perhaps this might help: cut-the-knot.org/arithmetic/NumbersDivideSums.shtml
$endgroup$
– crskhr
Jan 15 at 7:18
$begingroup$
Has to be a contest problem, indeed. Nevertheless, it is a good one.
$endgroup$
– Aaron
Jan 15 at 15:03
add a comment |
1
$begingroup$
Why $2006{}{}$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 7:14
$begingroup$
It's from a 2006 contest I guess. But I was told that it works for all $nge 3$.
$endgroup$
– Jerry Tao
Jan 15 at 7:18
2
$begingroup$
Perhaps this might help: cut-the-knot.org/arithmetic/NumbersDivideSums.shtml
$endgroup$
– crskhr
Jan 15 at 7:18
$begingroup$
Has to be a contest problem, indeed. Nevertheless, it is a good one.
$endgroup$
– Aaron
Jan 15 at 15:03
1
1
$begingroup$
Why $2006{}{}$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 7:14
$begingroup$
Why $2006{}{}$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 7:14
$begingroup$
It's from a 2006 contest I guess. But I was told that it works for all $nge 3$.
$endgroup$
– Jerry Tao
Jan 15 at 7:18
$begingroup$
It's from a 2006 contest I guess. But I was told that it works for all $nge 3$.
$endgroup$
– Jerry Tao
Jan 15 at 7:18
2
2
$begingroup$
Perhaps this might help: cut-the-knot.org/arithmetic/NumbersDivideSums.shtml
$endgroup$
– crskhr
Jan 15 at 7:18
$begingroup$
Perhaps this might help: cut-the-knot.org/arithmetic/NumbersDivideSums.shtml
$endgroup$
– crskhr
Jan 15 at 7:18
$begingroup$
Has to be a contest problem, indeed. Nevertheless, it is a good one.
$endgroup$
– Aaron
Jan 15 at 15:03
$begingroup$
Has to be a contest problem, indeed. Nevertheless, it is a good one.
$endgroup$
– Aaron
Jan 15 at 15:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer is no, and for any $n$, not just $n=2006$.
Suppose $a_1<a_2<cdots<a_n$ be a such numbers, with the desired property. Let $S=sum_{k=1}^n a_k$. Then observe that, $a_n+a_{n-1},a_n+a_{n-2},cdots,a_n+a_1$ all divide $S$, and
$$
frac{S}{a_n+a_1}>frac{S}{a_n+a_2}>cdots>frac{S}{a_n+a_{n-1}} geq 2.
$$
Hence,
$$
frac{S}{a_n+a_1}geq n+1 implies a_1+cdots+a_{n-1}geq na_n+(n+1)a_1,
$$
which is a clear contradiction.
answered Jan 15 at 14:58
AaronAaron
1,894415
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The answer is no, and for any $n$, not just $n=2006$.
Suppose $a_1<a_2<cdots<a_n$ be a such numbers, with the desired property. Let $S=sum_{k=1}^n a_k$. Then observe that, $a_n+a_{n-1},a_n+a_{n-2},cdots,a_n+a_1$ all divide $S$, and
$$
frac{S}{a_n+a_1}>frac{S}{a_n+a_2}>cdots>frac{S}{a_n+a_{n-1}} geq 2.
$$
Hence,
$$
frac{S}{a_n+a_1}geq n+1 implies a_1+cdots+a_{n-1}geq na_n+(n+1)a_1,
$$
which is a clear contradiction.
answered Jan 15 at 14:58
AaronAaron
1,894415
$endgroup$
add a comment |
$begingroup$
The answer is no, and for any $n$, not just $n=2006$.
Suppose $a_1<a_2<cdots<a_n$ be a such numbers, with the desired property. Let $S=sum_{k=1}^n a_k$. Then observe that, $a_n+a_{n-1},a_n+a_{n-2},cdots,a_n+a_1$ all divide $S$, and
$$
frac{S}{a_n+a_1}>frac{S}{a_n+a_2}>cdots>frac{S}{a_n+a_{n-1}} geq 2.
$$
Hence,
$$
frac{S}{a_n+a_1}geq n+1 implies a_1+cdots+a_{n-1}geq na_n+(n+1)a_1,
$$
which is a clear contradiction.
answered Jan 15 at 14:58
AaronAaron
1,894415
$endgroup$
add a comment |
$begingroup$
The answer is no, and for any $n$, not just $n=2006$.
Suppose $a_1<a_2<cdots<a_n$ be a such numbers, with the desired property. Let $S=sum_{k=1}^n a_k$. Then observe that, $a_n+a_{n-1},a_n+a_{n-2},cdots,a_n+a_1$ all divide $S$, and
$$
frac{S}{a_n+a_1}>frac{S}{a_n+a_2}>cdots>frac{S}{a_n+a_{n-1}} geq 2.
$$
Hence,
$$
frac{S}{a_n+a_1}geq n+1 implies a_1+cdots+a_{n-1}geq na_n+(n+1)a_1,
$$
which is a clear contradiction.
answered Jan 15 at 14:58
AaronAaron
1,894415
$endgroup$
The answer is no, and for any $n$, not just $n=2006$.
Suppose $a_1<a_2<cdots<a_n$ be a such numbers, with the desired property. Let $S=sum_{k=1}^n a_k$. Then observe that, $a_n+a_{n-1},a_n+a_{n-2},cdots,a_n+a_1$ all divide $S$, and
$$
frac{S}{a_n+a_1}>frac{S}{a_n+a_2}>cdots>frac{S}{a_n+a_{n-1}} geq 2.
$$
Hence,
$$
frac{S}{a_n+a_1}geq n+1 implies a_1+cdots+a_{n-1}geq na_n+(n+1)a_1,
$$
which is a clear contradiction.
answered Jan 15 at 14:58
AaronAaron
1,894415
answered Jan 15 at 14:58
AaronAaron
1,894415
answered Jan 15 at 14:58
AaronAaron
1,894415
answered Jan 15 at 14:58
AaronAaron
1,894415
1,894415
add a comment |
add a comment |
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1
$begingroup$
Why $2006{}{}$?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 7:14
$begingroup$
It's from a 2006 contest I guess. But I was told that it works for all $nge 3$.
$endgroup$
– Jerry Tao
Jan 15 at 7:18
2
$begingroup$
Perhaps this might help: cut-the-knot.org/arithmetic/NumbersDivideSums.shtml
$endgroup$
– crskhr
Jan 15 at 7:18
$begingroup$
Has to be a contest problem, indeed. Nevertheless, it is a good one.
$endgroup$
– Aaron
Jan 15 at 15:03