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Product of $k$ distinct positive integers is divisible by its sum
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Is it true that for every positive integer $k, n$ satisfying $2 leq k leq n$, there exist $n$ distinct positive integers such that the product of any $k$ integers selected from those $n$ integers is divisible by the sum of that $k$ integers?
It can be seen that the statement is true for $n = k$ with $k+1$ is not a prime (for example, choose $1, 2, ... , k$ , we have $k!$ is divisible by $1+2+..+k$),
however I cannot proof or find any $n$ different positive integers that satisfy the statement with $n > k$ or $k+1$ is a prime.
combinatorics number-theory elementary-number-theory
edited Aug 26 at 14:31
Asaf Karagila♦
294k31410737
asked Aug 26 at 11:12
Mathwriter
395
 |Â
show 6 more comments
up vote
2
down vote
favorite
Is it true that for every positive integer $k, n$ satisfying $2 leq k leq n$, there exist $n$ distinct positive integers such that the product of any $k$ integers selected from those $n$ integers is divisible by the sum of that $k$ integers?
It can be seen that the statement is true for $n = k$ with $k+1$ is not a prime (for example, choose $1, 2, ... , k$ , we have $k!$ is divisible by $1+2+..+k$),
however I cannot proof or find any $n$ different positive integers that satisfy the statement with $n > k$ or $k+1$ is a prime.
combinatorics number-theory elementary-number-theory
edited Aug 26 at 14:31
Asaf Karagila♦
294k31410737
asked Aug 26 at 11:12
Mathwriter
395
Where did you find this problem?
– Jam
Aug 26 at 11:14
@Jam I don't know. Was it already asked in this community?
– Mathwriter
Aug 26 at 11:18
1
@quasi I'm not sure those solutions both work? $(3+12)nmid3cdot12$ and $(4+20)nmid4cdot20$.
– Jam
Aug 26 at 13:02
1
$30,60,120$. $(30)(60)/(90)=20$, $(30)(120)/(150)=24$, $(60)(120)/(180)=40$.
– Gerry Myerson
Aug 26 at 13:07
1
Another one for $(k,n)=(2,3)$ is $15,30,60$.
– quasi
Aug 26 at 13:09
 |Â
show 6 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is it true that for every positive integer $k, n$ satisfying $2 leq k leq n$, there exist $n$ distinct positive integers such that the product of any $k$ integers selected from those $n$ integers is divisible by the sum of that $k$ integers?
It can be seen that the statement is true for $n = k$ with $k+1$ is not a prime (for example, choose $1, 2, ... , k$ , we have $k!$ is divisible by $1+2+..+k$),
however I cannot proof or find any $n$ different positive integers that satisfy the statement with $n > k$ or $k+1$ is a prime.
combinatorics number-theory elementary-number-theory
edited Aug 26 at 14:31
Asaf Karagila♦
294k31410737
asked Aug 26 at 11:12
Mathwriter
395
Is it true that for every positive integer $k, n$ satisfying $2 leq k leq n$, there exist $n$ distinct positive integers such that the product of any $k$ integers selected from those $n$ integers is divisible by the sum of that $k$ integers?
It can be seen that the statement is true for $n = k$ with $k+1$ is not a prime (for example, choose $1, 2, ... , k$ , we have $k!$ is divisible by $1+2+..+k$),
however I cannot proof or find any $n$ different positive integers that satisfy the statement with $n > k$ or $k+1$ is a prime.
combinatorics number-theory elementary-number-theory
edited Aug 26 at 14:31
Asaf Karagila♦
294k31410737
asked Aug 26 at 11:12
Mathwriter
395
edited Aug 26 at 14:31
Asaf Karagila♦
294k31410737
edited Aug 26 at 14:31
Asaf Karagila♦
294k31410737
edited Aug 26 at 14:31
Asaf Karagila♦
294k31410737
294k31410737
asked Aug 26 at 11:12
Mathwriter
395
asked Aug 26 at 11:12
Mathwriter
395
asked Aug 26 at 11:12
Mathwriter
395
395
Where did you find this problem?
– Jam
Aug 26 at 11:14
@Jam I don't know. Was it already asked in this community?
– Mathwriter
Aug 26 at 11:18
1
@quasi I'm not sure those solutions both work? $(3+12)nmid3cdot12$ and $(4+20)nmid4cdot20$.
– Jam
Aug 26 at 13:02
1
$30,60,120$. $(30)(60)/(90)=20$, $(30)(120)/(150)=24$, $(60)(120)/(180)=40$.
– Gerry Myerson
Aug 26 at 13:07
1
Another one for $(k,n)=(2,3)$ is $15,30,60$.
– quasi
Aug 26 at 13:09
 |Â
show 6 more comments
Where did you find this problem?
– Jam
Aug 26 at 11:14
@Jam I don't know. Was it already asked in this community?
– Mathwriter
Aug 26 at 11:18
1
@quasi I'm not sure those solutions both work? $(3+12)nmid3cdot12$ and $(4+20)nmid4cdot20$.
– Jam
Aug 26 at 13:02
1
$30,60,120$. $(30)(60)/(90)=20$, $(30)(120)/(150)=24$, $(60)(120)/(180)=40$.
– Gerry Myerson
Aug 26 at 13:07
1
Another one for $(k,n)=(2,3)$ is $15,30,60$.
– quasi
Aug 26 at 13:09
Where did you find this problem?
– Jam
Aug 26 at 11:14
Where did you find this problem?
– Jam
Aug 26 at 11:14
@Jam I don't know. Was it already asked in this community?
– Mathwriter
Aug 26 at 11:18
@Jam I don't know. Was it already asked in this community?
– Mathwriter
Aug 26 at 11:18
1
1
@quasi I'm not sure those solutions both work? $(3+12)nmid3cdot12$ and $(4+20)nmid4cdot20$.
– Jam
Aug 26 at 13:02
@quasi I'm not sure those solutions both work? $(3+12)nmid3cdot12$ and $(4+20)nmid4cdot20$.
– Jam
Aug 26 at 13:02
1
1
$30,60,120$. $(30)(60)/(90)=20$, $(30)(120)/(150)=24$, $(60)(120)/(180)=40$.
– Gerry Myerson
Aug 26 at 13:07
$30,60,120$. $(30)(60)/(90)=20$, $(30)(120)/(150)=24$, $(60)(120)/(180)=40$.
– Gerry Myerson
Aug 26 at 13:07
1
1
Another one for $(k,n)=(2,3)$ is $15,30,60$.
– quasi
Aug 26 at 13:09
Another one for $(k,n)=(2,3)$ is $15,30,60$.
– quasi
Aug 26 at 13:09
 |Â
show 6 more comments
2 Answers
2
active
oldest
votes
up vote
5
down vote
Suppose $a_1a_2cdots a_k$ is not divisible by $a_1+cdots+a_k$. You can fix it by choosing $d$ such that $(a_1d)(a_2d)cdots(a_kd)$ is divisible by $a_1d+cdots+a_kd=(a_1+cdots+a_k)d$. So for each set of $k$ of your $n$ numbers there a multiplier $d$ that fixes it. Take any common multiple of all these values of $d$, and multiply all the $a_i$ by it, and you've fixed everything.
answered Aug 26 at 13:14
Gerry Myerson
143k8145295
1
(+1) As a side-note, if I'm not mistaken, you can transform this answer into @quasi's answer by defining $d=a_1+ldots+a_k$, for each subset $a_1ldots a_k$. This would fulfil your criteria. And then the product of each $d$ over all subsets would be @quasi's "$a$" or your "common multiple".
– Jam
Aug 26 at 14:03
add a comment |Â
up vote
4
down vote
Fix positive integers $k,n$ with $2le k le n$.
Let $x_1,...,x_n$ be any $n$ distinct positive integers, and let $a$ be the product of all sums of $k$-element subsets of $x_1,...,x_n$.
Define $y_1,...,y_n$ by $y_i=ax_i$.
Then the product of any $k$ elements of $y_1,...,y_n$ is a multiple of $a^2$, hence is a multiple of the sum of those $k$ elements.
answered Aug 26 at 13:21
quasi
33.9k22461
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Suppose $a_1a_2cdots a_k$ is not divisible by $a_1+cdots+a_k$. You can fix it by choosing $d$ such that $(a_1d)(a_2d)cdots(a_kd)$ is divisible by $a_1d+cdots+a_kd=(a_1+cdots+a_k)d$. So for each set of $k$ of your $n$ numbers there a multiplier $d$ that fixes it. Take any common multiple of all these values of $d$, and multiply all the $a_i$ by it, and you've fixed everything.
answered Aug 26 at 13:14
Gerry Myerson
143k8145295
1
(+1) As a side-note, if I'm not mistaken, you can transform this answer into @quasi's answer by defining $d=a_1+ldots+a_k$, for each subset $a_1ldots a_k$. This would fulfil your criteria. And then the product of each $d$ over all subsets would be @quasi's "$a$" or your "common multiple".
– Jam
Aug 26 at 14:03
add a comment |Â
up vote
5
down vote
Suppose $a_1a_2cdots a_k$ is not divisible by $a_1+cdots+a_k$. You can fix it by choosing $d$ such that $(a_1d)(a_2d)cdots(a_kd)$ is divisible by $a_1d+cdots+a_kd=(a_1+cdots+a_k)d$. So for each set of $k$ of your $n$ numbers there a multiplier $d$ that fixes it. Take any common multiple of all these values of $d$, and multiply all the $a_i$ by it, and you've fixed everything.
answered Aug 26 at 13:14
Gerry Myerson
143k8145295
1
(+1) As a side-note, if I'm not mistaken, you can transform this answer into @quasi's answer by defining $d=a_1+ldots+a_k$, for each subset $a_1ldots a_k$. This would fulfil your criteria. And then the product of each $d$ over all subsets would be @quasi's "$a$" or your "common multiple".
– Jam
Aug 26 at 14:03
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Suppose $a_1a_2cdots a_k$ is not divisible by $a_1+cdots+a_k$. You can fix it by choosing $d$ such that $(a_1d)(a_2d)cdots(a_kd)$ is divisible by $a_1d+cdots+a_kd=(a_1+cdots+a_k)d$. So for each set of $k$ of your $n$ numbers there a multiplier $d$ that fixes it. Take any common multiple of all these values of $d$, and multiply all the $a_i$ by it, and you've fixed everything.
answered Aug 26 at 13:14
Gerry Myerson
143k8145295
Suppose $a_1a_2cdots a_k$ is not divisible by $a_1+cdots+a_k$. You can fix it by choosing $d$ such that $(a_1d)(a_2d)cdots(a_kd)$ is divisible by $a_1d+cdots+a_kd=(a_1+cdots+a_k)d$. So for each set of $k$ of your $n$ numbers there a multiplier $d$ that fixes it. Take any common multiple of all these values of $d$, and multiply all the $a_i$ by it, and you've fixed everything.
answered Aug 26 at 13:14
Gerry Myerson
143k8145295
answered Aug 26 at 13:14
Gerry Myerson
143k8145295
answered Aug 26 at 13:14
Gerry Myerson
143k8145295
answered Aug 26 at 13:14
Gerry Myerson
143k8145295
143k8145295
1
(+1) As a side-note, if I'm not mistaken, you can transform this answer into @quasi's answer by defining $d=a_1+ldots+a_k$, for each subset $a_1ldots a_k$. This would fulfil your criteria. And then the product of each $d$ over all subsets would be @quasi's "$a$" or your "common multiple".
– Jam
Aug 26 at 14:03
add a comment |Â
1
(+1) As a side-note, if I'm not mistaken, you can transform this answer into @quasi's answer by defining $d=a_1+ldots+a_k$, for each subset $a_1ldots a_k$. This would fulfil your criteria. And then the product of each $d$ over all subsets would be @quasi's "$a$" or your "common multiple".
– Jam
Aug 26 at 14:03
1
1
(+1) As a side-note, if I'm not mistaken, you can transform this answer into @quasi's answer by defining $d=a_1+ldots+a_k$, for each subset $a_1ldots a_k$. This would fulfil your criteria. And then the product of each $d$ over all subsets would be @quasi's "$a$" or your "common multiple".
– Jam
Aug 26 at 14:03
(+1) As a side-note, if I'm not mistaken, you can transform this answer into @quasi's answer by defining $d=a_1+ldots+a_k$, for each subset $a_1ldots a_k$. This would fulfil your criteria. And then the product of each $d$ over all subsets would be @quasi's "$a$" or your "common multiple".
– Jam
Aug 26 at 14:03
add a comment |Â
up vote
4
down vote
Fix positive integers $k,n$ with $2le k le n$.
Let $x_1,...,x_n$ be any $n$ distinct positive integers, and let $a$ be the product of all sums of $k$-element subsets of $x_1,...,x_n$.
Define $y_1,...,y_n$ by $y_i=ax_i$.
Then the product of any $k$ elements of $y_1,...,y_n$ is a multiple of $a^2$, hence is a multiple of the sum of those $k$ elements.
answered Aug 26 at 13:21
quasi
33.9k22461
add a comment |Â
up vote
4
down vote
Fix positive integers $k,n$ with $2le k le n$.
Let $x_1,...,x_n$ be any $n$ distinct positive integers, and let $a$ be the product of all sums of $k$-element subsets of $x_1,...,x_n$.
Define $y_1,...,y_n$ by $y_i=ax_i$.
Then the product of any $k$ elements of $y_1,...,y_n$ is a multiple of $a^2$, hence is a multiple of the sum of those $k$ elements.
answered Aug 26 at 13:21
quasi
33.9k22461
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Fix positive integers $k,n$ with $2le k le n$.
Let $x_1,...,x_n$ be any $n$ distinct positive integers, and let $a$ be the product of all sums of $k$-element subsets of $x_1,...,x_n$.
Define $y_1,...,y_n$ by $y_i=ax_i$.
Then the product of any $k$ elements of $y_1,...,y_n$ is a multiple of $a^2$, hence is a multiple of the sum of those $k$ elements.
answered Aug 26 at 13:21
quasi
33.9k22461
Fix positive integers $k,n$ with $2le k le n$.
Let $x_1,...,x_n$ be any $n$ distinct positive integers, and let $a$ be the product of all sums of $k$-element subsets of $x_1,...,x_n$.
Define $y_1,...,y_n$ by $y_i=ax_i$.
Then the product of any $k$ elements of $y_1,...,y_n$ is a multiple of $a^2$, hence is a multiple of the sum of those $k$ elements.
answered Aug 26 at 13:21
quasi
33.9k22461
answered Aug 26 at 13:21
quasi
33.9k22461
answered Aug 26 at 13:21
quasi
33.9k22461
answered Aug 26 at 13:21
quasi
33.9k22461
33.9k22461
add a comment |Â
add a comment |Â
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Where did you find this problem?
– Jam
Aug 26 at 11:14
@Jam I don't know. Was it already asked in this community?
– Mathwriter
Aug 26 at 11:18
1
@quasi I'm not sure those solutions both work? $(3+12)nmid3cdot12$ and $(4+20)nmid4cdot20$.
– Jam
Aug 26 at 13:02
1
$30,60,120$. $(30)(60)/(90)=20$, $(30)(120)/(150)=24$, $(60)(120)/(180)=40$.
– Gerry Myerson
Aug 26 at 13:07
1
Another one for $(k,n)=(2,3)$ is $15,30,60$.
– quasi
Aug 26 at 13:09