Given $n$ integers, is it always possible to choose $m$ from them so that their sum is a multiple of $m$?
The original question: given $6666$ integers, (positive, negative or $0$) is it always possible to choose $2018$ from them so that the chosen numbers add to a multiple of $2018$? (positive multiple, negative multiple or $0$) Prove or disprove.
More generally, for what $n$ and $m$, given $n$ integers, it is always possible to choose $m$ from them so that their sum is a multiple of $m$?
I tried for small $m$s to find a minimum $n$ that fits, and my research shows (in $n$ for $m$ form): $1$ for $1$, $3$ for $2$, $5$ for $3$, $7$ for $4$, $9$ for $5$, $11$ for $6$, $13$ for $7$, $15$ for $8$. My program takes already quite long for the last case ($15$ for $8$), so I terminated it.
My guess from these cases is that $2m - 1$ is the least fit for $m$, and therefore the original question is true since $6666 > 4035 = 2 cdot 2018 - 1$, but I apparently have no idea about the proof.
Disclaimer: this is not homework. I fail to do it cuz I'm bad.
algebra-precalculus
asked Nov 28 '18 at 10:50
L. F.
1264
add a comment |
The original question: given $6666$ integers, (positive, negative or $0$) is it always possible to choose $2018$ from them so that the chosen numbers add to a multiple of $2018$? (positive multiple, negative multiple or $0$) Prove or disprove.
More generally, for what $n$ and $m$, given $n$ integers, it is always possible to choose $m$ from them so that their sum is a multiple of $m$?
I tried for small $m$s to find a minimum $n$ that fits, and my research shows (in $n$ for $m$ form): $1$ for $1$, $3$ for $2$, $5$ for $3$, $7$ for $4$, $9$ for $5$, $11$ for $6$, $13$ for $7$, $15$ for $8$. My program takes already quite long for the last case ($15$ for $8$), so I terminated it.
My guess from these cases is that $2m - 1$ is the least fit for $m$, and therefore the original question is true since $6666 > 4035 = 2 cdot 2018 - 1$, but I apparently have no idea about the proof.
Disclaimer: this is not homework. I fail to do it cuz I'm bad.
algebra-precalculus
asked Nov 28 '18 at 10:50
L. F.
1264
5
A partial answer to your question is the theorem of Erdos-Ginzburg-Ziv: Given a list of $ 2n-1 $ integers, there exist $ n $ of them whose sum is divisible by $ n$. In particular, since $ 6666 > 2 times 2018 - 1$, such a choice is possible as you have already conjectured.
– hellHound
Nov 28 '18 at 11:12
@hellHound I never heard that theorem. Thank you so much!
– L. F.
Nov 28 '18 at 11:16
1
A proof of the Erdös-Ginzburg-Ziv theorem can be found in the second answer here: mathoverflow.net/questions/16721/egz-theorem-erdős-ginzburg-ziv .
– Christian Blatter
Nov 28 '18 at 14:19
add a comment |
The original question: given $6666$ integers, (positive, negative or $0$) is it always possible to choose $2018$ from them so that the chosen numbers add to a multiple of $2018$? (positive multiple, negative multiple or $0$) Prove or disprove.
More generally, for what $n$ and $m$, given $n$ integers, it is always possible to choose $m$ from them so that their sum is a multiple of $m$?
I tried for small $m$s to find a minimum $n$ that fits, and my research shows (in $n$ for $m$ form): $1$ for $1$, $3$ for $2$, $5$ for $3$, $7$ for $4$, $9$ for $5$, $11$ for $6$, $13$ for $7$, $15$ for $8$. My program takes already quite long for the last case ($15$ for $8$), so I terminated it.
My guess from these cases is that $2m - 1$ is the least fit for $m$, and therefore the original question is true since $6666 > 4035 = 2 cdot 2018 - 1$, but I apparently have no idea about the proof.
Disclaimer: this is not homework. I fail to do it cuz I'm bad.
algebra-precalculus
asked Nov 28 '18 at 10:50
L. F.
1264
The original question: given $6666$ integers, (positive, negative or $0$) is it always possible to choose $2018$ from them so that the chosen numbers add to a multiple of $2018$? (positive multiple, negative multiple or $0$) Prove or disprove.
More generally, for what $n$ and $m$, given $n$ integers, it is always possible to choose $m$ from them so that their sum is a multiple of $m$?
I tried for small $m$s to find a minimum $n$ that fits, and my research shows (in $n$ for $m$ form): $1$ for $1$, $3$ for $2$, $5$ for $3$, $7$ for $4$, $9$ for $5$, $11$ for $6$, $13$ for $7$, $15$ for $8$. My program takes already quite long for the last case ($15$ for $8$), so I terminated it.
My guess from these cases is that $2m - 1$ is the least fit for $m$, and therefore the original question is true since $6666 > 4035 = 2 cdot 2018 - 1$, but I apparently have no idea about the proof.
Disclaimer: this is not homework. I fail to do it cuz I'm bad.
algebra-precalculus
algebra-precalculus
asked Nov 28 '18 at 10:50
L. F.
1264
asked Nov 28 '18 at 10:50
L. F.
1264
edited Nov 28 '18 at 11:05
asked Nov 28 '18 at 10:50
L. F.
1264
asked Nov 28 '18 at 10:50
L. F.
1264
asked Nov 28 '18 at 10:50
L. F.
1264
1264
5
A partial answer to your question is the theorem of Erdos-Ginzburg-Ziv: Given a list of $ 2n-1 $ integers, there exist $ n $ of them whose sum is divisible by $ n$. In particular, since $ 6666 > 2 times 2018 - 1$, such a choice is possible as you have already conjectured.
– hellHound
Nov 28 '18 at 11:12
@hellHound I never heard that theorem. Thank you so much!
– L. F.
Nov 28 '18 at 11:16
1
A proof of the Erdös-Ginzburg-Ziv theorem can be found in the second answer here: mathoverflow.net/questions/16721/egz-theorem-erdős-ginzburg-ziv .
– Christian Blatter
Nov 28 '18 at 14:19
add a comment |
5
A partial answer to your question is the theorem of Erdos-Ginzburg-Ziv: Given a list of $ 2n-1 $ integers, there exist $ n $ of them whose sum is divisible by $ n$. In particular, since $ 6666 > 2 times 2018 - 1$, such a choice is possible as you have already conjectured.
– hellHound
Nov 28 '18 at 11:12
@hellHound I never heard that theorem. Thank you so much!
– L. F.
Nov 28 '18 at 11:16
1
A proof of the Erdös-Ginzburg-Ziv theorem can be found in the second answer here: mathoverflow.net/questions/16721/egz-theorem-erdős-ginzburg-ziv .
– Christian Blatter
Nov 28 '18 at 14:19
5
5
A partial answer to your question is the theorem of Erdos-Ginzburg-Ziv: Given a list of $ 2n-1 $ integers, there exist $ n $ of them whose sum is divisible by $ n$. In particular, since $ 6666 > 2 times 2018 - 1$, such a choice is possible as you have already conjectured.
– hellHound
Nov 28 '18 at 11:12
A partial answer to your question is the theorem of Erdos-Ginzburg-Ziv: Given a list of $ 2n-1 $ integers, there exist $ n $ of them whose sum is divisible by $ n$. In particular, since $ 6666 > 2 times 2018 - 1$, such a choice is possible as you have already conjectured.
– hellHound
Nov 28 '18 at 11:12
@hellHound I never heard that theorem. Thank you so much!
– L. F.
Nov 28 '18 at 11:16
@hellHound I never heard that theorem. Thank you so much!
– L. F.
Nov 28 '18 at 11:16
1
1
A proof of the Erdös-Ginzburg-Ziv theorem can be found in the second answer here: mathoverflow.net/questions/16721/egz-theorem-erdős-ginzburg-ziv .
– Christian Blatter
Nov 28 '18 at 14:19
A proof of the Erdös-Ginzburg-Ziv theorem can be found in the second answer here: mathoverflow.net/questions/16721/egz-theorem-erdős-ginzburg-ziv .
– Christian Blatter
Nov 28 '18 at 14:19
add a comment |
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5
A partial answer to your question is the theorem of Erdos-Ginzburg-Ziv: Given a list of $ 2n-1 $ integers, there exist $ n $ of them whose sum is divisible by $ n$. In particular, since $ 6666 > 2 times 2018 - 1$, such a choice is possible as you have already conjectured.
– hellHound
Nov 28 '18 at 11:12
@hellHound I never heard that theorem. Thank you so much!
– L. F.
Nov 28 '18 at 11:16
1
A proof of the Erdös-Ginzburg-Ziv theorem can be found in the second answer here: mathoverflow.net/questions/16721/egz-theorem-erdős-ginzburg-ziv .
– Christian Blatter
Nov 28 '18 at 14:19