Show that given a set of positive n integers, there exists a non-empty subset whose sum is divisible by n
$begingroup$
This is the question I've run into: Show that given a set of positive n integers, there exists a non-empty subset whose sum is divisible by n
I'm having trouble understanding how they came to the conclusion
the part that I'm having trouble understanding is how subtracting the two subsets results in a sum that's divisible by n given that the two subsets have the same remainder
pigeonhole-principle
$endgroup$
add a comment |
$begingroup$
This is the question I've run into: Show that given a set of positive n integers, there exists a non-empty subset whose sum is divisible by n
I'm having trouble understanding how they came to the conclusion
the part that I'm having trouble understanding is how subtracting the two subsets results in a sum that's divisible by n given that the two subsets have the same remainder
pigeonhole-principle
$endgroup$
$begingroup$
If $xequiv y pmod n$ then $x-y equiv 0 pmod n$
$endgroup$
– steven gregory
May 6 '18 at 3:27
add a comment |
$begingroup$
This is the question I've run into: Show that given a set of positive n integers, there exists a non-empty subset whose sum is divisible by n
I'm having trouble understanding how they came to the conclusion
the part that I'm having trouble understanding is how subtracting the two subsets results in a sum that's divisible by n given that the two subsets have the same remainder
pigeonhole-principle
$endgroup$
This is the question I've run into: Show that given a set of positive n integers, there exists a non-empty subset whose sum is divisible by n
I'm having trouble understanding how they came to the conclusion
the part that I'm having trouble understanding is how subtracting the two subsets results in a sum that's divisible by n given that the two subsets have the same remainder
pigeonhole-principle
pigeonhole-principle
asked May 6 '18 at 3:12
testinggnitset sertestinggnitset ser
465
465
$begingroup$
If $xequiv y pmod n$ then $x-y equiv 0 pmod n$
$endgroup$
– steven gregory
May 6 '18 at 3:27
add a comment |
$begingroup$
If $xequiv y pmod n$ then $x-y equiv 0 pmod n$
$endgroup$
– steven gregory
May 6 '18 at 3:27
$begingroup$
If $xequiv y pmod n$ then $x-y equiv 0 pmod n$
$endgroup$
– steven gregory
May 6 '18 at 3:27
$begingroup$
If $xequiv y pmod n$ then $x-y equiv 0 pmod n$
$endgroup$
– steven gregory
May 6 '18 at 3:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Via modular arithmetic this is very easy, and something you can check (if you're familiar with this). Otherwise, one can show the following result via the division algorithm (i.e. just straight division with quotient and remainder):
If $a$ and $b$ have the same remainder when divided by $n$, then $a-b$ is divisible by $n$.
The proof is as follows. Let $r$ be the common remainder of $a$ and $b$ when divided by $n$. The division algorithm gives integers $q,p$ such that $a=qn+r$ and $b=pn+r$. Thus, we get $$a-b=(qn+r)-(pn+r)=(qn-pn)+(r-r)=(q-p)n$$
so $n$ divides $a-b$.
Note that the above result is more general, and can be applied to your situation by having $a=S_j$ and $b=S_i$. It doesn't really matter that the $S_i$ and $S_j$ are sums of integers when applying the statement I just proved; as far as we care they are just some integers $a$ and $b$.
$endgroup$
$begingroup$
But this is $a-b$ not $a+b$
$endgroup$
– kelalaka
Jan 3 at 16:41
$begingroup$
@kelalaka yes, and this is what we want for this question. We want $a-b$.
$endgroup$
– Dave
Jan 3 at 17:43
$begingroup$
I'm confused. The OP requestedsum
is divisible but all the answers are $a-b$. I'm trying to prove $a+b$.
$endgroup$
– kelalaka
Jan 3 at 17:47
$begingroup$
@kelalaka I think you should read the proof in the question more carefully. Each of the $S_i$'s are sums of the elements in the set, so $S_j-S_i$ is also a sum of elements in the set when $ineq j$. Then in my answer I denote $a=S_j$ and $b=S_i$, so $a-b$ is a sum of elements in the given set.
$endgroup$
– Dave
Jan 3 at 17:52
add a comment |
$begingroup$
Let $p<q$, if $S_p = sum_{i=1}^p s_i equiv r pmod{n}$
and $S_q = sum_{i=1}^q s_i equiv r pmod{n}$
then we have $S_q-S_p=sum_{i=p+1}^q s_i equiv r-r equiv 0 pmod{n}$
where we have use the property that if $a equiv b pmod{n}$ and $c equiv d pmod{n}$, then we have $a-c equiv b-d pmod{n}$
If you are not familiar with modulo arithmetic, view it as if $a = nk+r$ and $c = nt+r$, then $a-c = n(k-t)$.
$endgroup$
$begingroup$
thanks for taking time to write this answer too! you can probably tell my math skills aren't that great so the bottom one made more sense to me for now with a bit more explanation without the modular arithmetic route :|
$endgroup$
– testinggnitset ser
May 6 '18 at 3:37
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Via modular arithmetic this is very easy, and something you can check (if you're familiar with this). Otherwise, one can show the following result via the division algorithm (i.e. just straight division with quotient and remainder):
If $a$ and $b$ have the same remainder when divided by $n$, then $a-b$ is divisible by $n$.
The proof is as follows. Let $r$ be the common remainder of $a$ and $b$ when divided by $n$. The division algorithm gives integers $q,p$ such that $a=qn+r$ and $b=pn+r$. Thus, we get $$a-b=(qn+r)-(pn+r)=(qn-pn)+(r-r)=(q-p)n$$
so $n$ divides $a-b$.
Note that the above result is more general, and can be applied to your situation by having $a=S_j$ and $b=S_i$. It doesn't really matter that the $S_i$ and $S_j$ are sums of integers when applying the statement I just proved; as far as we care they are just some integers $a$ and $b$.
$endgroup$
$begingroup$
But this is $a-b$ not $a+b$
$endgroup$
– kelalaka
Jan 3 at 16:41
$begingroup$
@kelalaka yes, and this is what we want for this question. We want $a-b$.
$endgroup$
– Dave
Jan 3 at 17:43
$begingroup$
I'm confused. The OP requestedsum
is divisible but all the answers are $a-b$. I'm trying to prove $a+b$.
$endgroup$
– kelalaka
Jan 3 at 17:47
$begingroup$
@kelalaka I think you should read the proof in the question more carefully. Each of the $S_i$'s are sums of the elements in the set, so $S_j-S_i$ is also a sum of elements in the set when $ineq j$. Then in my answer I denote $a=S_j$ and $b=S_i$, so $a-b$ is a sum of elements in the given set.
$endgroup$
– Dave
Jan 3 at 17:52
add a comment |
$begingroup$
Via modular arithmetic this is very easy, and something you can check (if you're familiar with this). Otherwise, one can show the following result via the division algorithm (i.e. just straight division with quotient and remainder):
If $a$ and $b$ have the same remainder when divided by $n$, then $a-b$ is divisible by $n$.
The proof is as follows. Let $r$ be the common remainder of $a$ and $b$ when divided by $n$. The division algorithm gives integers $q,p$ such that $a=qn+r$ and $b=pn+r$. Thus, we get $$a-b=(qn+r)-(pn+r)=(qn-pn)+(r-r)=(q-p)n$$
so $n$ divides $a-b$.
Note that the above result is more general, and can be applied to your situation by having $a=S_j$ and $b=S_i$. It doesn't really matter that the $S_i$ and $S_j$ are sums of integers when applying the statement I just proved; as far as we care they are just some integers $a$ and $b$.
$endgroup$
$begingroup$
But this is $a-b$ not $a+b$
$endgroup$
– kelalaka
Jan 3 at 16:41
$begingroup$
@kelalaka yes, and this is what we want for this question. We want $a-b$.
$endgroup$
– Dave
Jan 3 at 17:43
$begingroup$
I'm confused. The OP requestedsum
is divisible but all the answers are $a-b$. I'm trying to prove $a+b$.
$endgroup$
– kelalaka
Jan 3 at 17:47
$begingroup$
@kelalaka I think you should read the proof in the question more carefully. Each of the $S_i$'s are sums of the elements in the set, so $S_j-S_i$ is also a sum of elements in the set when $ineq j$. Then in my answer I denote $a=S_j$ and $b=S_i$, so $a-b$ is a sum of elements in the given set.
$endgroup$
– Dave
Jan 3 at 17:52
add a comment |
$begingroup$
Via modular arithmetic this is very easy, and something you can check (if you're familiar with this). Otherwise, one can show the following result via the division algorithm (i.e. just straight division with quotient and remainder):
If $a$ and $b$ have the same remainder when divided by $n$, then $a-b$ is divisible by $n$.
The proof is as follows. Let $r$ be the common remainder of $a$ and $b$ when divided by $n$. The division algorithm gives integers $q,p$ such that $a=qn+r$ and $b=pn+r$. Thus, we get $$a-b=(qn+r)-(pn+r)=(qn-pn)+(r-r)=(q-p)n$$
so $n$ divides $a-b$.
Note that the above result is more general, and can be applied to your situation by having $a=S_j$ and $b=S_i$. It doesn't really matter that the $S_i$ and $S_j$ are sums of integers when applying the statement I just proved; as far as we care they are just some integers $a$ and $b$.
$endgroup$
Via modular arithmetic this is very easy, and something you can check (if you're familiar with this). Otherwise, one can show the following result via the division algorithm (i.e. just straight division with quotient and remainder):
If $a$ and $b$ have the same remainder when divided by $n$, then $a-b$ is divisible by $n$.
The proof is as follows. Let $r$ be the common remainder of $a$ and $b$ when divided by $n$. The division algorithm gives integers $q,p$ such that $a=qn+r$ and $b=pn+r$. Thus, we get $$a-b=(qn+r)-(pn+r)=(qn-pn)+(r-r)=(q-p)n$$
so $n$ divides $a-b$.
Note that the above result is more general, and can be applied to your situation by having $a=S_j$ and $b=S_i$. It doesn't really matter that the $S_i$ and $S_j$ are sums of integers when applying the statement I just proved; as far as we care they are just some integers $a$ and $b$.
answered May 6 '18 at 3:24
DaveDave
9,06311033
9,06311033
$begingroup$
But this is $a-b$ not $a+b$
$endgroup$
– kelalaka
Jan 3 at 16:41
$begingroup$
@kelalaka yes, and this is what we want for this question. We want $a-b$.
$endgroup$
– Dave
Jan 3 at 17:43
$begingroup$
I'm confused. The OP requestedsum
is divisible but all the answers are $a-b$. I'm trying to prove $a+b$.
$endgroup$
– kelalaka
Jan 3 at 17:47
$begingroup$
@kelalaka I think you should read the proof in the question more carefully. Each of the $S_i$'s are sums of the elements in the set, so $S_j-S_i$ is also a sum of elements in the set when $ineq j$. Then in my answer I denote $a=S_j$ and $b=S_i$, so $a-b$ is a sum of elements in the given set.
$endgroup$
– Dave
Jan 3 at 17:52
add a comment |
$begingroup$
But this is $a-b$ not $a+b$
$endgroup$
– kelalaka
Jan 3 at 16:41
$begingroup$
@kelalaka yes, and this is what we want for this question. We want $a-b$.
$endgroup$
– Dave
Jan 3 at 17:43
$begingroup$
I'm confused. The OP requestedsum
is divisible but all the answers are $a-b$. I'm trying to prove $a+b$.
$endgroup$
– kelalaka
Jan 3 at 17:47
$begingroup$
@kelalaka I think you should read the proof in the question more carefully. Each of the $S_i$'s are sums of the elements in the set, so $S_j-S_i$ is also a sum of elements in the set when $ineq j$. Then in my answer I denote $a=S_j$ and $b=S_i$, so $a-b$ is a sum of elements in the given set.
$endgroup$
– Dave
Jan 3 at 17:52
$begingroup$
But this is $a-b$ not $a+b$
$endgroup$
– kelalaka
Jan 3 at 16:41
$begingroup$
But this is $a-b$ not $a+b$
$endgroup$
– kelalaka
Jan 3 at 16:41
$begingroup$
@kelalaka yes, and this is what we want for this question. We want $a-b$.
$endgroup$
– Dave
Jan 3 at 17:43
$begingroup$
@kelalaka yes, and this is what we want for this question. We want $a-b$.
$endgroup$
– Dave
Jan 3 at 17:43
$begingroup$
I'm confused. The OP requested
sum
is divisible but all the answers are $a-b$. I'm trying to prove $a+b$.$endgroup$
– kelalaka
Jan 3 at 17:47
$begingroup$
I'm confused. The OP requested
sum
is divisible but all the answers are $a-b$. I'm trying to prove $a+b$.$endgroup$
– kelalaka
Jan 3 at 17:47
$begingroup$
@kelalaka I think you should read the proof in the question more carefully. Each of the $S_i$'s are sums of the elements in the set, so $S_j-S_i$ is also a sum of elements in the set when $ineq j$. Then in my answer I denote $a=S_j$ and $b=S_i$, so $a-b$ is a sum of elements in the given set.
$endgroup$
– Dave
Jan 3 at 17:52
$begingroup$
@kelalaka I think you should read the proof in the question more carefully. Each of the $S_i$'s are sums of the elements in the set, so $S_j-S_i$ is also a sum of elements in the set when $ineq j$. Then in my answer I denote $a=S_j$ and $b=S_i$, so $a-b$ is a sum of elements in the given set.
$endgroup$
– Dave
Jan 3 at 17:52
add a comment |
$begingroup$
Let $p<q$, if $S_p = sum_{i=1}^p s_i equiv r pmod{n}$
and $S_q = sum_{i=1}^q s_i equiv r pmod{n}$
then we have $S_q-S_p=sum_{i=p+1}^q s_i equiv r-r equiv 0 pmod{n}$
where we have use the property that if $a equiv b pmod{n}$ and $c equiv d pmod{n}$, then we have $a-c equiv b-d pmod{n}$
If you are not familiar with modulo arithmetic, view it as if $a = nk+r$ and $c = nt+r$, then $a-c = n(k-t)$.
$endgroup$
$begingroup$
thanks for taking time to write this answer too! you can probably tell my math skills aren't that great so the bottom one made more sense to me for now with a bit more explanation without the modular arithmetic route :|
$endgroup$
– testinggnitset ser
May 6 '18 at 3:37
add a comment |
$begingroup$
Let $p<q$, if $S_p = sum_{i=1}^p s_i equiv r pmod{n}$
and $S_q = sum_{i=1}^q s_i equiv r pmod{n}$
then we have $S_q-S_p=sum_{i=p+1}^q s_i equiv r-r equiv 0 pmod{n}$
where we have use the property that if $a equiv b pmod{n}$ and $c equiv d pmod{n}$, then we have $a-c equiv b-d pmod{n}$
If you are not familiar with modulo arithmetic, view it as if $a = nk+r$ and $c = nt+r$, then $a-c = n(k-t)$.
$endgroup$
$begingroup$
thanks for taking time to write this answer too! you can probably tell my math skills aren't that great so the bottom one made more sense to me for now with a bit more explanation without the modular arithmetic route :|
$endgroup$
– testinggnitset ser
May 6 '18 at 3:37
add a comment |
$begingroup$
Let $p<q$, if $S_p = sum_{i=1}^p s_i equiv r pmod{n}$
and $S_q = sum_{i=1}^q s_i equiv r pmod{n}$
then we have $S_q-S_p=sum_{i=p+1}^q s_i equiv r-r equiv 0 pmod{n}$
where we have use the property that if $a equiv b pmod{n}$ and $c equiv d pmod{n}$, then we have $a-c equiv b-d pmod{n}$
If you are not familiar with modulo arithmetic, view it as if $a = nk+r$ and $c = nt+r$, then $a-c = n(k-t)$.
$endgroup$
Let $p<q$, if $S_p = sum_{i=1}^p s_i equiv r pmod{n}$
and $S_q = sum_{i=1}^q s_i equiv r pmod{n}$
then we have $S_q-S_p=sum_{i=p+1}^q s_i equiv r-r equiv 0 pmod{n}$
where we have use the property that if $a equiv b pmod{n}$ and $c equiv d pmod{n}$, then we have $a-c equiv b-d pmod{n}$
If you are not familiar with modulo arithmetic, view it as if $a = nk+r$ and $c = nt+r$, then $a-c = n(k-t)$.
answered May 6 '18 at 3:17
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
$begingroup$
thanks for taking time to write this answer too! you can probably tell my math skills aren't that great so the bottom one made more sense to me for now with a bit more explanation without the modular arithmetic route :|
$endgroup$
– testinggnitset ser
May 6 '18 at 3:37
add a comment |
$begingroup$
thanks for taking time to write this answer too! you can probably tell my math skills aren't that great so the bottom one made more sense to me for now with a bit more explanation without the modular arithmetic route :|
$endgroup$
– testinggnitset ser
May 6 '18 at 3:37
$begingroup$
thanks for taking time to write this answer too! you can probably tell my math skills aren't that great so the bottom one made more sense to me for now with a bit more explanation without the modular arithmetic route :|
$endgroup$
– testinggnitset ser
May 6 '18 at 3:37
$begingroup$
thanks for taking time to write this answer too! you can probably tell my math skills aren't that great so the bottom one made more sense to me for now with a bit more explanation without the modular arithmetic route :|
$endgroup$
– testinggnitset ser
May 6 '18 at 3:37
add a comment |
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$begingroup$
If $xequiv y pmod n$ then $x-y equiv 0 pmod n$
$endgroup$
– steven gregory
May 6 '18 at 3:27