Divisor Function of Sums in Fractions
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I have a question that I've been working on for a while now. It says, "Let $A=0,1,2,dots,2018$. Prove that $forall ninmathbbN,existsa_0,a_1,a_2,dots,a_2018subseteqmathbbN$, $$(a_0<a_1<dots<a_2018)wedgefrac(sum^2018_k=0a_k)div2019n+sum^2018_k=0(a_kdiv2019)=1."$$ I have no clue which $a_k$ to construct, and how they'll operate in that nasty fraction. I have been trying to find an upper bound for $a_2018$ and working from there, but I seem to be getting nowhere. I also tried setting the numerator equal to the denominator (since the fraction is equal to one) and working from there, but again, I didn't get very far. Help would be appreciated.
summation divisibility natural-numbers
asked Aug 8 at 18:56
Mr. Frothingslosh
796
 |Â
show 3 more comments
up vote
-1
down vote
favorite
I have a question that I've been working on for a while now. It says, "Let $A=0,1,2,dots,2018$. Prove that $forall ninmathbbN,existsa_0,a_1,a_2,dots,a_2018subseteqmathbbN$, $$(a_0<a_1<dots<a_2018)wedgefrac(sum^2018_k=0a_k)div2019n+sum^2018_k=0(a_kdiv2019)=1."$$ I have no clue which $a_k$ to construct, and how they'll operate in that nasty fraction. I have been trying to find an upper bound for $a_2018$ and working from there, but I seem to be getting nowhere. I also tried setting the numerator equal to the denominator (since the fraction is equal to one) and working from there, but again, I didn't get very far. Help would be appreciated.
summation divisibility natural-numbers
asked Aug 8 at 18:56
Mr. Frothingslosh
796
What is the source of the problem? The use of numbers such as 2018 makes it look like an exam or contest question.
– Carl Mummert
Aug 8 at 22:44
The source of the problem is a problem set. I've been working on this question for forever now. It's the only one that I haven't gotten yet
– Mr. Frothingslosh
Aug 8 at 23:02
I've been trying to find what I can do with $a_2018$, since $sum^n_i=0=fracn(n+1)2$, but I haven't gotten very far.
– Mr. Frothingslosh
Aug 8 at 23:04
What does $(div2019)$ mean? The last fraction is not a sentence, so how can it be part of a conjunction?
– Ross Millikan
Aug 9 at 14:15
$a div 2019$ means that $a=2019q+r$ for some unique $q$ and $r$, and $div$ is the $q$; it's part of the division theorem. For example, for $4 div2$, we know that 2 divides 4 twice, so the answer is $q=2$.
– Mr. Frothingslosh
Aug 9 at 15:07
 |Â
show 3 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have a question that I've been working on for a while now. It says, "Let $A=0,1,2,dots,2018$. Prove that $forall ninmathbbN,existsa_0,a_1,a_2,dots,a_2018subseteqmathbbN$, $$(a_0<a_1<dots<a_2018)wedgefrac(sum^2018_k=0a_k)div2019n+sum^2018_k=0(a_kdiv2019)=1."$$ I have no clue which $a_k$ to construct, and how they'll operate in that nasty fraction. I have been trying to find an upper bound for $a_2018$ and working from there, but I seem to be getting nowhere. I also tried setting the numerator equal to the denominator (since the fraction is equal to one) and working from there, but again, I didn't get very far. Help would be appreciated.
summation divisibility natural-numbers
asked Aug 8 at 18:56
Mr. Frothingslosh
796
I have a question that I've been working on for a while now. It says, "Let $A=0,1,2,dots,2018$. Prove that $forall ninmathbbN,existsa_0,a_1,a_2,dots,a_2018subseteqmathbbN$, $$(a_0<a_1<dots<a_2018)wedgefrac(sum^2018_k=0a_k)div2019n+sum^2018_k=0(a_kdiv2019)=1."$$ I have no clue which $a_k$ to construct, and how they'll operate in that nasty fraction. I have been trying to find an upper bound for $a_2018$ and working from there, but I seem to be getting nowhere. I also tried setting the numerator equal to the denominator (since the fraction is equal to one) and working from there, but again, I didn't get very far. Help would be appreciated.
summation divisibility natural-numbers
asked Aug 8 at 18:56
Mr. Frothingslosh
796
edited Aug 10 at 15:01
asked Aug 8 at 18:56
Mr. Frothingslosh
796
asked Aug 8 at 18:56
Mr. Frothingslosh
796
asked Aug 8 at 18:56
Mr. Frothingslosh
796
796
What is the source of the problem? The use of numbers such as 2018 makes it look like an exam or contest question.
– Carl Mummert
Aug 8 at 22:44
The source of the problem is a problem set. I've been working on this question for forever now. It's the only one that I haven't gotten yet
– Mr. Frothingslosh
Aug 8 at 23:02
I've been trying to find what I can do with $a_2018$, since $sum^n_i=0=fracn(n+1)2$, but I haven't gotten very far.
– Mr. Frothingslosh
Aug 8 at 23:04
What does $(div2019)$ mean? The last fraction is not a sentence, so how can it be part of a conjunction?
– Ross Millikan
Aug 9 at 14:15
$a div 2019$ means that $a=2019q+r$ for some unique $q$ and $r$, and $div$ is the $q$; it's part of the division theorem. For example, for $4 div2$, we know that 2 divides 4 twice, so the answer is $q=2$.
– Mr. Frothingslosh
Aug 9 at 15:07
 |Â
show 3 more comments
What is the source of the problem? The use of numbers such as 2018 makes it look like an exam or contest question.
– Carl Mummert
Aug 8 at 22:44
The source of the problem is a problem set. I've been working on this question for forever now. It's the only one that I haven't gotten yet
– Mr. Frothingslosh
Aug 8 at 23:02
I've been trying to find what I can do with $a_2018$, since $sum^n_i=0=fracn(n+1)2$, but I haven't gotten very far.
– Mr. Frothingslosh
Aug 8 at 23:04
What does $(div2019)$ mean? The last fraction is not a sentence, so how can it be part of a conjunction?
– Ross Millikan
Aug 9 at 14:15
$a div 2019$ means that $a=2019q+r$ for some unique $q$ and $r$, and $div$ is the $q$; it's part of the division theorem. For example, for $4 div2$, we know that 2 divides 4 twice, so the answer is $q=2$.
– Mr. Frothingslosh
Aug 9 at 15:07
What is the source of the problem? The use of numbers such as 2018 makes it look like an exam or contest question.
– Carl Mummert
Aug 8 at 22:44
What is the source of the problem? The use of numbers such as 2018 makes it look like an exam or contest question.
– Carl Mummert
Aug 8 at 22:44
The source of the problem is a problem set. I've been working on this question for forever now. It's the only one that I haven't gotten yet
– Mr. Frothingslosh
Aug 8 at 23:02
The source of the problem is a problem set. I've been working on this question for forever now. It's the only one that I haven't gotten yet
– Mr. Frothingslosh
Aug 8 at 23:02
I've been trying to find what I can do with $a_2018$, since $sum^n_i=0=fracn(n+1)2$, but I haven't gotten very far.
– Mr. Frothingslosh
Aug 8 at 23:04
I've been trying to find what I can do with $a_2018$, since $sum^n_i=0=fracn(n+1)2$, but I haven't gotten very far.
– Mr. Frothingslosh
Aug 8 at 23:04
What does $(div2019)$ mean? The last fraction is not a sentence, so how can it be part of a conjunction?
– Ross Millikan
Aug 9 at 14:15
What does $(div2019)$ mean? The last fraction is not a sentence, so how can it be part of a conjunction?
– Ross Millikan
Aug 9 at 14:15
$a div 2019$ means that $a=2019q+r$ for some unique $q$ and $r$, and $div$ is the $q$; it's part of the division theorem. For example, for $4 div2$, we know that 2 divides 4 twice, so the answer is $q=2$.
– Mr. Frothingslosh
Aug 9 at 15:07
$a div 2019$ means that $a=2019q+r$ for some unique $q$ and $r$, and $div$ is the $q$; it's part of the division theorem. For example, for $4 div2$, we know that 2 divides 4 twice, so the answer is $q=2$.
– Mr. Frothingslosh
Aug 9 at 15:07
 |Â
show 3 more comments
1 Answer
1
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0
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I figured out my own question: Choose $a_k=2019k+n$. Then, in the nemerator, we have that $$left(sum^2018_k=0a_kright)div2019=left(sum^2018_k=0(2019k+n)right)div2019 =sum^2018_k=0k+n.$$ In the denominator, we get $$n+sum^2018_k=0(a_kdiv2019)=n+sum^2018_k=0big((2019k+n)div2019big)=n+sum^2018_k=0k.$$ It's now easy to see that the ratio between the numerator and denominator is $1$.
answered Aug 10 at 14:57
Mr. Frothingslosh
796
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I figured out my own question: Choose $a_k=2019k+n$. Then, in the nemerator, we have that $$left(sum^2018_k=0a_kright)div2019=left(sum^2018_k=0(2019k+n)right)div2019 =sum^2018_k=0k+n.$$ In the denominator, we get $$n+sum^2018_k=0(a_kdiv2019)=n+sum^2018_k=0big((2019k+n)div2019big)=n+sum^2018_k=0k.$$ It's now easy to see that the ratio between the numerator and denominator is $1$.
answered Aug 10 at 14:57
Mr. Frothingslosh
796
add a comment |Â
up vote
0
down vote
I figured out my own question: Choose $a_k=2019k+n$. Then, in the nemerator, we have that $$left(sum^2018_k=0a_kright)div2019=left(sum^2018_k=0(2019k+n)right)div2019 =sum^2018_k=0k+n.$$ In the denominator, we get $$n+sum^2018_k=0(a_kdiv2019)=n+sum^2018_k=0big((2019k+n)div2019big)=n+sum^2018_k=0k.$$ It's now easy to see that the ratio between the numerator and denominator is $1$.
answered Aug 10 at 14:57
Mr. Frothingslosh
796
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I figured out my own question: Choose $a_k=2019k+n$. Then, in the nemerator, we have that $$left(sum^2018_k=0a_kright)div2019=left(sum^2018_k=0(2019k+n)right)div2019 =sum^2018_k=0k+n.$$ In the denominator, we get $$n+sum^2018_k=0(a_kdiv2019)=n+sum^2018_k=0big((2019k+n)div2019big)=n+sum^2018_k=0k.$$ It's now easy to see that the ratio between the numerator and denominator is $1$.
answered Aug 10 at 14:57
Mr. Frothingslosh
796
I figured out my own question: Choose $a_k=2019k+n$. Then, in the nemerator, we have that $$left(sum^2018_k=0a_kright)div2019=left(sum^2018_k=0(2019k+n)right)div2019 =sum^2018_k=0k+n.$$ In the denominator, we get $$n+sum^2018_k=0(a_kdiv2019)=n+sum^2018_k=0big((2019k+n)div2019big)=n+sum^2018_k=0k.$$ It's now easy to see that the ratio between the numerator and denominator is $1$.
answered Aug 10 at 14:57
Mr. Frothingslosh
796
answered Aug 10 at 14:57
Mr. Frothingslosh
796
answered Aug 10 at 14:57
Mr. Frothingslosh
796
answered Aug 10 at 14:57
Mr. Frothingslosh
796
796
add a comment |Â
add a comment |Â
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What is the source of the problem? The use of numbers such as 2018 makes it look like an exam or contest question.
– Carl Mummert
Aug 8 at 22:44
The source of the problem is a problem set. I've been working on this question for forever now. It's the only one that I haven't gotten yet
– Mr. Frothingslosh
Aug 8 at 23:02
I've been trying to find what I can do with $a_2018$, since $sum^n_i=0=fracn(n+1)2$, but I haven't gotten very far.
– Mr. Frothingslosh
Aug 8 at 23:04
What does $(div2019)$ mean? The last fraction is not a sentence, so how can it be part of a conjunction?
– Ross Millikan
Aug 9 at 14:15
$a div 2019$ means that $a=2019q+r$ for some unique $q$ and $r$, and $div$ is the $q$; it's part of the division theorem. For example, for $4 div2$, we know that 2 divides 4 twice, so the answer is $q=2$.
– Mr. Frothingslosh
Aug 9 at 15:07