Mixing
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For which $n$ is $37|1underbrace0cdots 0_textn-times1underbrace0cdots 0_textn-times1$
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I want to understand the solution to part b) of this problem that I have posted below. Part a) was solved by showing that $x^2+x+1 | x^2n+x^n+1$ iff $3 nmid n$. Since we are using the expression of part a) evaluated at $x=10$ to prove the divisibility of $37|1underbrace0cdots 0_textn-times1underbrace0cdots 0_textn-times1$ it seems to me that we can only conclude for which $n$, $3cdot 37|1underbrace0cdots 0_textn-times1underbrace0cdots 0_textn-times1$. Since the digital sum on the right is always divisible by $3$ how can we conclude for which $n$ the expression is divisible by $37$
Solution
elementary-number-theory proof-explanation
asked Aug 2 at 15:43
john fowles
1,066817
 |Â
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up vote
2
down vote
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I want to understand the solution to part b) of this problem that I have posted below. Part a) was solved by showing that $x^2+x+1 | x^2n+x^n+1$ iff $3 nmid n$. Since we are using the expression of part a) evaluated at $x=10$ to prove the divisibility of $37|1underbrace0cdots 0_textn-times1underbrace0cdots 0_textn-times1$ it seems to me that we can only conclude for which $n$, $3cdot 37|1underbrace0cdots 0_textn-times1underbrace0cdots 0_textn-times1$. Since the digital sum on the right is always divisible by $3$ how can we conclude for which $n$ the expression is divisible by $37$
Solution
elementary-number-theory proof-explanation
asked Aug 2 at 15:43
john fowles
1,066817
1
What's your question. You asked a question. Explained everything you knew. And posted an answer that gave the complete answer in coherent explanation. So.... what's your question.
– fleablood
Aug 2 at 15:47
1
I said that my interpretation of the solution was we only proved divisibility by $111$ not $37$, which is what the question is asking
– john fowles
Aug 2 at 15:49
3
$37|111$ so if $111|M$ then $37|M$.
– fleablood
Aug 2 at 15:49
1
Did you do something wrong? Shouldn't it be "$x^2+x+1 nmid x^2n+x^n+1$ if and only if $3mid n$"? For every integer $n$ divisible by $3$, $$x^2n+x^n+1equiv 3pmodx^2+x+1,.$$
– Batominovski
Aug 2 at 15:52
@Batominovski In the book it has $x^2+x+1 | x^2n+x^n+1$ iff $3nmid n$. The solution proved it by checking each case: $n=3k$, $n=3k+1$ and $n=3k+2$. Using the property that $x^3-1 | x^3m-1$, and factoring using difference of cubes.
– john fowles
Aug 2 at 15:56
 |Â
show 5 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I want to understand the solution to part b) of this problem that I have posted below. Part a) was solved by showing that $x^2+x+1 | x^2n+x^n+1$ iff $3 nmid n$. Since we are using the expression of part a) evaluated at $x=10$ to prove the divisibility of $37|1underbrace0cdots 0_textn-times1underbrace0cdots 0_textn-times1$ it seems to me that we can only conclude for which $n$, $3cdot 37|1underbrace0cdots 0_textn-times1underbrace0cdots 0_textn-times1$. Since the digital sum on the right is always divisible by $3$ how can we conclude for which $n$ the expression is divisible by $37$
Solution
elementary-number-theory proof-explanation
asked Aug 2 at 15:43
john fowles
1,066817
I want to understand the solution to part b) of this problem that I have posted below. Part a) was solved by showing that $x^2+x+1 | x^2n+x^n+1$ iff $3 nmid n$. Since we are using the expression of part a) evaluated at $x=10$ to prove the divisibility of $37|1underbrace0cdots 0_textn-times1underbrace0cdots 0_textn-times1$ it seems to me that we can only conclude for which $n$, $3cdot 37|1underbrace0cdots 0_textn-times1underbrace0cdots 0_textn-times1$. Since the digital sum on the right is always divisible by $3$ how can we conclude for which $n$ the expression is divisible by $37$
Solution
elementary-number-theory proof-explanation
asked Aug 2 at 15:43
john fowles
1,066817
edited Aug 2 at 16:00
asked Aug 2 at 15:43
john fowles
1,066817
asked Aug 2 at 15:43
john fowles
1,066817
asked Aug 2 at 15:43
john fowles
1,066817
1,066817
1
What's your question. You asked a question. Explained everything you knew. And posted an answer that gave the complete answer in coherent explanation. So.... what's your question.
– fleablood
Aug 2 at 15:47
1
I said that my interpretation of the solution was we only proved divisibility by $111$ not $37$, which is what the question is asking
– john fowles
Aug 2 at 15:49
3
$37|111$ so if $111|M$ then $37|M$.
– fleablood
Aug 2 at 15:49
1
Did you do something wrong? Shouldn't it be "$x^2+x+1 nmid x^2n+x^n+1$ if and only if $3mid n$"? For every integer $n$ divisible by $3$, $$x^2n+x^n+1equiv 3pmodx^2+x+1,.$$
– Batominovski
Aug 2 at 15:52
@Batominovski In the book it has $x^2+x+1 | x^2n+x^n+1$ iff $3nmid n$. The solution proved it by checking each case: $n=3k$, $n=3k+1$ and $n=3k+2$. Using the property that $x^3-1 | x^3m-1$, and factoring using difference of cubes.
– john fowles
Aug 2 at 15:56
 |Â
show 5 more comments
1
What's your question. You asked a question. Explained everything you knew. And posted an answer that gave the complete answer in coherent explanation. So.... what's your question.
– fleablood
Aug 2 at 15:47
1
I said that my interpretation of the solution was we only proved divisibility by $111$ not $37$, which is what the question is asking
– john fowles
Aug 2 at 15:49
3
$37|111$ so if $111|M$ then $37|M$.
– fleablood
Aug 2 at 15:49
1
Did you do something wrong? Shouldn't it be "$x^2+x+1 nmid x^2n+x^n+1$ if and only if $3mid n$"? For every integer $n$ divisible by $3$, $$x^2n+x^n+1equiv 3pmodx^2+x+1,.$$
– Batominovski
Aug 2 at 15:52
@Batominovski In the book it has $x^2+x+1 | x^2n+x^n+1$ iff $3nmid n$. The solution proved it by checking each case: $n=3k$, $n=3k+1$ and $n=3k+2$. Using the property that $x^3-1 | x^3m-1$, and factoring using difference of cubes.
– john fowles
Aug 2 at 15:56
1
1
What's your question. You asked a question. Explained everything you knew. And posted an answer that gave the complete answer in coherent explanation. So.... what's your question.
– fleablood
Aug 2 at 15:47
What's your question. You asked a question. Explained everything you knew. And posted an answer that gave the complete answer in coherent explanation. So.... what's your question.
– fleablood
Aug 2 at 15:47
1
1
I said that my interpretation of the solution was we only proved divisibility by $111$ not $37$, which is what the question is asking
– john fowles
Aug 2 at 15:49
I said that my interpretation of the solution was we only proved divisibility by $111$ not $37$, which is what the question is asking
– john fowles
Aug 2 at 15:49
3
3
$37|111$ so if $111|M$ then $37|M$.
– fleablood
Aug 2 at 15:49
$37|111$ so if $111|M$ then $37|M$.
– fleablood
Aug 2 at 15:49
1
1
Did you do something wrong? Shouldn't it be "$x^2+x+1 nmid x^2n+x^n+1$ if and only if $3mid n$"? For every integer $n$ divisible by $3$, $$x^2n+x^n+1equiv 3pmodx^2+x+1,.$$
– Batominovski
Aug 2 at 15:52
Did you do something wrong? Shouldn't it be "$x^2+x+1 nmid x^2n+x^n+1$ if and only if $3mid n$"? For every integer $n$ divisible by $3$, $$x^2n+x^n+1equiv 3pmodx^2+x+1,.$$
– Batominovski
Aug 2 at 15:52
@Batominovski In the book it has $x^2+x+1 | x^2n+x^n+1$ iff $3nmid n$. The solution proved it by checking each case: $n=3k$, $n=3k+1$ and $n=3k+2$. Using the property that $x^3-1 | x^3m-1$, and factoring using difference of cubes.
– john fowles
Aug 2 at 15:56
@Batominovski In the book it has $x^2+x+1 | x^2n+x^n+1$ iff $3nmid n$. The solution proved it by checking each case: $n=3k$, $n=3k+1$ and $n=3k+2$. Using the property that $x^3-1 | x^3m-1$, and factoring using difference of cubes.
– john fowles
Aug 2 at 15:56
 |Â
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1
What's your question. You asked a question. Explained everything you knew. And posted an answer that gave the complete answer in coherent explanation. So.... what's your question.
– fleablood
Aug 2 at 15:47
1
I said that my interpretation of the solution was we only proved divisibility by $111$ not $37$, which is what the question is asking
– john fowles
Aug 2 at 15:49
3
$37|111$ so if $111|M$ then $37|M$.
– fleablood
Aug 2 at 15:49
1
Did you do something wrong? Shouldn't it be "$x^2+x+1 nmid x^2n+x^n+1$ if and only if $3mid n$"? For every integer $n$ divisible by $3$, $$x^2n+x^n+1equiv 3pmodx^2+x+1,.$$
– Batominovski
Aug 2 at 15:52
@Batominovski In the book it has $x^2+x+1 | x^2n+x^n+1$ iff $3nmid n$. The solution proved it by checking each case: $n=3k$, $n=3k+1$ and $n=3k+2$. Using the property that $x^3-1 | x^3m-1$, and factoring using difference of cubes.
– john fowles
Aug 2 at 15:56