Find the number of elements in set $A={1,2,ldots, N}$ such that the element has multiples in each row of a $N...
$begingroup$
For a given $N$, the $N times N$ grid looks like :
$1$ $qquad$ $N+1$ $qquad$ $2N+1$ $qquad$ $3N+1$ $qquad$...$qquad$ $N(N-1)+1$
$2$ $qquad$ $N+2$ $qquad$ $2N+2$ $qquad$ $3N+2$ $qquad$... $qquad$ $N(N-1)+2$
$vdots$
$k$ $qquad$ $N+k$ $qquad$ $2N+k$ $qquad$ $3N+k$ $qquad$... $qquad$ $N(N-1)+k$
$vdots$
$N$ $qquad$ $N+N$ $qquad$ $2N+N$ $qquad$ $3N+N$ $qquad$... $qquad$ $N(N-1)+N$
upon observation,I can see that the number of elements in A is the number of elements coprime with $N$.
that is the $varphi(N)$
But i am not able to prove this.
if anyone could prove this?
For example $N=3$
we have
$1$ $quad$ $4$ $quad$ $7$
$2$ $quad$ $5$ $quad$ $8$
$3$ $quad$ $6$ $quad$ $9$
and $varphi(3)$=$2$
alternative-proof totient-function
edited Jan 10 at 15:52
littleO
30.5k648111
asked Jan 10 at 15:23
sonoroussonorous
327
$endgroup$
add a comment |
$begingroup$
For a given $N$, the $N times N$ grid looks like :
$1$ $qquad$ $N+1$ $qquad$ $2N+1$ $qquad$ $3N+1$ $qquad$...$qquad$ $N(N-1)+1$
$2$ $qquad$ $N+2$ $qquad$ $2N+2$ $qquad$ $3N+2$ $qquad$... $qquad$ $N(N-1)+2$
$vdots$
$k$ $qquad$ $N+k$ $qquad$ $2N+k$ $qquad$ $3N+k$ $qquad$... $qquad$ $N(N-1)+k$
$vdots$
$N$ $qquad$ $N+N$ $qquad$ $2N+N$ $qquad$ $3N+N$ $qquad$... $qquad$ $N(N-1)+N$
upon observation,I can see that the number of elements in A is the number of elements coprime with $N$.
that is the $varphi(N)$
But i am not able to prove this.
if anyone could prove this?
For example $N=3$
we have
$1$ $quad$ $4$ $quad$ $7$
$2$ $quad$ $5$ $quad$ $8$
$3$ $quad$ $6$ $quad$ $9$
and $varphi(3)$=$2$
alternative-proof totient-function
edited Jan 10 at 15:52
littleO
30.5k648111
asked Jan 10 at 15:23
sonoroussonorous
327
$endgroup$
$begingroup$
for n=3, we have A={1,2,3} and in this set, we have 2 elements, 1, which has multiples in each row 1 2 3, and the next element in set A to have multiples in each row is 2, it has 4 in first row, 8 in second and 6 in third. we have thus 2 elements in set A which have multiples in each row of the N*N grid.
$endgroup$
– sonorous
Jan 10 at 15:32
add a comment |
$begingroup$
For a given $N$, the $N times N$ grid looks like :
$1$ $qquad$ $N+1$ $qquad$ $2N+1$ $qquad$ $3N+1$ $qquad$...$qquad$ $N(N-1)+1$
$2$ $qquad$ $N+2$ $qquad$ $2N+2$ $qquad$ $3N+2$ $qquad$... $qquad$ $N(N-1)+2$
$vdots$
$k$ $qquad$ $N+k$ $qquad$ $2N+k$ $qquad$ $3N+k$ $qquad$... $qquad$ $N(N-1)+k$
$vdots$
$N$ $qquad$ $N+N$ $qquad$ $2N+N$ $qquad$ $3N+N$ $qquad$... $qquad$ $N(N-1)+N$
upon observation,I can see that the number of elements in A is the number of elements coprime with $N$.
that is the $varphi(N)$
But i am not able to prove this.
if anyone could prove this?
For example $N=3$
we have
$1$ $quad$ $4$ $quad$ $7$
$2$ $quad$ $5$ $quad$ $8$
$3$ $quad$ $6$ $quad$ $9$
and $varphi(3)$=$2$
alternative-proof totient-function
edited Jan 10 at 15:52
littleO
30.5k648111
asked Jan 10 at 15:23
sonoroussonorous
327
$endgroup$
For a given $N$, the $N times N$ grid looks like :
$1$ $qquad$ $N+1$ $qquad$ $2N+1$ $qquad$ $3N+1$ $qquad$...$qquad$ $N(N-1)+1$
$2$ $qquad$ $N+2$ $qquad$ $2N+2$ $qquad$ $3N+2$ $qquad$... $qquad$ $N(N-1)+2$
$vdots$
$k$ $qquad$ $N+k$ $qquad$ $2N+k$ $qquad$ $3N+k$ $qquad$... $qquad$ $N(N-1)+k$
$vdots$
$N$ $qquad$ $N+N$ $qquad$ $2N+N$ $qquad$ $3N+N$ $qquad$... $qquad$ $N(N-1)+N$
upon observation,I can see that the number of elements in A is the number of elements coprime with $N$.
that is the $varphi(N)$
But i am not able to prove this.
if anyone could prove this?
For example $N=3$
we have
$1$ $quad$ $4$ $quad$ $7$
$2$ $quad$ $5$ $quad$ $8$
$3$ $quad$ $6$ $quad$ $9$
and $varphi(3)$=$2$
alternative-proof totient-function
alternative-proof totient-function
edited Jan 10 at 15:52
littleO
30.5k648111
asked Jan 10 at 15:23
sonoroussonorous
327
edited Jan 10 at 15:52
littleO
30.5k648111
asked Jan 10 at 15:23
sonoroussonorous
327
edited Jan 10 at 15:52
littleO
30.5k648111
edited Jan 10 at 15:52
littleO
30.5k648111
edited Jan 10 at 15:52
littleO
30.5k648111
30.5k648111
asked Jan 10 at 15:23
sonoroussonorous
327
asked Jan 10 at 15:23
sonoroussonorous
327
asked Jan 10 at 15:23
sonoroussonorous
327
327
$begingroup$
for n=3, we have A={1,2,3} and in this set, we have 2 elements, 1, which has multiples in each row 1 2 3, and the next element in set A to have multiples in each row is 2, it has 4 in first row, 8 in second and 6 in third. we have thus 2 elements in set A which have multiples in each row of the N*N grid.
$endgroup$
– sonorous
Jan 10 at 15:32
add a comment |
$begingroup$
for n=3, we have A={1,2,3} and in this set, we have 2 elements, 1, which has multiples in each row 1 2 3, and the next element in set A to have multiples in each row is 2, it has 4 in first row, 8 in second and 6 in third. we have thus 2 elements in set A which have multiples in each row of the N*N grid.
$endgroup$
– sonorous
Jan 10 at 15:32
$begingroup$
for n=3, we have A={1,2,3} and in this set, we have 2 elements, 1, which has multiples in each row 1 2 3, and the next element in set A to have multiples in each row is 2, it has 4 in first row, 8 in second and 6 in third. we have thus 2 elements in set A which have multiples in each row of the N*N grid.
$endgroup$
– sonorous
Jan 10 at 15:32
$begingroup$
for n=3, we have A={1,2,3} and in this set, we have 2 elements, 1, which has multiples in each row 1 2 3, and the next element in set A to have multiples in each row is 2, it has 4 in first row, 8 in second and 6 in third. we have thus 2 elements in set A which have multiples in each row of the N*N grid.
$endgroup$
– sonorous
Jan 10 at 15:32
add a comment |
1 Answer
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$begingroup$
For a given number $k in A$, consider the elements in each row $bmod k$.
If $k$ is coprime to $N$ we will have at least one with every residue in $Bbb {Z/Z}_k$, so there will be a $0$ and $k$ will have a multiple in the row.
If $k$ is not coprime to $N$ there is a row where the left hand element is not a multiple of $(k,N)$. All the elements of the row will be equivalent to this $mod (k,N)$, so will not be a multiple of $k$.
Hence the number of elements of $A$ that have a multiple in each row is $varphi(N)$
answered Jan 10 at 15:52
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
If k is coprime to N we will have at least one with every residue in Z/Zk, so there will be a 0 and k will have a multiple in the row. could you please elaborate this?
$endgroup$
– sonorous
Jan 10 at 16:01
$begingroup$
Let $N=6,k=5$. If we look at the elements in row $4$ but reduce them $bmod 5$ you get $4,0,1,2,3,4,5$ and there is a $0$. If we look at the elements in row $2$ you get $2,3,4,5,0.1.2$. If we take $k=3$ row $1$ will have $1,4,1,4,1,4$ and there is no zero. It is standard in cyclic groups that any number coprime to the order of the group generates it.
$endgroup$
– Ross Millikan
Jan 10 at 16:06
add a comment |
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1 Answer
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$begingroup$
For a given number $k in A$, consider the elements in each row $bmod k$.
If $k$ is coprime to $N$ we will have at least one with every residue in $Bbb {Z/Z}_k$, so there will be a $0$ and $k$ will have a multiple in the row.
If $k$ is not coprime to $N$ there is a row where the left hand element is not a multiple of $(k,N)$. All the elements of the row will be equivalent to this $mod (k,N)$, so will not be a multiple of $k$.
Hence the number of elements of $A$ that have a multiple in each row is $varphi(N)$
answered Jan 10 at 15:52
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
If k is coprime to N we will have at least one with every residue in Z/Zk, so there will be a 0 and k will have a multiple in the row. could you please elaborate this?
$endgroup$
– sonorous
Jan 10 at 16:01
$begingroup$
Let $N=6,k=5$. If we look at the elements in row $4$ but reduce them $bmod 5$ you get $4,0,1,2,3,4,5$ and there is a $0$. If we look at the elements in row $2$ you get $2,3,4,5,0.1.2$. If we take $k=3$ row $1$ will have $1,4,1,4,1,4$ and there is no zero. It is standard in cyclic groups that any number coprime to the order of the group generates it.
$endgroup$
– Ross Millikan
Jan 10 at 16:06
add a comment |
$begingroup$
For a given number $k in A$, consider the elements in each row $bmod k$.
If $k$ is coprime to $N$ we will have at least one with every residue in $Bbb {Z/Z}_k$, so there will be a $0$ and $k$ will have a multiple in the row.
If $k$ is not coprime to $N$ there is a row where the left hand element is not a multiple of $(k,N)$. All the elements of the row will be equivalent to this $mod (k,N)$, so will not be a multiple of $k$.
Hence the number of elements of $A$ that have a multiple in each row is $varphi(N)$
answered Jan 10 at 15:52
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
If k is coprime to N we will have at least one with every residue in Z/Zk, so there will be a 0 and k will have a multiple in the row. could you please elaborate this?
$endgroup$
– sonorous
Jan 10 at 16:01
$begingroup$
Let $N=6,k=5$. If we look at the elements in row $4$ but reduce them $bmod 5$ you get $4,0,1,2,3,4,5$ and there is a $0$. If we look at the elements in row $2$ you get $2,3,4,5,0.1.2$. If we take $k=3$ row $1$ will have $1,4,1,4,1,4$ and there is no zero. It is standard in cyclic groups that any number coprime to the order of the group generates it.
$endgroup$
– Ross Millikan
Jan 10 at 16:06
add a comment |
$begingroup$
For a given number $k in A$, consider the elements in each row $bmod k$.
If $k$ is coprime to $N$ we will have at least one with every residue in $Bbb {Z/Z}_k$, so there will be a $0$ and $k$ will have a multiple in the row.
If $k$ is not coprime to $N$ there is a row where the left hand element is not a multiple of $(k,N)$. All the elements of the row will be equivalent to this $mod (k,N)$, so will not be a multiple of $k$.
Hence the number of elements of $A$ that have a multiple in each row is $varphi(N)$
answered Jan 10 at 15:52
Ross MillikanRoss Millikan
301k24200375
$endgroup$
For a given number $k in A$, consider the elements in each row $bmod k$.
If $k$ is coprime to $N$ we will have at least one with every residue in $Bbb {Z/Z}_k$, so there will be a $0$ and $k$ will have a multiple in the row.
If $k$ is not coprime to $N$ there is a row where the left hand element is not a multiple of $(k,N)$. All the elements of the row will be equivalent to this $mod (k,N)$, so will not be a multiple of $k$.
Hence the number of elements of $A$ that have a multiple in each row is $varphi(N)$
answered Jan 10 at 15:52
Ross MillikanRoss Millikan
301k24200375
answered Jan 10 at 15:52
Ross MillikanRoss Millikan
301k24200375
answered Jan 10 at 15:52
Ross MillikanRoss Millikan
301k24200375
answered Jan 10 at 15:52
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
If k is coprime to N we will have at least one with every residue in Z/Zk, so there will be a 0 and k will have a multiple in the row. could you please elaborate this?
$endgroup$
– sonorous
Jan 10 at 16:01
$begingroup$
Let $N=6,k=5$. If we look at the elements in row $4$ but reduce them $bmod 5$ you get $4,0,1,2,3,4,5$ and there is a $0$. If we look at the elements in row $2$ you get $2,3,4,5,0.1.2$. If we take $k=3$ row $1$ will have $1,4,1,4,1,4$ and there is no zero. It is standard in cyclic groups that any number coprime to the order of the group generates it.
$endgroup$
– Ross Millikan
Jan 10 at 16:06
add a comment |
$begingroup$
If k is coprime to N we will have at least one with every residue in Z/Zk, so there will be a 0 and k will have a multiple in the row. could you please elaborate this?
$endgroup$
– sonorous
Jan 10 at 16:01
$begingroup$
Let $N=6,k=5$. If we look at the elements in row $4$ but reduce them $bmod 5$ you get $4,0,1,2,3,4,5$ and there is a $0$. If we look at the elements in row $2$ you get $2,3,4,5,0.1.2$. If we take $k=3$ row $1$ will have $1,4,1,4,1,4$ and there is no zero. It is standard in cyclic groups that any number coprime to the order of the group generates it.
$endgroup$
– Ross Millikan
Jan 10 at 16:06
$begingroup$
If k is coprime to N we will have at least one with every residue in Z/Zk, so there will be a 0 and k will have a multiple in the row. could you please elaborate this?
$endgroup$
– sonorous
Jan 10 at 16:01
$begingroup$
If k is coprime to N we will have at least one with every residue in Z/Zk, so there will be a 0 and k will have a multiple in the row. could you please elaborate this?
$endgroup$
– sonorous
Jan 10 at 16:01
$begingroup$
Let $N=6,k=5$. If we look at the elements in row $4$ but reduce them $bmod 5$ you get $4,0,1,2,3,4,5$ and there is a $0$. If we look at the elements in row $2$ you get $2,3,4,5,0.1.2$. If we take $k=3$ row $1$ will have $1,4,1,4,1,4$ and there is no zero. It is standard in cyclic groups that any number coprime to the order of the group generates it.
$endgroup$
– Ross Millikan
Jan 10 at 16:06
$begingroup$
Let $N=6,k=5$. If we look at the elements in row $4$ but reduce them $bmod 5$ you get $4,0,1,2,3,4,5$ and there is a $0$. If we look at the elements in row $2$ you get $2,3,4,5,0.1.2$. If we take $k=3$ row $1$ will have $1,4,1,4,1,4$ and there is no zero. It is standard in cyclic groups that any number coprime to the order of the group generates it.
$endgroup$
– Ross Millikan
Jan 10 at 16:06
add a comment |
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$begingroup$
for n=3, we have A={1,2,3} and in this set, we have 2 elements, 1, which has multiples in each row 1 2 3, and the next element in set A to have multiples in each row is 2, it has 4 in first row, 8 in second and 6 in third. we have thus 2 elements in set A which have multiples in each row of the N*N grid.
$endgroup$
– sonorous
Jan 10 at 15:32