Find the number of elements in set $A={1,2,ldots, N}$ such that the element has multiples in each row of a $N...












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 15:52









littleO

30.5k648111




30.5k648111










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $varphi(N)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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0












$begingroup$

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 $varphi(N)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















0












$begingroup$

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 $varphi(N)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

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 $varphi(N)$






share|cite|improve this answer









$endgroup$



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 $varphi(N)$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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