How many numbers less than $m$ and relatively prime to $n$, where $m>n$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Count of numbers where both $n$ and $n+1$ have all prime factors less than $x$Counting elements of reduced residue systems modulo one number which are smaller than anotherFind all integers less than $m$ that are relatively prime to itInequality $varphi(N)>pi(N)$?Euler's totient function for large numbersHow find a closed form for the numbers which are relatively prime to $10$,On factoring and integer given the value of its Euler's totient function.How to count the number of perfect square greater than $N$ and less than $N^2$ that are relatively prime to $N$?Problem about the set of relatively prime integersSum of the number of relatively prime integers up to $x$, $x-1$, $ldots$, $1$
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How many numbers less than $m$ and relatively prime to $n$, where $m>n$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Count of numbers where both $n$ and $n+1$ have all prime factors less than $x$Counting elements of reduced residue systems modulo one number which are smaller than anotherFind all integers less than $m$ that are relatively prime to itInequality $varphi(N)>pi(N)$?Euler's totient function for large numbersHow find a closed form for the numbers which are relatively prime to $10$,On factoring and integer given the value of its Euler's totient function.How to count the number of perfect square greater than $N$ and less than $N^2$ that are relatively prime to $N$?Problem about the set of relatively prime integersSum of the number of relatively prime integers up to $x$, $x-1$, $ldots$, $1$
$begingroup$
Let $m$ and $n$ be two integers such that $m>n$. Then find the number of integers less than $m$ and relatively prime to $n$.
I had come across a problem of this type with specific values for $m$ and $n$, unfortunately, the values I don't remember anymore. So I asked this general question. How to solve this type of question?
My thoughts are that we need to count the numbers. So we can count the number of integers less than $m$ and relatively prime to $m$ using Euler's totient function. Now among those numbers, also belongs the numbers which are relatively prime to $n$. We can also find the numbers less than $n$ and relatively prime to $n$. That will give a rough estimation. But how do we know about the actual number of such integers? Can anybody help me with this?
Thanks.
number-theory totient-function coprime
asked May 28 '17 at 15:13
Kushal BhuyanKushal Bhuyan
5,08021246
$endgroup$
|
show 2 more comments
$begingroup$
Let $m$ and $n$ be two integers such that $m>n$. Then find the number of integers less than $m$ and relatively prime to $n$.
I had come across a problem of this type with specific values for $m$ and $n$, unfortunately, the values I don't remember anymore. So I asked this general question. How to solve this type of question?
My thoughts are that we need to count the numbers. So we can count the number of integers less than $m$ and relatively prime to $m$ using Euler's totient function. Now among those numbers, also belongs the numbers which are relatively prime to $n$. We can also find the numbers less than $n$ and relatively prime to $n$. That will give a rough estimation. But how do we know about the actual number of such integers? Can anybody help me with this?
Thanks.
number-theory totient-function coprime
asked May 28 '17 at 15:13
Kushal BhuyanKushal Bhuyan
5,08021246
$endgroup$
1
$begingroup$
Note that $gcd(k,n) = gcd(k + n, n)$.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:23
$begingroup$
@DanielFischer can you elaborate a bit more on how this relation will help me, please?
$endgroup$
– Kushal Bhuyan
May 28 '17 at 15:29
$begingroup$
It tells you how many integers $an < k leqslant (a+1)n$ are coprime to $n$. If $m$ is a multiple of $n$, that finishes it. If $m = qcdot n + r$ with $0 < r < n$, the remaining part of the problem is difficult.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:32
$begingroup$
@DanielFischer if $m$ is not a multiple of $n$, then how difficult is the problem to solve
$endgroup$
– Kushal Bhuyan
May 28 '17 at 16:41
3
$begingroup$
Let us define $Phi(n,m)$ as the number of positive integers coprime to $n$ and not exceeding $m$. Then the above gives us $$Phi(n,m) = biggllfloor fracmnbiggrrfloorcdot varphi(n) + Phibiggl(n, m - ncdot biggllfloor fracmnbiggrrfloorBiggr).$$ Assuming one knows $varphi(n)$, the problem is then to find $Phi(n,r)$ for $0 leqslant r < n$. If the numbers coprime to $n$ were evenly distributed, then $Phi(n,r)$ would be close to $fracrn-1cdot varphi(n)$. But since the numbers coprime to $n$ are in general not very close to being evenly distributed,
$endgroup$
– Daniel Fischer
May 28 '17 at 17:35
|
show 2 more comments
$begingroup$
Let $m$ and $n$ be two integers such that $m>n$. Then find the number of integers less than $m$ and relatively prime to $n$.
I had come across a problem of this type with specific values for $m$ and $n$, unfortunately, the values I don't remember anymore. So I asked this general question. How to solve this type of question?
My thoughts are that we need to count the numbers. So we can count the number of integers less than $m$ and relatively prime to $m$ using Euler's totient function. Now among those numbers, also belongs the numbers which are relatively prime to $n$. We can also find the numbers less than $n$ and relatively prime to $n$. That will give a rough estimation. But how do we know about the actual number of such integers? Can anybody help me with this?
Thanks.
number-theory totient-function coprime
asked May 28 '17 at 15:13
Kushal BhuyanKushal Bhuyan
5,08021246
$endgroup$
Let $m$ and $n$ be two integers such that $m>n$. Then find the number of integers less than $m$ and relatively prime to $n$.
I had come across a problem of this type with specific values for $m$ and $n$, unfortunately, the values I don't remember anymore. So I asked this general question. How to solve this type of question?
My thoughts are that we need to count the numbers. So we can count the number of integers less than $m$ and relatively prime to $m$ using Euler's totient function. Now among those numbers, also belongs the numbers which are relatively prime to $n$. We can also find the numbers less than $n$ and relatively prime to $n$. That will give a rough estimation. But how do we know about the actual number of such integers? Can anybody help me with this?
Thanks.
number-theory totient-function coprime
number-theory totient-function coprime
asked May 28 '17 at 15:13
Kushal BhuyanKushal Bhuyan
5,08021246
asked May 28 '17 at 15:13
Kushal BhuyanKushal Bhuyan
5,08021246
asked May 28 '17 at 15:13
Kushal BhuyanKushal Bhuyan
5,08021246
asked May 28 '17 at 15:13
Kushal BhuyanKushal Bhuyan
5,08021246
asked May 28 '17 at 15:13
Kushal BhuyanKushal Bhuyan
5,08021246
5,08021246
1
$begingroup$
Note that $gcd(k,n) = gcd(k + n, n)$.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:23
$begingroup$
@DanielFischer can you elaborate a bit more on how this relation will help me, please?
$endgroup$
– Kushal Bhuyan
May 28 '17 at 15:29
$begingroup$
It tells you how many integers $an < k leqslant (a+1)n$ are coprime to $n$. If $m$ is a multiple of $n$, that finishes it. If $m = qcdot n + r$ with $0 < r < n$, the remaining part of the problem is difficult.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:32
$begingroup$
@DanielFischer if $m$ is not a multiple of $n$, then how difficult is the problem to solve
$endgroup$
– Kushal Bhuyan
May 28 '17 at 16:41
3
$begingroup$
Let us define $Phi(n,m)$ as the number of positive integers coprime to $n$ and not exceeding $m$. Then the above gives us $$Phi(n,m) = biggllfloor fracmnbiggrrfloorcdot varphi(n) + Phibiggl(n, m - ncdot biggllfloor fracmnbiggrrfloorBiggr).$$ Assuming one knows $varphi(n)$, the problem is then to find $Phi(n,r)$ for $0 leqslant r < n$. If the numbers coprime to $n$ were evenly distributed, then $Phi(n,r)$ would be close to $fracrn-1cdot varphi(n)$. But since the numbers coprime to $n$ are in general not very close to being evenly distributed,
$endgroup$
– Daniel Fischer
May 28 '17 at 17:35
|
show 2 more comments
1
$begingroup$
Note that $gcd(k,n) = gcd(k + n, n)$.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:23
$begingroup$
@DanielFischer can you elaborate a bit more on how this relation will help me, please?
$endgroup$
– Kushal Bhuyan
May 28 '17 at 15:29
$begingroup$
It tells you how many integers $an < k leqslant (a+1)n$ are coprime to $n$. If $m$ is a multiple of $n$, that finishes it. If $m = qcdot n + r$ with $0 < r < n$, the remaining part of the problem is difficult.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:32
$begingroup$
@DanielFischer if $m$ is not a multiple of $n$, then how difficult is the problem to solve
$endgroup$
– Kushal Bhuyan
May 28 '17 at 16:41
3
$begingroup$
Let us define $Phi(n,m)$ as the number of positive integers coprime to $n$ and not exceeding $m$. Then the above gives us $$Phi(n,m) = biggllfloor fracmnbiggrrfloorcdot varphi(n) + Phibiggl(n, m - ncdot biggllfloor fracmnbiggrrfloorBiggr).$$ Assuming one knows $varphi(n)$, the problem is then to find $Phi(n,r)$ for $0 leqslant r < n$. If the numbers coprime to $n$ were evenly distributed, then $Phi(n,r)$ would be close to $fracrn-1cdot varphi(n)$. But since the numbers coprime to $n$ are in general not very close to being evenly distributed,
$endgroup$
– Daniel Fischer
May 28 '17 at 17:35
1
1
$begingroup$
Note that $gcd(k,n) = gcd(k + n, n)$.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:23
$begingroup$
Note that $gcd(k,n) = gcd(k + n, n)$.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:23
$begingroup$
@DanielFischer can you elaborate a bit more on how this relation will help me, please?
$endgroup$
– Kushal Bhuyan
May 28 '17 at 15:29
$begingroup$
@DanielFischer can you elaborate a bit more on how this relation will help me, please?
$endgroup$
– Kushal Bhuyan
May 28 '17 at 15:29
$begingroup$
It tells you how many integers $an < k leqslant (a+1)n$ are coprime to $n$. If $m$ is a multiple of $n$, that finishes it. If $m = qcdot n + r$ with $0 < r < n$, the remaining part of the problem is difficult.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:32
$begingroup$
It tells you how many integers $an < k leqslant (a+1)n$ are coprime to $n$. If $m$ is a multiple of $n$, that finishes it. If $m = qcdot n + r$ with $0 < r < n$, the remaining part of the problem is difficult.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:32
$begingroup$
@DanielFischer if $m$ is not a multiple of $n$, then how difficult is the problem to solve
$endgroup$
– Kushal Bhuyan
May 28 '17 at 16:41
$begingroup$
@DanielFischer if $m$ is not a multiple of $n$, then how difficult is the problem to solve
$endgroup$
– Kushal Bhuyan
May 28 '17 at 16:41
3
3
$begingroup$
Let us define $Phi(n,m)$ as the number of positive integers coprime to $n$ and not exceeding $m$. Then the above gives us $$Phi(n,m) = biggllfloor fracmnbiggrrfloorcdot varphi(n) + Phibiggl(n, m - ncdot biggllfloor fracmnbiggrrfloorBiggr).$$ Assuming one knows $varphi(n)$, the problem is then to find $Phi(n,r)$ for $0 leqslant r < n$. If the numbers coprime to $n$ were evenly distributed, then $Phi(n,r)$ would be close to $fracrn-1cdot varphi(n)$. But since the numbers coprime to $n$ are in general not very close to being evenly distributed,
$endgroup$
– Daniel Fischer
May 28 '17 at 17:35
$begingroup$
Let us define $Phi(n,m)$ as the number of positive integers coprime to $n$ and not exceeding $m$. Then the above gives us $$Phi(n,m) = biggllfloor fracmnbiggrrfloorcdot varphi(n) + Phibiggl(n, m - ncdot biggllfloor fracmnbiggrrfloorBiggr).$$ Assuming one knows $varphi(n)$, the problem is then to find $Phi(n,r)$ for $0 leqslant r < n$. If the numbers coprime to $n$ were evenly distributed, then $Phi(n,r)$ would be close to $fracrn-1cdot varphi(n)$. But since the numbers coprime to $n$ are in general not very close to being evenly distributed,
$endgroup$
– Daniel Fischer
May 28 '17 at 17:35
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If primes $p_1, p_2$ are the only prime factors of $n$, then we get numbers relatively prime to $n$ and less than $m$ are (assuming $p_1, p_2$ are not factors of $m$):
$$m - lfloorfracmp_1rfloor - lfloorfracmp_2rfloor + lfloorfracmp_1p_2rfloor$$
We can have a generalized expression for any number of prime factors of $n$ using PIE.
answered May 28 '17 at 20:01
skusku
1,013311
$endgroup$
add a comment |
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$begingroup$
If primes $p_1, p_2$ are the only prime factors of $n$, then we get numbers relatively prime to $n$ and less than $m$ are (assuming $p_1, p_2$ are not factors of $m$):
$$m - lfloorfracmp_1rfloor - lfloorfracmp_2rfloor + lfloorfracmp_1p_2rfloor$$
We can have a generalized expression for any number of prime factors of $n$ using PIE.
answered May 28 '17 at 20:01
skusku
1,013311
$endgroup$
add a comment |
$begingroup$
If primes $p_1, p_2$ are the only prime factors of $n$, then we get numbers relatively prime to $n$ and less than $m$ are (assuming $p_1, p_2$ are not factors of $m$):
$$m - lfloorfracmp_1rfloor - lfloorfracmp_2rfloor + lfloorfracmp_1p_2rfloor$$
We can have a generalized expression for any number of prime factors of $n$ using PIE.
answered May 28 '17 at 20:01
skusku
1,013311
$endgroup$
add a comment |
$begingroup$
If primes $p_1, p_2$ are the only prime factors of $n$, then we get numbers relatively prime to $n$ and less than $m$ are (assuming $p_1, p_2$ are not factors of $m$):
$$m - lfloorfracmp_1rfloor - lfloorfracmp_2rfloor + lfloorfracmp_1p_2rfloor$$
We can have a generalized expression for any number of prime factors of $n$ using PIE.
answered May 28 '17 at 20:01
skusku
1,013311
$endgroup$
If primes $p_1, p_2$ are the only prime factors of $n$, then we get numbers relatively prime to $n$ and less than $m$ are (assuming $p_1, p_2$ are not factors of $m$):
$$m - lfloorfracmp_1rfloor - lfloorfracmp_2rfloor + lfloorfracmp_1p_2rfloor$$
We can have a generalized expression for any number of prime factors of $n$ using PIE.
answered May 28 '17 at 20:01
skusku
1,013311
edited May 28 '17 at 20:08
answered May 28 '17 at 20:01
skusku
1,013311
answered May 28 '17 at 20:01
skusku
1,013311
answered May 28 '17 at 20:01
skusku
1,013311
1,013311
add a comment |
add a comment |
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1
$begingroup$
Note that $gcd(k,n) = gcd(k + n, n)$.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:23
$begingroup$
@DanielFischer can you elaborate a bit more on how this relation will help me, please?
$endgroup$
– Kushal Bhuyan
May 28 '17 at 15:29
$begingroup$
It tells you how many integers $an < k leqslant (a+1)n$ are coprime to $n$. If $m$ is a multiple of $n$, that finishes it. If $m = qcdot n + r$ with $0 < r < n$, the remaining part of the problem is difficult.
$endgroup$
– Daniel Fischer
May 28 '17 at 15:32
$begingroup$
@DanielFischer if $m$ is not a multiple of $n$, then how difficult is the problem to solve
$endgroup$
– Kushal Bhuyan
May 28 '17 at 16:41
3
$begingroup$
Let us define $Phi(n,m)$ as the number of positive integers coprime to $n$ and not exceeding $m$. Then the above gives us $$Phi(n,m) = biggllfloor fracmnbiggrrfloorcdot varphi(n) + Phibiggl(n, m - ncdot biggllfloor fracmnbiggrrfloorBiggr).$$ Assuming one knows $varphi(n)$, the problem is then to find $Phi(n,r)$ for $0 leqslant r < n$. If the numbers coprime to $n$ were evenly distributed, then $Phi(n,r)$ would be close to $fracrn-1cdot varphi(n)$. But since the numbers coprime to $n$ are in general not very close to being evenly distributed,
$endgroup$
– Daniel Fischer
May 28 '17 at 17:35