Checking if the Product of n Integers is Divisible by Prime N
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Given $n$ integers, $x_1, ... , x_n$, is there some well-known procedure or algorithm that checks if the product $x_1 * ... * x_n$ is divisible by some arbitrary prime $N$ using minimal space?
Since the product and $N$ can be arbitrarily large, I don't think you can just simply use modular arithmetic to see if the remainder is $0$. I was thinking maybe convert the product to its prime factorization, and if $N$ isn't one of the primes used, then we know the product cannot be divisible by $N$, right? Does this seem like something that could work?
prime-numbers divisibility prime-factorization
asked Nov 19 at 0:26
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755
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up vote
1
down vote
favorite
Given $n$ integers, $x_1, ... , x_n$, is there some well-known procedure or algorithm that checks if the product $x_1 * ... * x_n$ is divisible by some arbitrary prime $N$ using minimal space?
Since the product and $N$ can be arbitrarily large, I don't think you can just simply use modular arithmetic to see if the remainder is $0$. I was thinking maybe convert the product to its prime factorization, and if $N$ isn't one of the primes used, then we know the product cannot be divisible by $N$, right? Does this seem like something that could work?
prime-numbers divisibility prime-factorization
asked Nov 19 at 0:26
Axioms
755
2
You don't need to do the multiplication, you can just check to see if $p$ divides any of $x_1,cdots, x_n$. Factoring is almost certainly a bad idea...that's very costly and unnecessary.
– lulu
Nov 19 at 0:28
1
Just trial divide each $x_i$ by $p$. Then $N$ is divisible by $p$ if and only at least one $x_i$ is.
– Hagen von Eitzen
Nov 19 at 0:28
What about writing down the prime decomposition of $x_i$ first, hence deriving the prime decomposition of $x_1x_2dotsm x_n$? Then proceed as you have written.
– YiFan
Nov 19 at 0:29
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $n$ integers, $x_1, ... , x_n$, is there some well-known procedure or algorithm that checks if the product $x_1 * ... * x_n$ is divisible by some arbitrary prime $N$ using minimal space?
Since the product and $N$ can be arbitrarily large, I don't think you can just simply use modular arithmetic to see if the remainder is $0$. I was thinking maybe convert the product to its prime factorization, and if $N$ isn't one of the primes used, then we know the product cannot be divisible by $N$, right? Does this seem like something that could work?
prime-numbers divisibility prime-factorization
asked Nov 19 at 0:26
Axioms
755
Given $n$ integers, $x_1, ... , x_n$, is there some well-known procedure or algorithm that checks if the product $x_1 * ... * x_n$ is divisible by some arbitrary prime $N$ using minimal space?
Since the product and $N$ can be arbitrarily large, I don't think you can just simply use modular arithmetic to see if the remainder is $0$. I was thinking maybe convert the product to its prime factorization, and if $N$ isn't one of the primes used, then we know the product cannot be divisible by $N$, right? Does this seem like something that could work?
prime-numbers divisibility prime-factorization
prime-numbers divisibility prime-factorization
asked Nov 19 at 0:26
Axioms
755
asked Nov 19 at 0:26
Axioms
755
asked Nov 19 at 0:26
Axioms
755
asked Nov 19 at 0:26
Axioms
755
asked Nov 19 at 0:26
Axioms
755
755
2
You don't need to do the multiplication, you can just check to see if $p$ divides any of $x_1,cdots, x_n$. Factoring is almost certainly a bad idea...that's very costly and unnecessary.
– lulu
Nov 19 at 0:28
1
Just trial divide each $x_i$ by $p$. Then $N$ is divisible by $p$ if and only at least one $x_i$ is.
– Hagen von Eitzen
Nov 19 at 0:28
What about writing down the prime decomposition of $x_i$ first, hence deriving the prime decomposition of $x_1x_2dotsm x_n$? Then proceed as you have written.
– YiFan
Nov 19 at 0:29
add a comment |
2
You don't need to do the multiplication, you can just check to see if $p$ divides any of $x_1,cdots, x_n$. Factoring is almost certainly a bad idea...that's very costly and unnecessary.
– lulu
Nov 19 at 0:28
1
Just trial divide each $x_i$ by $p$. Then $N$ is divisible by $p$ if and only at least one $x_i$ is.
– Hagen von Eitzen
Nov 19 at 0:28
What about writing down the prime decomposition of $x_i$ first, hence deriving the prime decomposition of $x_1x_2dotsm x_n$? Then proceed as you have written.
– YiFan
Nov 19 at 0:29
2
2
You don't need to do the multiplication, you can just check to see if $p$ divides any of $x_1,cdots, x_n$. Factoring is almost certainly a bad idea...that's very costly and unnecessary.
– lulu
Nov 19 at 0:28
You don't need to do the multiplication, you can just check to see if $p$ divides any of $x_1,cdots, x_n$. Factoring is almost certainly a bad idea...that's very costly and unnecessary.
– lulu
Nov 19 at 0:28
1
1
Just trial divide each $x_i$ by $p$. Then $N$ is divisible by $p$ if and only at least one $x_i$ is.
– Hagen von Eitzen
Nov 19 at 0:28
Just trial divide each $x_i$ by $p$. Then $N$ is divisible by $p$ if and only at least one $x_i$ is.
– Hagen von Eitzen
Nov 19 at 0:28
What about writing down the prime decomposition of $x_i$ first, hence deriving the prime decomposition of $x_1x_2dotsm x_n$? Then proceed as you have written.
– YiFan
Nov 19 at 0:29
What about writing down the prime decomposition of $x_i$ first, hence deriving the prime decomposition of $x_1x_2dotsm x_n$? Then proceed as you have written.
– YiFan
Nov 19 at 0:29
add a comment |
1 Answer
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Note that primes are prime: that is, if $p mid a b$ then $p mid a$ or $p mid b$. (This is Euclid's lemma.) Inductively, then, if $p mid x_1 x_2 dots x_n$ then $p$ divides some $x_i$. So just check each of the $x_i$ in turn. Modular arithmetic is fine for that.
answered Nov 19 at 0:32
Patrick Stevens
28.2k52874
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Note that primes are prime: that is, if $p mid a b$ then $p mid a$ or $p mid b$. (This is Euclid's lemma.) Inductively, then, if $p mid x_1 x_2 dots x_n$ then $p$ divides some $x_i$. So just check each of the $x_i$ in turn. Modular arithmetic is fine for that.
answered Nov 19 at 0:32
Patrick Stevens
28.2k52874
add a comment |
up vote
3
down vote
Note that primes are prime: that is, if $p mid a b$ then $p mid a$ or $p mid b$. (This is Euclid's lemma.) Inductively, then, if $p mid x_1 x_2 dots x_n$ then $p$ divides some $x_i$. So just check each of the $x_i$ in turn. Modular arithmetic is fine for that.
answered Nov 19 at 0:32
Patrick Stevens
28.2k52874
add a comment |
up vote
3
down vote
up vote
3
down vote
Note that primes are prime: that is, if $p mid a b$ then $p mid a$ or $p mid b$. (This is Euclid's lemma.) Inductively, then, if $p mid x_1 x_2 dots x_n$ then $p$ divides some $x_i$. So just check each of the $x_i$ in turn. Modular arithmetic is fine for that.
answered Nov 19 at 0:32
Patrick Stevens
28.2k52874
Note that primes are prime: that is, if $p mid a b$ then $p mid a$ or $p mid b$. (This is Euclid's lemma.) Inductively, then, if $p mid x_1 x_2 dots x_n$ then $p$ divides some $x_i$. So just check each of the $x_i$ in turn. Modular arithmetic is fine for that.
answered Nov 19 at 0:32
Patrick Stevens
28.2k52874
answered Nov 19 at 0:32
Patrick Stevens
28.2k52874
answered Nov 19 at 0:32
Patrick Stevens
28.2k52874
answered Nov 19 at 0:32
Patrick Stevens
28.2k52874
28.2k52874
add a comment |
add a comment |
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You don't need to do the multiplication, you can just check to see if $p$ divides any of $x_1,cdots, x_n$. Factoring is almost certainly a bad idea...that's very costly and unnecessary.
– lulu
Nov 19 at 0:28
1
Just trial divide each $x_i$ by $p$. Then $N$ is divisible by $p$ if and only at least one $x_i$ is.
– Hagen von Eitzen
Nov 19 at 0:28
What about writing down the prime decomposition of $x_i$ first, hence deriving the prime decomposition of $x_1x_2dotsm x_n$? Then proceed as you have written.
– YiFan
Nov 19 at 0:29