Mixing
Product of n integers of AP is divisible by $n!$
$begingroup$
Prove that the product of the $n$ integers of an arithmetic progression of $n$ terms is divisible by $n!$ if common difference is relatively prime to $n!$.
First part of the question was to prove for d=1 which I was able to do using binomial coefficient. In this question though I am absolutely stumped.
number-theory divisibility
asked Jan 29 at 7:45
sn24sn24
34716
$endgroup$
add a comment |
$begingroup$
Prove that the product of the $n$ integers of an arithmetic progression of $n$ terms is divisible by $n!$ if common difference is relatively prime to $n!$.
First part of the question was to prove for d=1 which I was able to do using binomial coefficient. In this question though I am absolutely stumped.
number-theory divisibility
asked Jan 29 at 7:45
sn24sn24
34716
$endgroup$
add a comment |
$begingroup$
Prove that the product of the $n$ integers of an arithmetic progression of $n$ terms is divisible by $n!$ if common difference is relatively prime to $n!$.
First part of the question was to prove for d=1 which I was able to do using binomial coefficient. In this question though I am absolutely stumped.
number-theory divisibility
asked Jan 29 at 7:45
sn24sn24
34716
$endgroup$
Prove that the product of the $n$ integers of an arithmetic progression of $n$ terms is divisible by $n!$ if common difference is relatively prime to $n!$.
First part of the question was to prove for d=1 which I was able to do using binomial coefficient. In this question though I am absolutely stumped.
number-theory divisibility
number-theory divisibility
asked Jan 29 at 7:45
sn24sn24
34716
asked Jan 29 at 7:45
sn24sn24
34716
asked Jan 29 at 7:45
sn24sn24
34716
asked Jan 29 at 7:45
sn24sn24
34716
asked Jan 29 at 7:45
sn24sn24
34716
34716
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Assume that the result is true for $d=1$, so $n!$ divides $(a+1)cdots(a+n)$ for any $a$. In modular mathematics, this is written as $(a+1)cdots(a+n)equiv 0pmod{n!}$.
If $d$ is coprime to $n!$, then it is invertible, and has some inverse $c$, $cdequiv 1pmod{n!}$. So $$(a+d)(a+2d)cdots(a+nd)equiv d^n(ac+1)(ac+2)cdots(ac+n)equiv 0pmod{n!}$$
edited Jan 29 at 15:17
J. W. Tanner
4,0711320
answered Jan 29 at 9:40
ChrystomathChrystomath
1,898513
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Assume that the result is true for $d=1$, so $n!$ divides $(a+1)cdots(a+n)$ for any $a$. In modular mathematics, this is written as $(a+1)cdots(a+n)equiv 0pmod{n!}$.
If $d$ is coprime to $n!$, then it is invertible, and has some inverse $c$, $cdequiv 1pmod{n!}$. So $$(a+d)(a+2d)cdots(a+nd)equiv d^n(ac+1)(ac+2)cdots(ac+n)equiv 0pmod{n!}$$
edited Jan 29 at 15:17
J. W. Tanner
4,0711320
answered Jan 29 at 9:40
ChrystomathChrystomath
1,898513
$endgroup$
add a comment |
$begingroup$
Assume that the result is true for $d=1$, so $n!$ divides $(a+1)cdots(a+n)$ for any $a$. In modular mathematics, this is written as $(a+1)cdots(a+n)equiv 0pmod{n!}$.
If $d$ is coprime to $n!$, then it is invertible, and has some inverse $c$, $cdequiv 1pmod{n!}$. So $$(a+d)(a+2d)cdots(a+nd)equiv d^n(ac+1)(ac+2)cdots(ac+n)equiv 0pmod{n!}$$
edited Jan 29 at 15:17
J. W. Tanner
4,0711320
answered Jan 29 at 9:40
ChrystomathChrystomath
1,898513
$endgroup$
add a comment |
$begingroup$
Assume that the result is true for $d=1$, so $n!$ divides $(a+1)cdots(a+n)$ for any $a$. In modular mathematics, this is written as $(a+1)cdots(a+n)equiv 0pmod{n!}$.
If $d$ is coprime to $n!$, then it is invertible, and has some inverse $c$, $cdequiv 1pmod{n!}$. So $$(a+d)(a+2d)cdots(a+nd)equiv d^n(ac+1)(ac+2)cdots(ac+n)equiv 0pmod{n!}$$
edited Jan 29 at 15:17
J. W. Tanner
4,0711320
answered Jan 29 at 9:40
ChrystomathChrystomath
1,898513
$endgroup$
Assume that the result is true for $d=1$, so $n!$ divides $(a+1)cdots(a+n)$ for any $a$. In modular mathematics, this is written as $(a+1)cdots(a+n)equiv 0pmod{n!}$.
If $d$ is coprime to $n!$, then it is invertible, and has some inverse $c$, $cdequiv 1pmod{n!}$. So $$(a+d)(a+2d)cdots(a+nd)equiv d^n(ac+1)(ac+2)cdots(ac+n)equiv 0pmod{n!}$$
edited Jan 29 at 15:17
J. W. Tanner
4,0711320
answered Jan 29 at 9:40
ChrystomathChrystomath
1,898513
edited Jan 29 at 15:17
J. W. Tanner
4,0711320
edited Jan 29 at 15:17
J. W. Tanner
4,0711320
edited Jan 29 at 15:17
J. W. Tanner
4,0711320
4,0711320
answered Jan 29 at 9:40
ChrystomathChrystomath
1,898513
answered Jan 29 at 9:40
ChrystomathChrystomath
1,898513
answered Jan 29 at 9:40
ChrystomathChrystomath
1,898513
1,898513
add a comment |
add a comment |
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