Least Common Multiple and the product of a sequence of consecutive integers

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Let $x>0, n>0$ be integers.

Let LCM$(x+1, x+2, dots, x+n)$ be the least common multiple of $x+1, x+2, dots, x+n$.

It seems to me that:

$$frac(x+1)(x+2)timesdotstimes(x+n)textLCM(x+1, x+2, dots, x+n) le (n-1)!$$

Here is my thinking:

Let $p$ be any prime that divides $(x+1)timesdotstimes(x+n)$.

Let $x+i$ be the integer in $x+1,dots, x+n$ that is divisible by the highest power of $p$.

All integers in the sequence $x+1, dots, x+n$ that are divisible by $p$ will have the form $x+i+ap$ or $x+i-ap$ where $a>0$ is an integer where $x+1 le x+i - ap le x+n$ or $x+1 le x+i+ap le x+n$.

This means that the highest $a$ will be $lfloorfracn-1prfloor$

So, it follows that the maximum power of $p$ that divides $dfrac(x+1)timesdotstimes(x+n)textLCM(x+1,dots,x+n)$ is the same as the maximum power of $p$ that divides $(n-1)!$

asked Sep 6 at 5:23

Larry Freeman

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1
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favorite

Let $x>0, n>0$ be integers.

Let LCM$(x+1, x+2, dots, x+n)$ be the least common multiple of $x+1, x+2, dots, x+n$.

It seems to me that:

$$frac(x+1)(x+2)timesdotstimes(x+n)textLCM(x+1, x+2, dots, x+n) le (n-1)!$$

Here is my thinking:

Let $p$ be any prime that divides $(x+1)timesdotstimes(x+n)$.

Let $x+i$ be the integer in $x+1,dots, x+n$ that is divisible by the highest power of $p$.

All integers in the sequence $x+1, dots, x+n$ that are divisible by $p$ will have the form $x+i+ap$ or $x+i-ap$ where $a>0$ is an integer where $x+1 le x+i - ap le x+n$ or $x+1 le x+i+ap le x+n$.

This means that the highest $a$ will be $lfloorfracn-1prfloor$

So, it follows that the maximum power of $p$ that divides $dfrac(x+1)timesdotstimes(x+n)textLCM(x+1,dots,x+n)$ is the same as the maximum power of $p$ that divides $(n-1)!$

asked Sep 6 at 5:23

Larry Freeman

3,01021136

add a commentÂ |Â

up vote
1
down vote

favorite

Let $x>0, n>0$ be integers.

Let LCM$(x+1, x+2, dots, x+n)$ be the least common multiple of $x+1, x+2, dots, x+n$.

It seems to me that:

$$frac(x+1)(x+2)timesdotstimes(x+n)textLCM(x+1, x+2, dots, x+n) le (n-1)!$$

Here is my thinking:

Let $p$ be any prime that divides $(x+1)timesdotstimes(x+n)$.

Let $x+i$ be the integer in $x+1,dots, x+n$ that is divisible by the highest power of $p$.

All integers in the sequence $x+1, dots, x+n$ that are divisible by $p$ will have the form $x+i+ap$ or $x+i-ap$ where $a>0$ is an integer where $x+1 le x+i - ap le x+n$ or $x+1 le x+i+ap le x+n$.

This means that the highest $a$ will be $lfloorfracn-1prfloor$

So, it follows that the maximum power of $p$ that divides $dfrac(x+1)timesdotstimes(x+n)textLCM(x+1,dots,x+n)$ is the same as the maximum power of $p$ that divides $(n-1)!$

asked Sep 6 at 5:23

Larry Freeman

3,01021136

Let $x>0, n>0$ be integers.

Let LCM$(x+1, x+2, dots, x+n)$ be the least common multiple of $x+1, x+2, dots, x+n$.

It seems to me that:

$$frac(x+1)(x+2)timesdotstimes(x+n)textLCM(x+1, x+2, dots, x+n) le (n-1)!$$

Here is my thinking:

Let $p$ be any prime that divides $(x+1)timesdotstimes(x+n)$.

Let $x+i$ be the integer in $x+1,dots, x+n$ that is divisible by the highest power of $p$.

All integers in the sequence $x+1, dots, x+n$ that are divisible by $p$ will have the form $x+i+ap$ or $x+i-ap$ where $a>0$ is an integer where $x+1 le x+i - ap le x+n$ or $x+1 le x+i+ap le x+n$.

This means that the highest $a$ will be $lfloorfracn-1prfloor$

So, it follows that the maximum power of $p$ that divides $dfrac(x+1)timesdotstimes(x+n)textLCM(x+1,dots,x+n)$ is the same as the maximum power of $p$ that divides $(n-1)!$

elementary-number-theory inequality prime-numbers least-common-multiple

asked Sep 6 at 5:23

Larry Freeman

3,01021136

asked Sep 6 at 5:23

Larry Freeman

3,01021136

asked Sep 6 at 5:23

Larry Freeman

3,01021136

asked Sep 6 at 5:23

Larry Freeman

3,01021136

asked Sep 6 at 5:23

Larry Freeman

3,01021136

add a commentÂ |Â

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