powers and factorials

**thippli** · January 22nd 2009, 10:52 PM

let $p, r_1 ,r_2,s_1,s_2$ be positive integers with
$p>1 ; r_1 < r_2$ and $s_1<s_2$ then

(1) $p^{r_1}+ p^{r_2}= p^{s_1}+ p^{s_2}$
if and only if $r_i = s_i$ , $i=1,2$ .

(2) $r_1r_1!+ r_2 r_2!= s_1 s_1!+ s_2s_2!$
if and only if $r_i = s_i$ , $i=1,2$ .

is it correct ?

the one way in each case is obvious. the other way...?

plz help.

**Laurent** · January 24th 2009, 03:22 AM

Originally Posted by thippli

let $p, r_1 ,r_2,s_1,s_2$ be positive integers with
$p>1 ; r_1 < r_2$ and $s_1<s_2$ then

(1) $p^{r_1}+ p^{r_2}= p^{s_1}+ p^{s_2}$
if and only if $r_i = s_i$ , $i=1,2$ .

Note that $p^{r_1}+p^{r_2}=p^{r_1}(1+p^{r_2-r_1})$ , and $1+p^{r_2-r_1}$ is not divisible by $p$ (except if $r_1=r_2$ and $p\neq 2$ , but remember $r_1<r_2$ ). As a consequence, $r_1$ is the largest number $m$ such $p^m$ divides $p^{r_1}+p^{r_2}$ . Can you see why this answers your question ?

(2) $r_1r_1!+ r_2 r_2!= s_1 s_1!+ s_2s_2!$
if and only if $r_i = s_i$ , $i=1,2$ .

You can solve this by the same method like above. Remember that $r_2!=r_1! \times (r_1+1)(r_1+2)\cdots (r_2-1)r_2$ .

**thippli** · February 2nd 2009, 03:21 AM

First answer is very clear. Thank you very much !

But I think the second won't work , because

$r_1r_1 ! + r_2r_2! = r_1 r_1!(1+ \frac{r_2} {r_1} \frac{r_2!}{r_1!})$ .

Here $r_1$ may divide $r_2$ . In this case how can we say $r_1r_1!$ is the largest integer which divides
$<br /> 1+ \frac{r_2} {r_1} \frac{r_2!}{r_1!}<br />$ ?

**Laurent** · February 3rd 2009, 03:25 AM

Originally Posted by thippli

First answer is very clear. Thank you very much !

But I think the second won't work , because

$r_1r_1 ! + r_2r_2! = r_1 r_1!(1+ c)$ .

Here $r_1$ may divide $r_2$ . In this case how can we say $r_1r_1!$ is the largest integer which divides
$<br /> 1+ \frac{r_2} {r_1} \frac{r_2!}{r_1!}<br />$ ?

You don't even know if $\frac{r_2} {r_1} \frac{r_2!}{r_1!}$ is an integer, so this is not the right factorization. What works however is:
$r_1 r_1!+r_2 r_2!=r_1!\left(r_1+r_2\frac{r_2!}{r_1!}\right)=r_1 !\left(r_1+r_2(r_1+1)(r_1+2)\cdots r_2\right).$
Notice indeed that $r_1+1$ does not divide the second factor, so that $r_1$ is the largest $n$ such that $n!$ divides $r_1r_1!+r_2r_2!$ .

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