+0

# go to the web address below

0
899
13

visit this to view the question

and melody whatever you asked regarding the last question is right(i mean whatever query you asked about the last question is true)

the question(starting part) is actually 3^2n+2 but i thought you may understand it as 3^2n + 2 so i wrote like (3^2n)9

its one and the same way anyway

Jan 5, 2016

#9
+10

Bertie

I don't have a complete solution either but this is what I have so far.

$$\displaystyle f(n) = 9^{n+1}-8n-9$$,

so

$$\displaystyle f(n+1) = 9^{n+2}-8(n+1)-9$$ ,

in which case

$$\displaystyle \frac{f(n+1)}{f(n)}=\frac{9-\{8(n+1)+9\}/9^{n+1}}{1-\{8n+9\}/9^{n+1}}$$,

having divided top and bottom by $$\displaystyle 9^{(n+1)}$$.

That means that for increasing n, $$\displaystyle f(n+1) \approx 9f(n) \text{ or } f(n+1) \approx 2.08^{3}f(n)$$ ,

and that implies that for each increase of 1 in n, m will increase by 3.

That effect seems to kick in early,

f(1) = 64 = 2^6, so m = 6

f(2) = 704 = 2^9 + 192, so m = 9

f(3) = 6528 = 2^12 + 2432 so m = 12

f(4) = 59008 = 2^15 + 26420 so m = 15

That in turn suggests the rule

m = 3n+3.

I've not yet convinced myself whether or not that this is true for all n.

-Bertie

Jan 7, 2016

#1
0

3^2n +2 = k

3^2n = k-2

log_3(k-2)=2n

1/2(log_3(k-2)) = n

k is whatever the expression is equal to initally. you didn't say it was equal to anything at first

Jan 5, 2016
#2
0

thats not even the question

i told you to go the link

Jan 6, 2016
#3
+5

The address that you have provided is not correct. :(

Did you check it for yourself?

Anyway I am starting to feel sorry for you - you must be getting very frustrated by now.

[Aditya has been PMing me with this question for the better part of a day. ]

People can answer it here I guess.

if n belongs to natural numbers then the highest integer m such that 2^m divides (3^2n)9-8n-9

if n belongs to natural numbers then the highest integer m such that

2^m        divides     3^(2n+2)  -8n -9

I have looked at this question and have not made any immediate progress.

I am too busy to look properly now.

I will however draw attention to your question, so that other mathematicians will pay attention to it.

-----------------------------------------------------------------------

I have called for more help here:

Jan 6, 2016
edited by Melody  Jan 6, 2016
edited by Melody  Jan 6, 2016
edited by Melody  Jan 6, 2016
#4
0

THE STATEMENT IS FALSE. NO SOLUTIONS EXIST!.

Jan 6, 2016
#5
+5

Your statement may be true but it is valueless since it is comes with no proof or supporting argument!

Jan 6, 2016
#6
+5

I think that guest #4 is correct

http://www.wolframalpha.com/input/?i=%283%5E%282n%2B2%29++-8n+-9%29%2F2%5Ek+%3Dg+

I would still be interested in hearing a supporting argument. :)

Jan 7, 2016
#7
+10

sorry i have no closed solution. Here my way so far.

$$\small{ \begin{array}{rcl} 3^{2n+2} -8n -9 &=& 3^{2n}\cdot 3^2-8n-9 \\ &=& 3^{2n}\cdot 9-8n-9 \\ &=& (3^2)^n\cdot 9-8n-9 \\ &=& 9^n\cdot 9-8n-9 \\ &=& 9^{n+1}-8n-9 \\ &=& (1+8)^{n+1}-8n-9 \\ &=& -8n-9 + (1+8)^{n+1}\\ &=& -8n-9 + \underbrace{\binom{n+1}{0}+ \binom{n+1}{1}\cdot 8}_{=8n+9} + \binom{n+1}{2}\cdot 8^2 + \binom{n+1}{3}\cdot 8^3 + \binom{n+1}{4}\cdot 8^4 +\cdots \\ && + \binom{n+1}{n-1}\cdot 8^{n-1} + \binom{n+1}{n}\cdot 8^{n} + \binom{n+1}{n+1}\cdot 8^{n+1} \\\\ &=& \binom{n+1}{2}\cdot 8^2 + \binom{n+1}{3}\cdot 8^3 + \binom{n+1}{4}\cdot 8^4 +\cdots + \binom{n+1}{n-1}\cdot 8^{n-1} + \binom{n+1}{n}\cdot 8^{n} + \binom{n+1}{n+1}\cdot 8^{n+1} \\\\ &=& \binom{n+1}{2}\cdot 2^6 + \binom{n+1}{3}\cdot 2^9 + \binom{n+1}{4}\cdot 2^{12} +\cdots + \binom{n+1}{n-1}\cdot 2^{3n-3} + \binom{n+1}{n}\cdot 2^{3n} + \binom{n+1}{n+1}\cdot 2^{3n+3} \\ \end{array} }$$

It is not easy to factorize the binoms $$\binom{n+1}{2},~\binom{n+1}{3},~\binom{n+1}{3}\cdots$$  to get the exponent of the prim number 2 for every n

but m mimimum is 6, because we have $$2^6$$

It seems that (no proof !!!!)

$$\begin{array}{rcl} \binom{n+1}{2}\cdot 2^6 &=& 2^m\cdot p \\\\ (n+1)\cdot n \cdot 2^5 &=& 2^m\cdot p \\ \end{array}$$

Example:

$$\begin{array}{|r|c|l|} \hline n & m & p \\ \hline 1 & 6 & 1 \\ 2 & 6 & 3 \\ \hline 3 & 7 & 3 \\ 4 & 7 & 5 \\ \hline 5 & 6 & 15 \\ 6 & 6 & 21 \\ \hline 7 & 8 & 7 \\ 8 & 8 & 9 \\ \hline 9 & 6 & 45 \\ 10 & 6 & 55 \\ \hline 11 & 7 & 33 \\ 12 & 7 & 39 \\ \hline 13 & 6 & 91 \\ 14 & 6 & 105 \\ \hline 15 & 9 & 15 \\ 16 & 9 & 17 \\ \hline 17 & 6 & 153 \\ 18 & 6 & 171 \\ \hline \cdots \\ \hline \end{array}$$ Jan 7, 2016
edited by heureka  Jan 7, 2016
#8
+5

Here's my attempt (basically the same as heureka's): I tried to upload this from my phone yesterday (I was away from my computer), but with no success!

Jan 7, 2016
#9
+10

Bertie

I don't have a complete solution either but this is what I have so far.

$$\displaystyle f(n) = 9^{n+1}-8n-9$$,

so

$$\displaystyle f(n+1) = 9^{n+2}-8(n+1)-9$$ ,

in which case

$$\displaystyle \frac{f(n+1)}{f(n)}=\frac{9-\{8(n+1)+9\}/9^{n+1}}{1-\{8n+9\}/9^{n+1}}$$,

having divided top and bottom by $$\displaystyle 9^{(n+1)}$$.

That means that for increasing n, $$\displaystyle f(n+1) \approx 9f(n) \text{ or } f(n+1) \approx 2.08^{3}f(n)$$ ,

and that implies that for each increase of 1 in n, m will increase by 3.

That effect seems to kick in early,

f(1) = 64 = 2^6, so m = 6

f(2) = 704 = 2^9 + 192, so m = 9

f(3) = 6528 = 2^12 + 2432 so m = 12

f(4) = 59008 = 2^15 + 26420 so m = 15

That in turn suggests the rule

m = 3n+3.

I've not yet convinced myself whether or not that this is true for all n.

-Bertie

Guest Jan 7, 2016
#10
+5

OK., so that doesn't work.

Back to the drawing board.

-Bertie

Jan 7, 2016
#11
0

Thanks Heureka, Alan and Bertie,

You are all terrific.   Jan 8, 2016
#12
+10

Bertie

Updating my earlier posts #9 and #10, the sequence for m is

6,9,12,15, (as  in #9)

then 19 (an increase of 4 which caused me to write #10 because I thought #9 was wrong),

but it then continues

22,25,28,31,34, (steps of 3)

38 (step of 4)

41,44,47,50,53, (steps of 3)

57 (step of 4)

60,63,66,69,72, (steps of 3)

76 (step of 4).

i.e. five steps of 3 then one step of 4.

Whether or not this pattern continues indefinitely I don't know.

-Bertie

Jan 8, 2016
#13
0

Thanks for the update Bertie :)

Jan 9, 2016