n belongs to natural no's, then the highest integer m such that 2^m divides 3^2n+2-8n-9 is

=9(9^n)-8n-9

9(8+1)^n -8n - 9

9(nc_{0}8^{n} +nc_{1}8^{n-1} + ...........+nc_{n}8^{0}) -8n -9

9(8x+1) -8n-9 (let the remaining terms be expressed as x(ie.nc_{0}8^{n-1},nc_{1}8^{n-2} etc.)

72x+9-8n-9

72x-8n

8(9x-n) divisible by 2^{m}

2^{m }=8

m=3

aditya@calc.com
Jan 11, 2016

#2**+15 **

Hallo aditya.

Example: If n = 1 we have:

\(3^{2n+2} - 8n -9 = 3^{2\cdot 1+2} - 8\cdot 1 -9 = 3^4-8-9 = 81-17 = 64 ={\color{red}2^6}\)

\(2^6 \text{ can be devide by } 2^m \text{, if }m=6.\\ So~the~mimimum~ m~ is ~ 6.\\ \text{The maximum m must be equal or greater than the minimum m }.\\ \text{So m is }6\text{ or greater}.\)

m is not a constant.

heureka
Jan 11, 2016

#2**+15 **

Best Answer

Hallo aditya.

Example: If n = 1 we have:

\(3^{2n+2} - 8n -9 = 3^{2\cdot 1+2} - 8\cdot 1 -9 = 3^4-8-9 = 81-17 = 64 ={\color{red}2^6}\)

\(2^6 \text{ can be devide by } 2^m \text{, if }m=6.\\ So~the~mimimum~ m~ is ~ 6.\\ \text{The maximum m must be equal or greater than the minimum m }.\\ \text{So m is }6\text{ or greater}.\)

m is not a constant.

heureka
Jan 11, 2016

#3**+5 **

Hi Aditya, your answer looks really interesting so I will take a look at your working :)

n belongs to natural no's, then the highest integer m such that 2^m divides 3^2n+2-8n-9 is

=9(9^n)-8n-9

9(8+1)^n -8n - 9

9(nc08n +nc18n-1 + ...........+ncn80) -8n -9

9(8x+1) -8n-9 (let the remaining terms be expressed as x(ie.nc08n-1,nc18n-2 etc.)

72x+9-8n-9

72x-8n

8(9x-n) divisible by 2m

2m =8

m=3

\(3^{2n+2}-8n-9\\ =9*3^{2n}-8n-9\\ =9*9^n-8n-9\\ =9*(8+1)^n-8n-9\\ =9\left[\begin{pmatrix}n\\0\end{pmatrix}8^0+\begin{pmatrix}n\\1\end{pmatrix}8^1+.......\begin{pmatrix}n\\{n-1}\end{pmatrix}8^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}8^n\right]-8^n-9\\ =9\left[1+n*8^1+.......\begin{pmatrix}n\\{n-1}\end{pmatrix}8^{n-1}+8^n\right]-8^n-9\\ =9\left[8n+.......\begin{pmatrix}n\\{n-1}\end{pmatrix}8^{n-1}+8^n\right]-8^n\\\)

I think that is correct but I do not think it simplifies to what you have there.......

In response to aditya below.

You want me to factor 8 out, ok I can do that.

\(=9*8\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-8 ^n\\ =2^3*9\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-2^{3n}\\ =2^3*9\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-2^3*2^{3n-3}\\ =2^3*\left[9\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-2^{3n-3}\right]\)

So if n>1 and a natural number then this expression is divisable by 8.

Note: if n=1 I don't think that this expression IS divisable by 8. because 8^(-1) is a fraction.

Heureka's simple substitution showed this to be incorrect so there must be something wrong

Melody
Jan 11, 2016

#4**0 **

answer to heureka's comment=

8 also divides 64

you may argue that 4, 2, 1 also divides 64

but i have already proved it above

answer to melody's comment=

when we take 8 common from (nc_{0}8^n +nc_{1}8^n-1 + ...........+nc_{n}8^0) we get (nc_{0}^8n-1,nc_{1}8^n-2....)

for eg- 8 common from (1(nc_{0}8^n) we get= 8(8^n-1)

nc_{1}8^n-1=8(nc_{1}8^n-2) etc

remaining same

what do you think

please reply back

exams are approaching

and binomial theorem carries 90 marks

aditya@calc.com
Jan 11, 2016

#5**0 **

At Aditya's request, I have continued my analysis on my post #3. BUT it is not correct.

There is an error somewhere. ://

Melody
Jan 12, 2016

#6**0 **

can you please explain where i went wrong in my recent explanation

i had a look at heureka's comment and replied to it

please tell

in the answer book given by IIT(indian institute of technology) which is without explanation, the answer is given as 3

so i thought i got ir right

please elaborate where i went wrong

aditya@calc.com
Jan 12, 2016