Questions of geometric progression.

[sankalpmittal]

Homework Statement

1. Prove that 1111111...65 times is not a prime number.

2. If a,b,c,d are in GP , then (a²+b²+c²)(b²+c²+d²) is :

(a) (ab+ac+bc)²
(b) (ac+cd+ad)²
(c) (ab+bc+cd)²
(d) None of the above

Homework Equations

Formulas of GP :

1.S_n = a(1-rⁿ)/(1-r)
2.a,b,c in GP , then b²=ac
3.a,b,c,d,e,...,z in GP , then
az=by=cx=dw =...

4.T_n = ar^n-1

The Attempt at a Solution

1. The number can be written as

N = 1+10 + 10² + 10³ +... + 10⁶⁴
N= (10⁶⁵-1)/9

Now how to proceed ?

2. Well I tried using b²=ac but to no avail.

Please help !

Thanks in advance. :)

 

[Simon Bridge]

http://en.wikipedia.org/wiki/Primality_test :)
Though I notice that 1/9 = 0.11111111...

b²=ac only applies if the GP has three terms {a,b,c}

if it has an odd number of terms then one of the products will be with itself.

the general rule is:
if $$a_1, a_2,\cdots,a_{n-1},a_n$$ is a GP, then $$ a_1 a_n = a_2 a_{n-1} = a_3 a_{n-2} = \cdots$$

So for just the four ... what does this become?

 

[ehild]

Homework Statement

1. Prove that 1111111...65 times is not a prime number.

The Attempt at a Solution

1. The number can be written as

N = 1+10 + 10² + 10³ +... + 10⁶⁴
N= (10⁶⁵-1)/9

Now how to proceed ?

65=5˙13, so the number can be also written as N= ((10⁵)¹³-1)/9. Can you factor it?

ehild

 

[sankalpmittal]

http://en.wikipedia.org/wiki/Primality_test :)
Though I notice that 1/9 = 0.11111111...

b²=ac only applies if the GP has three terms {a,b,c}

if it has an odd number of terms then one of the products will be with itself.

the general rule is:
if $$a_1, a_2,\cdots,a_{n-1},a_n$$ is a GP, then $$ a_1 a_n = a_2 a_{n-1} = a_3 a_{n-2} = \cdots$$

So for just the four ... what does this become?

Ok , first stick to first question. And I've got to use GP to prove that 11111...65 times isn't a prime number.

65=5˙13, so the number can be also written as N= ((10⁵)¹³-1)/9. Can you factor it?

ehild

Ah ! Do you mean using aⁿ-bⁿ formula ? Or you mean doing plain factorization ?

I apologize for such a delay. Things were slightly downhill for me.

 

[ehild]

Ah ! Do you mean using aⁿ-bⁿ formula ? Or you mean doing plain factorization ?

Yes, you always can factor out (a-b) from aⁿ-bⁿ,
10⁶⁵-1=((10⁵))¹³-1,
a=10⁵ b=1.

ehild

 

[sankalpmittal]

Yes, you always can factor out (a-b) from aⁿ-bⁿ,
10⁶⁵-1=((10⁵))¹³-1,
a=10⁵ b=1.

ehild

Woops !

I forgot the formula for aⁿ-bⁿ. Ok , so I'm doing simple factorization..

N= ((10¹³)⁵-1)/(10-1)

Now factorizing 10⁶⁵-1 by 10¹³-1 , I get : 10⁵² + 10³⁹ + 10²⁶ + 10¹³ + 1

So ,N= (10⁵² + 10³⁹ + 10²⁶ + 10¹³ + 1)(10¹³-1)/(10-1)

On further simplifying , I get :

N=(10¹² +10¹¹ + 10¹⁰ + 10⁹ + 10⁸ + 10⁷ + ... + 1)(10⁵² + 10³⁹ + 10²⁶ + 10¹³ + 1)

Thus N or given number is a multiplication of two non zero numbers. So it has to be composite number. Proved!

Thanks ehild and Simon !

(Wait , Am I methodically correct in question 1 ?)

All right now to second problem :

"If a,b,c,d are in GP , then (a^2+b^2+c^2)(b^2+c^2+d^2) is :

(a) (ab+ac+bc)^2
(b) (ac+cd+ad)^2
(c) (ab+bc+cd)^2
(d) None of the above"

So we have

ad=bc...

Now how to proceed further ?

 

[ehild]

Woops !

I forgot the formula for aⁿ-bⁿ. Ok , so I'm doing simple factorization..

N= ((10¹³)⁵-1)/(10-1)


Thus N or given number is a multiplication of two non zero numbers. So it has to be composite number. Proved!

It is OK, but I suggested the easier way:
N=\frac{10^{65}-1}{9}=\frac{(10^{5})^{13}-1}{9}=\frac{99999(10^{60}+10^{55}+...+1)}{9}=11111(10^{60}+10^{55}+...+1)

All right now to second problem :

"If a,b,c,d are in GP , then (a^2+b^2+c^2)(b^2+c^2+d^2) is :

(a) (ab+ac+bc)^2
(b) (ac+cd+ad)^2
(c) (ab+bc+cd)^2
(d) None of the above"

So we have

ad=bc...

Now how to proceed further ?

I would do it with "brute force" using that b=ar, c=ar², d=ar³, and write up the expressions in terms of a and r.

ehild

 

[sankalpmittal]

I would do it with "brute force" using that b=ar, c=ar², d=ar³, and write up the expressions in terms of a and r.

ehild

That is going to take a lot of time , right ? Does it not ? And that is making use of objective problem. What if it were not an objective problem. (Oops I just now understood that such problem can only be given as objective problem ! Sorry !)

Ok , but by so called "brute force" , I've to first find (a^2+b^2+c^2)(b^2+c^2+d^2) in terms of a and r. Then I've to solve for all the options also , in terms of a and r. Any other method ?

By brute force , I have :

(a²+b²+c²)(b²+c²+d²) = a²(1+r²+r⁴) a²r²(1+r²+r⁴) = a⁴r²(1+r²+r⁴)²

All right , now I'll have to do all this for all the options...

Well , I'm on it...

(a) (ab+ac+bc)² =( a²r + a²r² + a²r³)² = a⁴r²(1+r+r²)²

So option a is out...

(b)(ac+cd+ad)²
I can simplify it to ,

(a²r²+a²r⁵+ar³)²

a²r⁴(1+r³+r)²

b is out.

(c)(ab+bc+cd)²

(a²r+a²r³+a²r⁵)²

a⁴r²(1+r²+r⁴)²

So (c) is correct option. Yes ! Thanks ehild !

But do you know any other method ?

 

[ehild]

So (c) is correct option. Yes ! Thanks ehild !

But do you know any other method ?

You are unsatiable Was not it an easy and short solution?

ehild

 

[sankalpmittal]

You are unsatiable Was not it an easy and short solution?

ehild

Yes , it was a short solution though. I've heard that (a²+b²+c²)(b²+c²+d²) = (ab+bc+cd)² is an identity also. Lagrange's identity you call it ? Or is it Cauchy Schwarz inequality - as I can see here :
http://www.mathdb.org/notes_download/elementary/algebra/ae_A5b.pdf

Thanks ! Once again for help !

 

[ehild]

Yes , it was a short solution though. I've heard that (a²+b²+c²)(b²+c²+d²) = (ab+bc+cd)² is an identity also. Lagrange's identity you call it ? Or is it Cauchy Schwarz inequality - as I can see here :
http://www.mathdb.org/notes_download/elementary/algebra/ae_A5b.pdf

Thanks ! Once again for help !

It is an inequality in general,

(a²+b²+c²)(b²+c²+d²) ≥ (ab+bc+cd)².

I used to be a bad student, I did not like to memorize theorems, and if I did , I could not remember the name. I preferred to prove a theorem myself. It was fun, memorizing was difficult and boring

Anyway, thank you for the file, I saved it and try to remember it...

ehild

 

[sankalpmittal]

It is an inequality in general,

(a²+b²+c²)(b²+c²+d²) ≥ (ab+bc+cd)².

I used to be a bad student, I did not like to memorize theorems, and if I did , I could not remember the name. I preferred to prove a theorem myself. It was fun, memorizing was difficult and boring

Anyway, thank you for the file, I saved it and try to remember it...

ehild

I know proving is more fun , but I am preparing for engineering entrance exams. There , I have to do each question in 1 minute. So , my guess is that its better to use analytical skills to derive shortcut formulas by yourself and then memorize them. But I never get satisfied by a formula , if I do not know its proof.

That's Ok... But what I have doubt still is regarding :

It is OK, but I suggested the easier way:
N=\frac{10^{65}-1}{9}=\frac{(10^{5})^{13}-1}{9}=\frac{99999(10^{60}+10^{55}+...+1)}{9}=11111(10^{60}+10^{55}+...+1)

What did you take factor out of what ?

I would do it with "brute force" using that b=ar, c=ar², d=ar³, and write up the expressions in terms of a and r.

ehild

If I take 10⁵=x , say and then

((10⁵)¹³-1)/(10-1)

N = (x¹³-1)/9
N= (x-1)(x¹²+x¹¹+x¹⁰...+1)/9
N= (1+10+10²+10³+10⁴)(10⁶⁰+10⁵⁵ +...1)

Oh ! This is what you've got ! How did you do this in just one line !?

Thanks again for your endeavors , ehild !

 

[ehild]

I know proving is more fun , but I am preparing for engineering entrance exams. There , I have to do each question in 1 minute. So , my guess is that its better to use analytical skills to derive shortcut formulas by yourself and then memorize them. But I never get satisfied by a formula , if I do not know its proof.

It is very good. Never believe anything without understanding the proof or proving it yourself.

If I take 10⁵=x , say and then

((10⁵)¹³-1)/(10-1)
N = (x¹³-1)/9

Do not forget that 10⁵=100000 (5 zeros), and 100000-1=99999.
N= (x-1)(x¹²+x¹¹+x¹⁰...+1)/9=99999(...)/9=11111(...) (as 99999 divided by 9 is 11111...) Never forget what you learned in the first or second class .

ehild