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Questions of geometric progression.

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

    Formulas of GP :

    1.Sn = 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.Tn = arn-1

    3. 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. :)
     
  2. jcsd
  3. Sep 15, 2012 #2

    Simon Bridge

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    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?
     
  4. Sep 15, 2012 #3

    ehild

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

    ehild
     
  5. Sep 29, 2012 #4
    Ok , first stick to first question. And I've gotta use GP to prove that 11111.....65 times isn't a prime number.

    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.
     
  6. Sep 29, 2012 #5

    ehild

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

    ehild
     
  7. Sep 29, 2012 #6
    Woops ! :redface:

    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 !! :smile:

    (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 ?
     
    Last edited: Sep 29, 2012
  8. Sep 30, 2012 #7

    ehild

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    It is OK, but I suggested the easier way:
    [tex]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)[/tex]

    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
     
  9. Sep 30, 2012 #8
    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... :frown:

    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 ? :smile:
     
  10. Sep 30, 2012 #9

    ehild

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

    ehild
     
  11. Oct 1, 2012 #10
    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 !!:smile:
     
  12. Oct 1, 2012 #11

    ehild

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    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 :tongue2:

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

    ehild
     
  13. Oct 1, 2012 #12
    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 :

    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 !? :eek:

    Thanks again for your endeavors , ehild !!:smile:
     
  14. Oct 1, 2012 #13

    ehild

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



    Do not forget that 105=100000 (5 zeros), and 100000-1=99999. :biggrin:
    N= (x-1)(x12+x11+x10....+1)/9=99999(......)/9=11111(....) (as 99999 divided by 9 is 11111....) Never forget what you learnt in the first or second class :tongue2:.

    ehild
     
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