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Find the number of n that can be divided by 7

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Given that Un=4n6+n3+5. Find the number of n, where n = {1, 2, 3,... 2009} so that Un can be divided by 7


    2. Relevant equations
    Don't know


    3. The attempt at a solution
    Completely blank.....
     
  2. jcsd
  3. Apr 8, 2012 #2
    I'd say to use induction.
     
  4. Apr 8, 2012 #3

    micromass

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    Why don't you check a few n (say for n from 1 to 10) to see what you get?

    Another thing you should do is factorize your expression to see if you get something nice.
     
  5. Apr 8, 2012 #4

    Dick

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    Think about the problem mod 7. Check n=0,1,2,3,4,5,6.
     
  6. Apr 14, 2012 #5
    I only know mathematical induction is used for proofing. Can it be used to find number of n in this question?

    U1=10
    U2=269
    U3=2948
    U4=16453
    U5=62630
    U6=186845
    U7=470944
    U8=1049093
    U9=2126498
    U10=4001005

    I can't see any patterns there that can be used to determine which n can be divided by 7....
    I also can't factorize the expression. Or maybe you mean using perfect square?

    Actually I don't have any knowledge about mod but I'll give it a try.

    Let Un=a so for this question a = b (mod 7), where b must be 7k (multiplication of 7)

    This is as far as I can go :blushing:
     
  7. Apr 14, 2012 #6

    Dick

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    Ok, let's not use the word 'mod'. Can you show [itex]U_{n+7}-U_n[/itex] is divisible by 7? That means [itex]U_n[/itex] is divisible by 7 iff [itex]U_{n+7}[/itex] is divisible by 7.
     
  8. Apr 14, 2012 #7
    I can show that Un+7 - Unis divisible by 7 by expanding Un+7 and after subtraction, the terms left will all have 7 as factor.

    I have several things that I still not clear:
    1. How "Un+7 - Unis divisible by 7" be used to find number of n?
    2. Why you use Un+7 - Un? Is it because divisible by 7? If the question asking about divisibility by 5 then Un+5 - Un?

    Thanks
     
  9. Apr 17, 2012 #8

    Dick

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    Look. U_0 is not divisible by 7, right? That means U_7, U_14, U_21,... are not divisible by 7. Agree with that?
     
  10. Apr 17, 2012 #9
    Yes. So if we know which n produces Un that is divisible by 7, then Un+7 will also be divisible by 7 and they will form arithmetic sequence. Do i get your hint correctly? :biggrin:

    Then how to find the first n?

    Thanks
     
  11. Apr 17, 2012 #10

    Curious3141

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    Why not simply try testing U1 to U7? I did it, and it took me no more than 5 minutes with a calculator.

    EDIT: Or U0 to U6, as Dick suggested, which entails less work. In fact you can even test U-3 to U3, and that would involve the smallest numbers of all, and still give a complete result.
     
    Last edited: Apr 17, 2012
  12. Apr 17, 2012 #11

    Dick

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    There may not be any n. Check each point that might be a start of the arithmetic sequence. Like I said before, try n=0,1,2,3,4,5,6.
     
  13. Apr 17, 2012 #12
    Ah, I think I get it. Thanks a lot for the help :smile:
     
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