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anonymous

  • one year ago

Show that one and only one out of n,n+2 or n+4 is divisible by 3 , where n is any positive integer .

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  1. SolomonZelman
    • one year ago
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    Maybe this is what I can do: All of them can NOT be divisible by 3, because if you add a number divisible by 3 and a number not divisible by 3, we get a sum not divisible by 3. Thus if any of these is divisible by 3, then only one of them.

  2. anonymous
    • one year ago
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    sorry , but i am unable to understand the second line .

  3. SolomonZelman
    • one year ago
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    nevermind, we can use induction. Want to?

  4. SolomonZelman
    • one year ago
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    http://www.algebra.com/algebra/homework/real-numbers/real-numbers.faq.question.242615.html

  5. anonymous
    • one year ago
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    can't we do this through Euclid's division lemma ?

  6. SolomonZelman
    • one year ago
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    Honestly, I don't know that. Thanks for giving me something to look at:)

  7. anonymous
    • one year ago
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    in ur ans adding 1 to both sides , is that part of the induction rule or we just do it as a compulsory thing ?

  8. anonymous
    • one year ago
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    cause' in 10 std we haven't got induction.

  9. SolomonZelman
    • one year ago
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    At the final step when they had k+3=3m, they add 1, and they obtain k+4=3m+1. (Where m and k are natural numbers). So they are shoing the final case, where k+3 (or just k) is divisible by 3, then the rest are not divisible by 3. A precisely, at that step thaqt you asked, they showed that then k+4 is NOT divisible by 3, because it is same thing as 3m+1 which is not divisible by 3.

  10. SolomonZelman
    • one year ago
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    And Euclid's division lemma is something I have always been mentally doing, but never know it is called this way. LOL Happens to me sometimes:D

  11. anonymous
    • one year ago
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    can u then please explain me this que through Euclids division lemma, cause' this is in our course ?

  12. SolomonZelman
    • one year ago
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    I am not very sure about doing it. Don't want to get it wrong-:( I am guessing you would start from supposing that \(\large\color{black}{ \displaystyle n=3k+r }\) where m>k

  13. SolomonZelman
    • one year ago
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    i mean n>k

  14. SolomonZelman
    • one year ago
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    Sorry about that.

  15. SolomonZelman
    • one year ago
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    You can tag other people who are better than me though, like this @sonali46 . Good luck!

  16. SolomonZelman
    • one year ago
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    TO tag, just put the @ key in front of the person's username whom you desire to tag.

  17. anonymous
    • one year ago
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    where should i write this ?

  18. madhu.mukherjee.946
    • one year ago
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    We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q & one and only one of these possibilities can occur Case I : When n = 3q In this case, we have, n=3q, which is divisible by 3 n=3q = adding 2 on both sides n + 2 = 3q + 2 n + 2 leaves a remainder 2 when divided by 3 Therefore, n + 2 is not divisible by 3 n = 3q n + 4 = 3q + 4 = 3(q + 1) + 1 n + 4 leaves a remainder 1 when divided by 3 n + 4 is not divisible by 3 Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3 Case II : When n = 3q + 1 In this case, we have n = 3q +1 n leaves a reaminder 1 when divided by 3 n is not divisible by 3 n = 3q + 1 n + 2 = (3q + 1) + 2 = 3(q + 1) n + 2 is divisible by 3 n = 3q + 1 n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2 n + 4 leaves a remainder 2 when divided by 3 n + 4 is not divisible by 3 Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3 Case III : When n = 3q + 2 In this case, we have n = 3q + 2 n leaves remainder 2 when divided by 3 n is not divisible by 3 n = 3q + 2 n + 2 = 3q + 2 + 2 = 3(q + 1) + 1 n + 2 leaves remainder 1 when divided by 3 n + 2 is not divsible by 3 n = 3q + 2 n + 4 = 3q + 2 + 4 = 3(q + 2) n + 4 is divisible by 3 Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3

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