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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 .
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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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Maybe 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry , but i am unable to understand the second line .

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0nevermind, we can use induction. Want to?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0http://www.algebra.com/algebra/homework/realnumbers/realnumbers.faq.question.242615.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can't we do this through Euclid's division lemma ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Honestly, I don't know that. Thanks for giving me something to look at:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in ur ans adding 1 to both sides , is that part of the induction rule or we just do it as a compulsory thing ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cause' in 10 std we haven't got induction.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0At 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.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0And 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u then please explain me this que through Euclids division lemma, cause' this is in our course ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0I 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

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Sorry about that.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0You can tag other people who are better than me though, like this @sonali46 . Good luck!

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0TO tag, just put the @ key in front of the person's username whom you desire to tag.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where should i write this ?

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.0We 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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