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Natural number divisibility by 3

 
 
vinsan
 
Reply Thu 8 Sep, 2011 06:45 am
Hey frnds,

If x belongs to set of natural numbers then how can we prove that x^2 + 8 is divisble by 3 for any x which is not divisible by 3.

I need this to prove another math riddle I am solving currently.
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Type: Question • Score: 2 • Views: 7,489 • Replies: 7
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engineer
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  2  
Reply Thu 8 Sep, 2011 07:44 am
@vinsan,
X is a natural number not divisible by three therefore X can be written as

3n+1 or 3n+2 where n is a whole number.

The expression x^2 + 8 can be written in terms of n as either:

(3n+1)^2 + 8 = 9n^2 + 6n + 9 = 3(3n^2 + 2n +3) or
(3n+2)^2 + 8 = 9n^2 + 12n + 12 = 3(3n^2 + 4n + 4)

both of which are clearly divisible by three
anky2930
 
  0  
Reply Wed 25 Apr, 2012 03:16 am
Nice answer and explanation,Its the best solution for this question and I have just solved this question and found the answer same as vinsan that is 3(3n^2 + 2n +3).
Edit [Moderator]: Link removed
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rishi banerjee
 
  1  
Reply Wed 13 Jun, 2012 10:45 pm
@engineer,
Beatifully solved Nd explained. i also got the same answer (3n^2+2n+3)
0 Replies
 
jamesoconner
 
  1  
Reply Sat 30 Jun, 2012 05:56 am
@vinsan,
3,6,9,12,15,21,24,27,30......as on.
0 Replies
 
uvosky
 
  1  
Reply Fri 3 Aug, 2012 04:10 am
@vinsan,
Here is a variant of the argument given by ''engineer'' ;
Any two consecutive integer multiples of 3 differ by exactly 3 , example:-
.... -9 , -6, -3 , 0 , 3 , 6 ,9 .... etc. Now if x is an integer then x-1 , x , x+1 are three consecutive integers and so exactly one of them is divisible by 3 . Now as given x is not divisible by 3 , so exactly one of x-1 , x+1 is divisible
by 3 i.e. (x - 1) (x+1) is divisible by 3 . Now we write x^2 + 8 simply as
x^2 + 8 = x^2 - 1 +9 = (x - 1) (x+1) + 9 , and since 9 is divisible by 3 we get
x^2 + 8 is divisible by 3 for any integer x which is not divisible by 3 .
engineer
 
  1  
Reply Fri 3 Aug, 2012 06:22 am
@uvosky,
Clever. I like it.
uvosky
 
  1  
Reply Mon 6 Aug, 2012 05:17 am
@engineer,
thanks
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