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Hyadarani

Apr 15, 2014

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

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

Answers(3)

Answer

Syeda

Answer.

We applied Euclid Division algorithm on n and 3.
a = bq +r  on putting a = n and b = 3
n = 3q +r  , 0<r<3
i.e n = 3q   -------- (1),n = 3q +1 --------- (2), n = 3q +2  -----------(3)
n = 3q is divisible by 3
or n +2  = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.

Raghunath Reddy

Sol :
We applied Euclid Division algorithm on n and 3.
a = bq +r  on putting a = n and b = 3
n = 3q +r  , 0<r<3
i.e n = 3q   -------- (1),n = 3q +1 --------- (2), n = 3q +2  -----------(3)
n = 3q is divisible by 3
or n +2  = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.

Parthasaradhi M

Member since Apr 1, 2017

By Euclid division lemma:
Let ‘a’ be any positive integer such that  a = 3q +r,  where 0 ≤ r < 6 and q is not equal to zero
Thus a= 3q, a= 3q+1, a=3q+2

Cas: 1 when n=3q
n=3q 
n+2= 3q+1
n+4= 3q+4 = 3q+3+1=3(q+1)+1
From above n=3q is divisible by 3
Therefore, in this case only one out of n,n+2,n+4 is divisible by 3

Case 2: when n = 3q+1
n = 3q+1
n+2 = 3q+1+2 = 3q+3 = 3(q+1) 
n+4=3q+1+4=3q+3+2=3(3q+1)+2
From above n+2=3(q+1) is divisible by 3
Therefore, in this case only one out of n,n+2,n+4 is divisible by 3

Case 3: when n = 3q+2
n=3q+2
n+2=3q+2+2=3q+3+1=3(3q+1)+1
n+4=3q+2+4=3q+6=3(q+2)
From above n+4=3(q+2) is divisible by 3
Therefore, in this case only one out of n,n+2,n+4 is divisible by 3

 
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