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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.

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

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

Member since Apr 1, 2017

By Euclid division lemma: Let â€˜aâ€™ be any positive integer such that  a = 3q +r,  where 0 â‰¤ 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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