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Question

Show that one and only one out of n, n+1, n+4 is divisible by 3 where n is any positive integer.

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Solution

THE QUESTION MUST BE WRONG IT SHOULD BE LIKE n, n+2 , n+4 is divisible by 3

let n be any positive integer and b=3
n =3q+r
where q is the quotient and r is the remainder
0_ <r<3
so the remainders may be 0,1 and 2
so n may be in the form of 3q, 3q=1,3q+2


CASE-1

IF N=3q
n+4=3q+4
n+2=3q+2
here n is only divisible by 3

CASE 2
if n = 3q+1
n+4=3q+5
n+2=3q=3
here only n+2 is divisible by 3

CASE 3
IF n=3q+2
n+2=3q+4
n+4=3q+2+4
=3q+6
here only n+4 is divisible by 3



HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE


METHODE 2

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.

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