Question

Show that one and only one out of n, n+1 or n+2 is divisible by3, where n is any positive integer​

Answer 1

Answer:

Step-by-step explanation:

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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Answer 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.