Question

Use Euclid Division Lemma to prove that one and only one out of n,n+2and and n+4 is divisible by 3 where n is any positive integer

Answer 1

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.

Answer 2

Hey!

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By Euclid Division Lemma any natural number can be written as :-

b = aq + r

Where, r = 0, 1, 2, ...... (a-1), and q is the quotient.

Put a = 3: b = 3q + r and r = 0, 1, 2

Thus, any number is in the form of 3q, 3q+1, or 3q+2

CASE I : if n =3q

n is divisible by 3,

n+2 = 3q+2 is not divisible by 3.

n+4 = 3q+4 = 3(q+1)+1 is not divisible by 3.

CASE II : if n =3q+1

n = 3q+1 is not divisible by 3.

n+2 = 3q+1+2=3q+3 = 3(q+1) is divisible by 3.

n+4 = 3q+1+4 = 3q+5 = 3(q+1)+2 is not divisible by 3.

CASE III : if n = 3q+2

n =3q+2 is not divisible by 3.

n+2 = 3q+2+2 =3q+4 = 3(q+1)+1 is not divisible by 3.

n+4 = 3q+2+4 = 3q+6 = 3(q+2) is divisible by 3.

Thus, one and only one out of n , n+2, n+4 is divisible by 3.

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Hope it hepls :)