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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.

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.

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.

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 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

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

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

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

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