Question

Show that one and only one out of n n+2,n+4 is divisible by 3,where n is any positive integer.

Answer 1

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Let the 3 is divisible by n, n + 2 and n + 4.
Then, b = 3.

→ Using Euclid's Division Lemme .

=> 0 ≤ r < b.

=> 0 ≤ r < 3.

=> r = 0, 1, 2.


→ Taking r = 0.

→ a = bq + r.

=> a = 3q + 0.

=> a = 3q.

→ Taking r = 1.

=> a = 3q + 1.

→ Taking r = 2.

=> a = 3q + 2.

▶Now,

➡ Case 1 at n.

=> n = 3q = 3(q).

=> n = 3q + 1.

=> n = 3q + 2.


➡ Case 2 at n + 2.

=> n + 2 = 3q + 2.

=> n + 2 = 3q + 1 + 2 = 3q + 3 = 3( q + 1 ).

=> n + 4 = 3q + 1 + 4 = 3q + 5.


➡ Case 3 at n + 4.

=> n + 4 = 3q + 4.

=> n + 4 = 3q + 1 + 4 = 3q + 5.

=> n + 4 = 3q + 2 + 4 = 3q + 6 = 3( q + 2 ).


↪ Therefore,
In case 1, n is divisible by 3q.
In case 2, n + 2 is divisible by 3q + 1.
In case 3, n + 4 is divisible by 3q + 2.



✔✔ Hence, it is proved ✅✅.

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Answer 2

We know that any positive integer is of the form 3q,3q+1,3q+2 for some integer q

CASE 1

When n =3q , is divisible by 3

n+2= 3q+2 , is not divisible by 3 as it leaves a remainder 2

n+4 = 3q+4, 3q + 3 +1, 3(q+1)+1, not divisible by 3 as it leaves a remainder 1

Therefore n is divisible by 3 and n+2 and n+4 are not divisible by 3.

CASE 2

n= 3q +1 noy divisible by 3 because it leaves a remainder 1

n+2= 3q+1+2, 3q+3, 3(q+1) , divisible by 3

n + 4 = 3q+1+4 , 3q+5 , 3q +3+2,3(q+1)+2 ,not divisible by 3 because it leavesa remainder 2

Therefore n +2 is divisible by 3 and n and n+4 are not divisible by 3

CASE 3

n=3q+2 ,not divisible by 3 because it leaves a remainder 2

n+2= 3q+2+2 , 3q+4 , 3q + 3+1, 3 (q+1)+1 ,not divisible by 3 as it leaves a remainder 1

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

Therefore n+4 is divisible by 3 but n and n +2 are not divisible by 3

Therefor only one out of n , n+2, n+4 is divisible by 3

HOpE It hElpS U ALl !!!...:)