Show that one and only out of n,n+2 Or n+4 is divisible by 3 where n is any positive integer
Answers
Now, we have to show that this is the only number out of these three that is divided by three.
Clearly, division of n+2 will give remainder 2.
and division of n+4 will give remainder 1.
Case 2: Suppose n on division by 3 leaves remainder 1.
In this case n+2 will be divided by 3 and n+4 will leave remainder 2.
Case 3: Suppose n on division by 3 leaves remainder 2.
In this case, n+2 on division by 3 will leave remainder 1 and hence n+4 will be divided by 3.
Case 4: Suppose none of them is divided by 3. And say division of n leaves remainder 1.
It is clear that n+2 will be divided by 3 (same as Case 2). Hence our assumption in this case is wrong. Similarly when division of n leaves remainder 2, (n+4) will be divided by 3 (same as Case 3).
Considering these four cases, it is seen that one and only one of n, n+2 or n+4 is divided by 3.
Hope this helps. Thanks!
Answer:
step-by-step explanation:
Euclid's division Lemma any natural number can be written as: .
where r = 0, 1, 2,. and q is the quotient.
∵ Thus any number is in the form of 3q , 3q+1 or 3q+2.
→ Case I: if n =3q
⇒n = 3q = 3(q) 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.
Hence, it is solved.