Question

prove that one and only one out of an n + 2 and n + 4 is divisible by 3 Where n is any positive integer​

Answer 1

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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

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Answer 2

Step-by-step explanation:

Case i: Let n be an even positive integer.

When n=2q

In this case , we have  

n^2  −n= (2q) ^2 −2q

n^2  −n=4q^2 −2q

n^2  −n=2q(2q−1)

n^2  −n= 2r , where r=q(2q−1)

n^2  −n is divisible by 2 .

Case ii: Let n be an odd positive integer.

When n=2q+1

In this case  

n^2  −n= (2q+1)^2  −(2q+1)

n^2  −n=(2q+1)(2q+1−1)

n^2  −n =2q(2q+1)

n^2  −n =2r, where r=q(2q+1)

n^2 −n is divisible by 2.

∴ n^2 −n is divisible by 2 for every integer n