We are given that n is a positive integer and n² is divisible by 12. The prime factorization of 12 is 2² × 3. Since 12 divides n², both prime factors 2 and 3 must divide n².

Similarly, 3 is a prime factor of 12, so 3 divides n². Since 3 is prime, if 3 divides n², then 3 must also divide n.

We have established that both 2 and 3 divide n. Since 2 and 3 are distinct prime numbers (and hence coprime, with (2, 3) = 1), their product 2 × 3 = 6 must also divide n. Hence, n is divisible by 6.

if any se positive engineer and N2 is divisible by 12 then prove that n is divisible by 6

Answer: Hence proved that if $n$ is a positive integer and $n^2$ is divisible by $12$ , then $n$ is divisible by $6$ .

Step-by-step solution

Step 1: Express the given divisibility using prime factorization

We are given that $n$ is a positive integer and $n^2$ is divisible by $12$ . The prime factorization of $12$ is $2^2 \times 3$ . Since $12$ divides $n^2$ , both prime factors $2$ and $3$ must divide $n^2$ .

Step 2: Apply the Fundamental Theorem of Arithmetic for prime 2

Since $12$ divides $n^2$ , it follows that $2$ divides $n^2$ . By the Fundamental Theorem of Arithmetic, if a prime $p$ divides $a^2$ , then $p$ divides $a$ . Thus, $2$ divides $n$ .

Step 3: Apply the theorem for prime 3

Similarly, $3$ is a prime factor of $12$ , so $3$ divides $n^2$ . Since $3$ is prime, if $3$ divides $n^2$ , then $3$ must also divide $n$ .

Step 4: Conclude divisibility by 6

We have established that both $2$ and $3$ divide $n$ . Since $2$ and $3$ are distinct prime numbers (and hence coprime, with $\gcd(2, 3) = 1$ ), their product $2 \times 3 = 6$ must also divide $n$ . Hence, $n$ is divisible by $6$ .

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