Since 12 divides n², 2 is a prime dividing n². By the Fundamental Theorem of Arithmetic (Theorem 1.2: if a prime p divides a², then p divides a), since 2 is a prime number and 2 n², it follows that 2 must divide n.

Since 12 divides n², 2 is a prime dividing n². By the Fundamental Theorem of Arithmetic (Theorem 1.2: if a prime p divides a², then p divides a), since 2 is a prime number and 2 n², it follows that 2 must divide n.

Similarly, 3 is a prime factor of 12, so 3 divides n². Applying the same theorem (if a prime p divides a², then p divides a), since 3 is prime and 3 n², it follows that 3 must divide n.

We have shown that both 2 and 3 divide n. Since 2 and 3 are distinct prime numbers (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 $N^2$ is divisible by 12 then prove that n is divisible by 6

Answer: Hence, it is proved that $n$ is divisible by $6$ .

Step-by-step solution

Step 1: Analyze the divisibility of $n^2$

We are given that $n$ is a positive integer and $n^2$ is divisible by $12$ . Writing $12$ as a product of prime powers, we have $12 = 2^2 \times 3$ . Since $12$ divides $n^2$ , it follows that both $2$ and $3$ divide $n^2$ .

Step 2: Apply Theorem on divisibility by prime 2

Since $12$ divides $n^2$ , $2$ is a prime dividing $n^2$ . By the Fundamental Theorem of Arithmetic (Theorem 1.2: if a prime $p$ divides $a^2$ , then $p$ divides $a$ ), since $2$ is a prime number and $2 \mid n^2$ , it follows that $2$ must divide $n$ .

Step 3: Apply Theorem on divisibility by prime 3

Similarly, $3$ is a prime factor of $12$ , so $3$ divides $n^2$ . Applying the same theorem (if a prime $p$ divides $a^2$ , then $p$ divides $a$ ), since $3$ is prime and $3 \mid n^2$ , it follows that $3$ must divide $n$ .

Step 4: Conclude divisibility by 6

We have shown that both $2$ and $3$ divide $n$ . Since $2$ and $3$ are distinct prime numbers (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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if any se positive engineer and N2N^2N2 is divisible by 12 then prove that n is divisible by 6