A better way would’ve been that you attached a solution and we would’ve pointed out at which step you’re making the mistake but anyways, here is how I approach this:

I will start by finding the LCM of 24 and 108 and you might wonder why ?

Ans: **LCM** is a common strategy for solving problems that involve finding common multiples or common divisors of two or more numbers. In this particular problem, we are given that n^2 is a multiple of both 24 and 108. This means that n^2 must be divisible by any number that is divisible by both 24 and 108. The LCM of 24 and 108 is the smallest such number. Hence, by finding the LCM of 24 and 108, we can **determine the smallest possible value for n^2 that satisfies the given conditions**

LCM of 24 and 108 = 216

I will be doing the prime factorization of 216 next, why?

Ans: To find the divisors of this value and determine which integers are divisors of every integer n in S.

Prime factorization of 216 = 2^3 * 3^3

However, since n^2 is a perfect square and 216 is not a perfect square (because the exponents of its prime factors are odd), the smallest possible value for n^2 must be the smallest perfect square that is divisible by 216. This can be obtained by increasing the exponents of the prime factors to the next **even number**: 2^4 * 3^4 =1296. Therefore, the smallest possible value for n is \sqrt{1296} = 36

So every integer n in S must be divisible by **36** . Since **12** ( divisor/factors of 36 : **1** , **2** , **3** , **4** , **6** , **9** , **12** , **18** , and **36** ) is also a divisor of **36** , it follows that every integer n in S must also be divisible by **12** .

**Ans:** a) 12 and c) 36

How Greg approaches it : https://www.youtube.com/watch?app=desktop&v=GIir0YccY9U