Answer: idk if there’s any specific choice, but it should be the case.
Step-by-step explanation:
if n^2 is divisible by 3, n must also be divisible by three, since its squaring n by itself. probably a bad explanation. 3^2 is 9, 9^2 is 81, 12^144. all of them are divisible by three. Basically think of it as: 9^2 is (3x3)^2. 12 is (3x4)^2. So if n^2 is divisible by 3, n should also be divisible by 3. Sorry for the bad explanation. also I hope it’s right, sorry if it’s wrong.
Hello,
We can work in three different cases which cover all the possibilities for n
Case 1 - We can find k such that n = 3k-1
[tex]n^2=(3k-1)^2=9k^2-6k+1=3k(3k-2)+1\\\\n^2 \text{ is not divisible by 3}[/tex]
Case 2 - We can find k such that n = 3k
[tex]n^2=9k^2=3\times3k^2\\\\n^2 \text{ is divisible by 3}[/tex]
Case 3 - We can find k such that n = 3k+1
[tex]n^2=(3k+1)^2=9k^2+6k+1=3k(3k+2)+1\\\\n^2 \text{ is not divisible by 3}[/tex]
So, the only case where [tex]n^2[/tex] is divisible by 3 is actually when n is divisible by 3, so if [tex]n^2[/tex] is divisible by 3 , n is divisible by 3
Thanks
