Test any number. Let p = 5 for example. \(p^2 + 17 = 42\). 42 divided by 12 has a remainder of 6.

Ans: C

For the proof of why the remainder is always 6:

Let n be any prime number greater than 3. We can write n as \(n = E+ 1\) where E is even.

Then \(n^2 = (E + 1)^2 = E^2 + 2E + 1\).

The claim is \(E^2 + 2E\) will always have a factor of 12. We need to prove it has a factor of both 4 and 3.

The factor of 4 is easier to prove. Since \(E^2 + 2E = E*(2+E)\), both E and (2+E) have a factor of 2 so the product must have a factor of 4.

For the factor of 3, recall E + 1 is a prime number greater than 3 so it cannot be divisible by 3. Thus either E or E + 2 must be divisible by 3. Both are factors of the product, therefore in any case \(E^2 + 2E\) must be divisible by 3.

Finally, we proved \(E^2 + 2E\) is divisible by 12, then \(n^2 = E^2 + 2E + 1\) divided by 12 will have a remainder of 1 and \(n^2 + 17\) must have a remainder of 6.
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P^2+17/12, Remainder R=?
P^2+17=12a+R
P^2+17-12a=R
P is a prime number and greater than 3 and 12a could be any multiple of 12

Let's use a trial and error method
If P=5 and a=1,2,3
25+17-12=30 (Substituting a = 1)
25+17-24=18 (Substituting a = 2)
25+17-36=6 (Substituting a = 3)

Answer is 6. Option C

We see that p can be any prime number greater than 3, so it could be 5, 7, 11, 13, etc. Let’s choose the smallest possible value for p.

If p = 5, we have:

25 + 17 = 42

42/12 = 3 remainder 6

Answer: C
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Prime number greater than 3 = 5, 7, 11, 13………..
Suppose, P= 5,
P^2 + 17 = 5^2 + 17 = 25 + 17 = 42
Now 42/ 12 = 3 + 6/12, Remainder = 6(C)

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