hgp2k wrote:
Well I first started with something like this:
\(P^2/12 = P^2/3*(2^2)\)
As P is a prime number greater than 3, it must be odd, so is \(P^2\). Now any odd number divided by 3 either gives 1 or 2 as remainder. But we also know that product of any 2 even number is divisible by 4. So I was quite sure that this is going to give me remainder of 1 and not 2. I did some number plugging and it worked so I chose C as the answer.
But then I found something on internet, and then realized that I was going in right direction:
\(P^2 = (P+1)*(P-1) + 1\). As I stated above, \((P+1)*(P-1)\) is surely divisible by 4 as both these numbers are even. Also, one of these 2 numbers is divisible by 3, as these P-1, P and P+1 are consecutive number. Now we can definitely say that \((P+1)*(P-1)\) is divisible by 12, and thus \(P^2/12\) will always give remainder as 1.
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Consider KUDOS for good posts
This question also appears on the
OG 12th Ed albeit in a PS format.
It's PS Ques 23 on page 155 and goes "If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12 ?"
Obviously in a PS setting picking any prime and testing for its remainder when divided by 12 is the way to go.
OG goes into a theoretical discussion identical to Bunuel’s solution.
However I like the solution based on expressing p^2 as (p+1)*(p-1) + 1 that hgp2k mentions above A LOT MORE !
I usually don't get into the theoetical discussion of this problems with my students, but if I do in the future, I know I'll be using the above.
Thanks a bunch. Oh and since I can't buy you a drink, I gave you a kudos !
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Avijit Sarkar
Kaplan Instructor