VeritasPrepKarishma wrote:
Actually, you are using Binomial too.
How do you explain Remainder of\(\frac{(51^{25})}{13}\)= Remainder of\(\frac{(12^{25})}{13}\)?
Binomial leads to this equality!
I don't think I am using Binomial. Maybe you are correct.
However, I would try to prove the above point, not using Binomial. Let me know if I went wrong somewhere.
--> \(51^{25} = 13Q_1 + R_1\)
and \(12^{25} = 13Q_2 + R_2\)
with all the understood notations.
I want to prove that \(R_1\) = \(R_2\)
I will assume they are equal and replace the value of \(R_1\) in the first equation by that of \(R_2\)
Thus, \(51^{25} = 13Q_1 +12^{25} - 13Q_2\)
or \(51^{25}-12^{25} = 13(Q_1-Q_2)\)
Thus, if I could show that \(51^{25}-12^{25}\) is divisible by 13, my assumption would be correct.
Now,\(x^n-a^n\), is always divisible by (x-a), x and a are integers, n is odd.
Thus,\(51^{25}-12^{25}\) is always divisible by (51-12) = 39 --> 13*3. Thus, it is divisible by 13.
In the spirit of a healthy discussion, I would like to point out that you can certainly prove that the two are equal since they ARE equal. It can be done in many ways. The point is that when you see \(51^{25}\), what makes you think of \(12^{25}\)?
You think of it because 51 = 39 + 12. You separate out the part which is divisible by 13 and take control of the rest. Why?
Because \((39 + 12)^{25}\), when divided by 13 will have the same remainder as \(12^{25}\) because every term in this expansion is divisible by 39 except the last term which is \(12^{25}\). You understand this intuitively and that is all binomial is about.