• LOGIN
  • No products in the cart.

Profile Photo

Divisibility by 16 (Tomato Subjective 35)

(a) Prove that, for any odd integer n, \(n^4 \) when divided by 16 always leaves remainder 1.

(b) Hence or otherwise show that we cannot find integers \(n_1 , n_2 , … , n_8 \) such that \(n_1^4 + n_2^4 + … + n_8^4 = 1993 \).

Solution: For part (a) we consider n = 2k +1 and expand it’s fourth power binomially to get \((2k+1)^4 = \binom {4}{0} (2k)^4 +\binom {4}{1} (2k)^3 + \binom {4}{2} (2k)^2 + \binom {4}{3} (2k)^1 + \binom {4}{4} (2k)^0 = 16k^4 + 32k^3 + 24k^2 + 8k + 1 \)

Read More…

No comments, be the first one to comment !

    Leave a Reply

    Your email address will not be published. Required fields are marked *

    Login

    Register

    GOOGLECreate an Account
    X