Bunuel wrote:

For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2

B. 3

C. 4

D. 5

E. 6

Prime number greater than 3 can be expressed as \(6k + 1\) or \(6k -1\).

\(\frac{(6k ± 1)^3}{12} = \frac{6^3k^3 ± 3*6^2k^2 + 3*6k ± 1}{12} = \frac{0 + 0 + 18k ± 1}{12} = \frac{6k ± 1}{12}\)

Let’s check first case \(6k + 1\)

k=1 (7/12) rem=7

k=2 (13/12) rem =1

k=3 (19/12) rem = 7

k =4 (25/12) rem = 1

we have cycle of remainders \(7\) and \(1\).

Next \(6k – 1\)

k=1 (5/12) rem=5

k=2 (11/12) rem =11

k=3 (17/12) rem = 5

k =4 (23/12) rem = 11

again cycle of only two remainders \(5\) and \(11\).

The only remainders after division by 12 cube of prime number \(y > 3\) can have are \(1\), \(5\), \(7\) and \(11\).

Checking our options.

Answer D.