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.