Spartan85 wrote:
Given two positive integers A and B such that A > B, what is the remainder when the square of B is subtracted from the square of A and then divided by 15?
(1) When the sum of A and B is divided by 5, the remainder is 1
(2) When B is subtracted from A and then divided by 3, the remainder is 1.
Hi
(1) When the sum of A and B is divided by 5, the remainder is 1
A + B = 5x + 1. ------> A + B = 6, 11, 16 ... Putting this into remainders we can have (0 + 1), (1 + 0), (2 + 4), (4 + 2).
We are getting different values for A^2 - B^2 and remainders as a result. Insufficient.
(2) When B is subtracted from A and then divided by 3, the remainder is 1.
A - B = 3y + 1. Same logic as in previous case. Insufficient.
(1)&(2) Combining two:
\(A + B = 6,
11,
16, 21, 26, 31, 36 ....\)
\(A - B =
1,
4, 7, 10, 13, 16, 19, 22 ...\)
A + B = 11, A - B = 1 ----> A = 6, B = 5 ------->\(\frac{A^2 - B^2}{15} = \frac{36 - 25}{15} = \frac{16}{15}\) ----> remainder 1.
A + B = 16, A - B = 4 ----> A = 10, B = 6 -------> \(\frac{A^2 - B^2}{15} = \frac{100 - 36}{15} = \frac{64}{15}\) ----> remainder 4.
Insufficient
Another way to look at this:
\(A^2 - B^2 = (A - B)*(A + B) = (5x + 1)(3y+1) = 15xy + 5x + 3y + 1.\)
We can ingnore \(15xy\) which brings remainder \(0\). \(\frac{5x + 3y + 1}{15}\) will give us diferent remainders for different values of \(x\) and \(y\). Insufficient.
Answer E.