Bunuel
How many integer solutions exist for the equation 8x – 5y = 221 such that x*y < 0
A. 4
B. 5
C. 6
D. 7
E. 8
If x*y < 0, then x and y must have opposite signs (i.e., one is positive and the other is negative).
If y is positive, then -5y is negative. Therefore, 8x must be greater than 221 in order for 8x - 5y = 221. However, in that case, x will also be positive, and we won’t have x*y < 0.
If x (and hence 8x) is positive, then y must be negative and -5y will be positive. Therefore, 8x < 221. Furthermore, the units digit of x must be either 2 or 7 so that y could be an integer (notice that the units digit of 8x will be 6 and that of -5y will be 5). Therefore, x can be 2, 7, 12, 17, 22 or 27 (we stop at 27 since 27 * 8 = 216 but 32 * 8 = 256). So we see that x can have 6 values (and any one of these will yield a negative value for y).
Answer: C