**Problem : If \(\ a,b,c,d\) satisfy the equations**

$$a+7b+3c+5d=0,$$

$$8a+7b+6c+2d=-16,$$

$$2a+6b+4c+8d=16,$$

$$5a+3b+7c+d=-16,$$

then \(\ (a+d)(b+c)\) equals

\(\ (A)16 \quad (B)-16\quad (C)0 \quad\) *(D)none of the foregoing numbers*

**Solution: **

$$a+7b+3c+5d=0\dots(1),$$

$$8a+7b+6c+2d=-16\dots(2),$$

$$2a+6b+4c+8d=16\dots(3),$$

$$5a+3b+7c+d=-16\dots(4),$$

\(\ (1)-(3)\), and \(\ (2)-(4)\), we get

$$-a+b-c-3d=-16\dots(5),$$

$$3a+b-c+d=0\dots(6),$$

\(\ (6)-(5)\), we get

$$a+d=4\dots(7),$$

\(\ (2)+(3)\),we get

$$a+b+c+d=0\dots(8),$$

\(\ (8)-(7)\),we get

$$b+c=-4\dots(9),$$

\(\ (7)\times(9)\),we get

Therefore,$$(a+d)(b+c)=-16$$

Thus,\(\ (B)\) is the correct option.