Let `a, b, c, d` be distinct integers such that the equation `(x - a) (x -b) (x -c)(x-d) - 9 = 0` has an integer root 'r', then the value of `a +b+c+d - 4r` is equal to :

Answer

To solve the problem, we start with the equation given:

\[ (x - a)(x - b)(x - c)(x - d) - 9 = 0 \]

Since \( r \) is an integer root of this equation, we can substitute \( x = r \):

\[ (r - a)(r - b)(r - c)(r - d) - 9 = 0 \]

This simplifies to:

\[ (r - a)(r - b)(r - c)(r - d) = 9 \]

Next, we need to factor 9 into the product of four distinct integers. The only way to express 9 as a product of four distinct integers is:

\[ (-1) \times 1 \times (-3) \times 3 = 9 \]

Thus, we can assign:

\[ r - a = -1, \quad r - b = 1, \quad r - c = -3, \quad r - d = 3 \]

Now we can solve for \( a, b, c, \) and \( d \):

1. From \( r - a = -1 \):
\[ a = r + 1 \]

2. From \( r - b = 1 \):
\[ b = r - 1 \]

3. From \( r - c = -3 \):
\[ c = r + 3 \]

4. From \( r - d = 3 \):
\[ d = r - 3 \]

Now, we need to find \( a + b + c + d - 4r \):

\[ a + b + c + d = (r + 1) + (r - 1) + (r + 3) + (r - 3) \]

Simplifying this expression:

\[ = r + 1 + r - 1 + r + 3 + r - 3 \]
\[ = 4r \]

Thus, we have:

\[ a + b + c + d - 4r = 4r - 4r = 0 \]

Therefore, the final answer is:

\[ \boxed{0} \]