The number 1987 can be written as a three digit number $xyz$ in some base $b$.  If $x + y + z = 1 + 9 + 8 + 7$, determine all possible values of $x$, $y$, $z$, $b$.
1. **Express 1987 in base \( b \) as \( xyz \):**
   \[
   1987 = x \cdot b^2 + y \cdot b + z
   \]
   where \( x, y, z \) are the digits in base \( b \).

2. **Given condition:**
   \[
   x + y + z = 1 + 9 + 8 + 7 = 25
   \]

3. **Polynomial representation:**
   Let \( p(x) = ax^2 + cx + d \) where \( a, c, d \) are the digits of 1987 in base \( b \). Then:
   \[
   p(b) = 1987
   \]
   and
   \[
   p(1) = a + c + d = 25
   \]

4. **Divisibility condition:**
   \[
   (b-1) \mid p(b) - p(1) = 1987 - 25 = 1962
   \]
   Factorize 1962:
   \[
   1962 = 2 \cdot 3^2 \cdot 109
   \]

5. **Determine bounds for \( b \):**
   Since \( 1987 \) is a three-digit number in base \( b \):
   \[
   b^2 \leq 1987 < b^3
   \]
   Taking square and cube roots:
   \[
   \sqrt{1987} \approx 44.6 \quad \text{and} \quad \sqrt[3]{1987} \approx 12.6
   \]
   Therefore:
   \[
   12 < b < 45
   \]

6. **Possible values for \( b-1 \):**
   \[
   11 < b-1 < 44
   \]
   The only divisor of 1962 within this range is 18. Thus:
   \[
   b-1 = 18 \implies b = 19
   \]

7. **Verify base 19:**
   \[
   1987 = x \cdot 19^2 + y \cdot 19 + z
   \]
   Calculate:
   \[
   19^2 = 361
   \]
   \[
   1987 = x \cdot 361 + y \cdot 19 + z
   \]
   Find \( x \):
   \[
   x = \left\lfloor \frac{1987}{361} \right\rfloor = 5
   \]
   Remaining:
   \[
   1987 - 5 \cdot 361 = 1987 - 1805 = 182
   \]
   Find \( y \):
   \[
   y = \left\lfloor \frac{182}{19} \right\rfloor = 9
   \]
   Remaining:
   \[
   182 - 9 \cdot 19 = 182 - 171 = 11
   \]
   Thus:
   \[
   z = 11
   \]

8. **Sum of digits:**
   \[
   x + y + z = 5 + 9 + 11 = 25
   \]
   This satisfies the given condition.

The final answer is \( \boxed{ x = 5 } \), \( y = 9 \), \( z = 11 \), \( b = 19 \).