The correct answer is option (E):
9
Let N be a two-digit or three-digit integer such that 10 < N < 501.
We are given that P is the product of the digits of N, and S is the sum of the digits of N.
We want to find the number of integers N in the given range for which P + S = N.
Case 1: Two-digit numbers (10 < N < 100)
Let N = 10a + b, where a and b are digits (1 <= a <= 9, 0 <= b <= 9).
P = a * b, S = a + b.
We want P + S = N, so ab + a + b = 10a + b.
ab = 9a
Since a is not zero, we can divide by a:
b = 9
So, the two-digit numbers are of the form 19, 29, 39, 49, 59, 69, 79, 89, 99.
Since N is between 10 and 99, we have 19, 29, 39, 49, 59, 69, 79, 89, 99. There are 9 such numbers.
For example, if N = 19, then P = 1 * 9 = 9, S = 1 + 9 = 10. Then P + S = 9 + 10 = 19 = N.
If N = 29, then P = 2 * 9 = 18, S = 2 + 9 = 11. Then P + S = 18 + 11 = 29 = N.
Case 2: Three-digit numbers (100 <= N < 501)
Let N = 100a + 10b + c, where a, b, and c are digits (1 <= a <= 4, 0 <= b <= 9, 0 <= c <= 9).
P = a * b * c, S = a + b + c.
We want P + S = N, so abc + a + b + c = 100a + 10b + c.
abc + a + b = 100a + 10b
abc = 99a + 9b
Since a >= 1, we can write abc = 9(11a + b).
If a = 1, then bc = 99 + 9b. But if b = 0, then c is undefined. Also if b = 9, bc >= 99, c >= 11, which is not possible.
If a = 2, then bc = 198 + 9b. Not possible.
If a = 3, then bc = 297 + 9b. Not possible.
If a = 4, then bc = 396 + 9b. Not possible.
Therefore, there are no solutions for three-digit numbers.
Combining the results, there are 9 such integers (19, 29, 39, 49, 59, 69, 79, 89, 99).
Final Answer: The final answer is $\boxed{9}$