How many three-digit palindromes (numbers that read the same forward and backward) satisfy the following property: the sum of three things -- the hundreds digit, the units digit, and the product of the units and tens digits -- is eight more than the tens digit?

Let the 3-digit palindrome be 101a+10b. a+a+ab=8+b, so 2a+ab-b=8.
2a+b(a-1)=8; b=2(4-a)/(a-1). From this we know that a > 1. And we know 4-a > 0 so a < 4.
This limits a to 2 and 3 which make b=4 and 1.
The palindromes are 242 and 313.