Let's analyze the given information and set up an equation to find the desired number of palindromes.
Three-digit palindrome: We can represent the number as abc=100a+10b+c, where a is the hundreds digit, b is the tens digit, and c is the units digit (since it's a palindrome, a=c ).
Sum of digits and product: The problem states that the sum of the hundreds digit, units digit, and the product of the units and tens digits is eight more than the tens digit. Translating this into math, we get: a+c+bc=b+8.
Solving the equation: Since a=c, we can substitute to get: $$ a + a + ab = b + 8.$$ Combining like terms: $$ 2a + ab = b + 8.$$ Factoring out b from the left side: b(a+1)=b+8.
Now we can analyze the solutions for b:
If b=0, then the equation is always true regardless of a. However, a three-digit palindrome cannot have a leading zero (hundreds digit cannot be zero). So, b=0.
If b=0, then we can divide both sides by b to get: a+1=1+b8. Since a is a digit (0-9), the expression on the right side must be an integer. This is only possible when b is a divisor of 8.
Possible values for b are: b=1,2,4,8. Let's check each case to see if it leads to a valid three-digit palindrome:
b=1: In this case, a=0 (which we cannot have).
b=2: Here, a=3, resulting in the palindrome 323.
b=4: We get a=1, which leads to the palindrome 141.
b=8: This results in a=−1, which is not a valid digit.
Therefore, there are only two three-digit palindromes that satisfy the given property: 323 and 141.
