Question

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?

Answer 1

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.

Answer 2

There are 3 three-digit palindromes satisfy the property.

242 , 313 and 404

Explanation:

Consider the provided information.

It is given that 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 number is "aba" (last and first digit will be same, As it is palindromes)

Then we have:

a + a + ba = b + 8

2a + ba - b = 8

Substitute a = 1 in 2a + ba - b = 8

2 + b - b = 8

2 ≠ 8 so 'a" can't be 1.

Substitute a = 2 in 2a + ba - b = 8

4 + 2b - b = 8

b = 8- 4

b = 4

So, the first number is: 242

Substitute a = 3 in 2a + ba - b = 8

6 + 3b - b = 8

2b = 2

b = 1

So, the second number is: 313

Substitute a = 4 in 2a + ba - b = 8

8 + 4b - b = 8

3b = 8-8

b = 0

So, the third number is: 404

Substitute a = 5 in 2a + ba - b = 8

10 + 5b - b = 8

4b = -2

b should be whole number so no other numbers are possible.

Hence, there are 3 three-digit palindromes satisfy the property.

242 , 313 and 404