It is a five digit whole number

1. It is a palindrome
2. It's hundreds digit is a multiple of 5
3. The product of its one digit and 8 is 12 greater than the square of its ones digit
4. The product of its thousands digit and 18 is 45 greater than the square of it's thousands digit
5. It is divisible by 8
6. One of its digits is a square number
7. The product of two of its digits is 6
8. The sum of its digits is 10
9. Its tens digit is 3

Name two 2-diget factors whose product is graeter than

200 but less than 600

To find the five-digit whole number that satisfies all the given conditions, we can use the following steps:

Step 1: Based on condition 2, we know that the hundreds digit is a multiple of 5. The only multiple of 5 that is also a single digit is 5 itself. So the hundreds digit is 5.

Step 2: Based on condition 9, we know that the tens digit is 3.

Step 3: Based on condition 8, we know that the sum of all the digits is 10.

Since we already have the hundreds digit as 5 and the tens digit as 3, the sum of the remaining three digits must be 10 - 5 - 3 = 2.

Let's consider the remaining three digits as a, b, and c.

Step 4: Based on condition 7, we know that the product of two of its digits is 6. Since 6 is the only single-digit number whose product is 6, it can only be obtained as a product of 2 and 3. Therefore, two of the digits must be 2 and 3.

Now we have the digits: 5, 3, 2, and 2.

Step 5: Based on condition 6, one of its digits must be a square number. The only square number among our current digits is 9 (which is 3 squared). Therefore, the last digit must be 9.

Now we have the digits: 5, 3, 2, 2, and 9.

Step 6: Based on condition 5, the number must be divisible by 8. To determine if a number is divisible by 8, we need to check if the last three digits form a number divisible by 8. In this case, the last three digits are 2, 2, and 9. The number formed by these last three digits is 229. Checking if 229 is divisible by 8, we find that it is not. Since we have already used all the digits, we need to go back and change one of the digits.

Considering condition 3, the product of its one digit and 8 is 12 greater than the square of its ones digit. Let's assume the ones digit is 9 (based on the previous reasoning) and the one digit is a. According to condition 3, 8a = 9^2 + 12.

Simplifying this equation, we get 8a = 81 + 12, which means 8a = 93. However, there is no single-digit whole number that satisfies this equation. Hence, the assumed ones digit of 9 is incorrect.

Let's go back and consider a different ones digit. Let's assume the ones digit is 2. With this assumption, we have 8a = 2^2 + 12, which simplifies to 8a = 16 + 12 and further simplifies to 8a = 28. Solving for 'a,' we divide both sides of the equation by 8, giving us a = 3.

Now we have the digits: 5, 3, 2, 3, and 2.

Step 7: Based on condition 4, the product of its thousands digit and 18 is 45 greater than the square of its thousands digit. The only single-digit number that satisfies this condition is 5. 5^2 + 45 = 70, which is not divisible by 18. Therefore, the assumed thousands digit of 5 is incorrect.

Let's go back again and consider a different thousands digit. Let's assume the thousands digit is 2. With this assumption, we have 2 * 18 = 4^2 + 45, which simplifies to 36 = 29. This equation is not valid. Hence, the assumed thousands digit of 2 is incorrect.

Let's try the thousands digit of 3. With this assumption, we have 3 * 18 = 3^2 + 45, which simplifies to 54 = 54. Finally, this equation is valid!

Now we have the digits: 3, 5, 2, 3, and 2.

Putting all the digits together, the five-digit whole number that satisfies all the given conditions is 35,323.