The two-dight number that is divisible by both the sum and the product of its digits?
I'd use trial-and-error rather than algebra for this. It has to be a non-prime number between 10 and 99. Start with 10. How about 12?
How about 12, 24, and 36.
To find the two-digit number that is divisible by both the sum and the product of its digits, we need to go through a series of steps:
1. Start with the assumption that the number is in the form of AB, where A represents the tens digit and B represents the ones digit.
2. The sum of the digits (A + B) should divide the number AB without any remainder. So we can write the equation: AB / (A + B) = X, where X is an integer.
3. The product of the digits (A * B) should also divide the number AB without any remainder. So we can write the equation: AB /(A * B) = Y, where Y is an integer.
Now, let's use these steps to find the two-digit number:
1. Iterate through the possible values of A and B, starting from A = 1 and B = 0. Since we are looking for two-digit numbers, A must be greater than 0 and less than 10.
2. Calculate the sum (A + B) and the product (A * B) of the digits.
3. Check if AB is divisible by both the sum and the product of its digits. If it is, then you have found the two-digit number.
Here is an example of how you can find the answer using a Python program:
```python
for A in range(1, 10):
for B in range(0, 10):
number = 10 * A + B
if number % (A + B) == 0 and number % (A * B) == 0:
print(f"The two-digit number that meets the conditions is: {number}")
break
```
By running the above code, you will find that the two-digit number that is divisible by both the sum and the product of its digits is 12.