**Problem Analysis**

Let's assume the two-digit number is represented as "10a + b," where "a" is the tens digit and "b" is the units digit.

According to the problem, the product of the digits is 35. So, we have the equation:

a * b = 35

When 18 is added to the number, the digits interchange their places. This implies that the new number is "10b + a". So, we have the equation:

(10a + b) + 18 = 10b + a

Simplifying the equation, we get:

9a - 9b = -18
a - b = -2

Now we have two equations:

1. a * b = 35
2. a - b = -2

We can solve these two equations to find the values of "a" and "b" and hence determine the two-digit number.

**Solution**

1. Let's list down all the pairs of numbers whose product is 35:
- 1 * 35 = 35
- 5 * 7 = 35

2. Now, let's try each pair of numbers and check if their difference is -2:
- For 1 and 35: 1 - 35 = -34 (not equal to -2)
- For 5 and 7: 5 - 7 = -2 (equal to -2)

3. Since the difference of 5 and 7 is -2, we can conclude that "a" equals 7 and "b" equals 5.

4. Therefore, the two-digit number is 75.

5. To verify our solution, let's substitute the values of "a" and "b" in the second equation:
- 7 - 5 = -2 (which is true)

6. Additionally, let's check if the digits interchange their places when 18 is added to the number:
- 75 + 18 = 93 (the digits have indeed interchanged)

7. Hence, the two-digit number that satisfies the given conditions is 75.

**Final Answer**
The required two-digit number is 75.

let unit digit of number be Y &Tens digit be X.so,Required number=10X + YATQ,XY=35 Y=35/X .........(1)also,10X + Y + 18 =10Y + X9(Y - X) =18Y - X = 235/X - X = 2 ..USING(1)X^2 + 2X - 35 = 0(X - 5) (X + 7) = 0SO,X= 5 ;X WILL NEVER EQUAL (-7) AS DIGIT IS NEVER NEGATIVE. NOW,Y = 35/X =35/5 =7So, Required number =10 X + Y =10×5+7 =57.