The sum of the three digits equals 36 times of the number? 

Let’s assume the three digits of the number to be x, y, and z, where x is the hundreds digit, y is the tens digit, and z is the units digit.

From the given conditions, we can form the following equations:

x + y + z = 36x
This is based on the given condition that the sum of the three digits equals 36 times of the number.

7x + 9 = 5(y + z)
This is based on the given condition that seven times the left digit plus 9 is equal to 5 times the sum of the two other digits.

8y – 9 = x + z
This is based on the given condition that 8 times the second digit minus 9 is equal to the sum of the first and third digits.

We can simplify equation 1 as:

y + z = 35x

We can substitute this value of y + z in equation 2 as:

7x + 9 = 5(35x – y)
7x + 9 = 175x – 5y
5y = 168x – 9
y = (168x – 9)/5

Since y is a digit, it must be an integer. Therefore, 168x – 9 must be divisible by 5. The only value of x that satisfies this condition is x = 2.

Substituting x = 2 in the equation for y, we get:

y = (168(2) – 9)/5 = 67

Substituting x = 2 and y = 67 in equation 3, we get:

8(67) – 9 = 2 + z
535 = 2 + z
z = 533

However, this value of z is not a single digit. Therefore, there is no three-digit number that satisfies all the given conditions.

Hence, there is no solution to this problem.