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.

239315 37468brilliantly insightful post. If only it was as easy to implement some with the solutions as it was to read and nod my head at each of your points 284329