The sum of the digits of a positive three digit number Is 19. The tens digits is four less than twice the hundreds digit. The number is decreased by 99 when the digits are reversed. Find the number

call it

A*100 + B*10 + C
then we know

A + B + C = 19

B + 4 = 2 A

C*100 + B*10 + A + 99 = A*100 + B*10 + C

now plug and chug

685

Let's break down the given information step by step to find the number.

Step 1: The sum of the digits of a positive three-digit number is 19.
Let's assume the three-digit number is ABC, where A represents the hundreds digit, B represents the tens digit, and C represents the units digit. From the given information, we know that:
A + B + C = 19 (equation 1)

Step 2: The tens digit is four less than twice the hundreds digit.
This can be expressed as:
B = 2A - 4 (equation 2)

Step 3: The number is decreased by 99 when the digits are reversed.
If we reverse the digits of ABC to form another three-digit number, we get CBA. From this, we can form the following equation:
ABC - CBA = 99 (equation 3)

Step 4: Solve the equations simultaneously to find the values of A, B, and C.
Now, let's solve equations 1, 2, and 3 together to find the values of A, B, and C. Rearranging equation 3, we have:
(100A + 10B + C) - (100C + 10B + A) = 99
99A - 99C = 99
Dividing both sides of the equation by 99, we get:
A - C = 1 (equation 4)

Subtracting equation 4 from equation 1, we obtain:
(A + B + C) - (A - C) = 19 - 1
2C = 18
C = 9

Substituting the value of C into equation 4, we have:
A - 9 = 1
A = 10

Substituting the values of A and C into equation 2, we get:
B = 2(10) - 4
B = 16

Therefore, the three-digit number is 109.

So, the number is 109.

To solve this problem, we will use algebraic equations to represent the given conditions and find the unknown number. Let's break down the information provided:

1. "The sum of the digits of a positive three-digit number is 19."
Let's assume the hundreds digit of the number is "x," the tens digit is "y," and the ones digit is "z." We can express this information as an equation:
x + y + z = 19.

2. "The tens digit is four less than twice the hundreds digit."
This condition can be represented as:
y = 2x - 4.

3. "The number is decreased by 99 when the digits are reversed."
We can assume the original three-digit number is 100x + 10y + z. When its digits are reversed, the new number becomes 100z + 10y + x. According to the condition, the original number decreased by 99 when the digits were reversed. This can be written as:
100x + 10y + z - (100z + 10y + x) = 99.

Now, we have a system of three equations:
equation 1: x + y + z = 19
equation 2: y = 2x - 4
equation 3: 99 = 99x - 99z

Using these equations, we can solve for the values of x, y, and z, which will give us the desired three-digit number.

Solving equation 2 for x (in terms of y):
x = (y + 4) / 2

Substituting this value of x into equation 1:
((y + 4) / 2) + y + z = 19
(y + 4 + 2y + 2z) / 2 = 19
3y + 2z + 4 = 38
3y + 2z = 34

Now, we have a simplified version of equation 1: 3y + 2z = 34.

Next, substituting x = (y + 4) / 2 into equation 3:
99((y + 4) / 2) - 99z = 99
99y + 396 - 99z = 99
99y - 99z = -297

Now, we have a simplified version of equation 3: 99y - 99z = -297.

We now have a system of two equations in two variables:
equation 4: 3y + 2z = 34
equation 5: 99y - 99z = -297

We can solve this system of equations using any suitable method such as substitution or elimination.

To solve using substitution, isolate one variable in terms of the other from equation 5:
y = (99z - 297) / 99
y = z - 3

Substituting this expression for y into equation 4:
3(z - 3) + 2z = 34
3z - 9 + 2z = 34
5z - 9 = 34
5z = 43
z = 43 / 5
z ≈ 8.6

Since z represents a digit, it must be a whole number. Therefore, we round down z to 8:
z = 8

Substituting this value of z into equation 4:
3y + 2(8) = 34
3y + 16 = 34
3y = 18
y = 18 / 3
y = 6

Substituting the values of y and z back into equation 2 to find x:
6 = 2x - 4
2x = 10
x = 10 / 2
x = 5

Hence, the hundreds digit (x) is 5, the tens digit (y) is 6, and the ones digit (z) is 8.

Therefore, the original three-digit number is 5 * 100 + 6 * 10 + 8 = 568.