the winninq three-digit number in a drawing for a door prize had the same units and hundreds digit. the product of its units and tens digits was 5 less than twice the sum of its digits. its middle digit was one more than twice its units digit. What was the number?

To find the three-digit number in this problem, we'll use algebraic equations to represent the given information.

Let's assume the three-digit number is represented as "ABC", where A is the hundreds digit, B is the tens digit, and C is the units digit.

1. The digits in the number are the same, meaning A = C.
2. The product of the units and tens digits is 5 less than twice the sum of the digits, which can be expressed as B*C = 2*(A + B + C) - 5.
3. The middle digit is one more than twice the units digit, represented by B = 2*C + 1.

Now let's solve these equations step by step:

From equation 1, we have A = C. So we can replace A with C in equations 2 and 3 to simplify them:

B*C = 2*(A + B + C) - 5 ---> (2*C + 1) * C = 2*(C + B + C) - 5

Expanding the right side of the equation:

(2C^2 + C) = 2(2C + B) - 5
2C^2 + C = 4C + 2B - 5

We can replace B with the given equation B = 2C + 1:

2C^2 + C = 4C + 2(2C + 1) - 5
2C^2 + C = 4C + 4C + 2 - 5
2C^2 + C = 8C - 3

First, let's rearrange equation 3:

2C^2 + C - 8C + 3 = 0
2C^2 - 7C + 3 = 0

To solve this quadratic equation, we can use factoring or the quadratic formula:

Using factoring, we can rewrite the equation as:

(2C - 1)(C - 3) = 0

So, two possible solutions here: C = 1/2 or C = 3.

Since C represents the units digit, it must be a positive integer, so C = 3.

Now we know C = 3, we can substitute this value into equation 2:

B*3 = 2*(A + B + 3) - 5
3B = 2A + 2B + 6 - 5
B = 2A + 1

Now, we use equation 1 to substitute C with 3:

A = C = 3

We have our three equations:
A = 3
B = 2A + 1
C = 3

From these equations, we can find the values of A, B, and C:

A = 3
B = 2(3) + 1 = 7
C = 3

So, the three-digit number is 377.