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.