in a 3 digit number, the tens digit is twice the hundreds digit and 1/2 the units digit. if the digits are reversed, the new number is 49 more than 3 times that of the original number

t = 2h ... 2t = u ... 4h = u

3*(100h + 10t + u) + 49 = 100u + 10t + h

3*(100h + 20h + 4h) + 49 = 400h + 20h + h

solve for h , then find t and u

To find the original 3-digit number, let's break down the information given step by step:

1. Let's assume the hundreds digit is "x", the tens digit is "2x", and the units digit is "0.5x" (half of the units digit).

2. The original number can be represented as 100x + 10(2x) + 0.5x.

3. Simplifying, we get 100x + 20x + 0.5x = 120.5x.

4. Now, let's consider the reversed number. It can be represented as 100(0.5x) + 10(2x) + x.

5. Simplifying this expression, we get 50x + 20x + x = 71x.

6. According to the problem, the reversed number is 49 more than 3 times the original number. So, we can set up the equation: 71x = 3(120.5x) + 49.

7. Solving the equation, we get 71x = 361.5x + 49.

8. Let's now solve for "x":
- Subtracting 361.5x from both sides: 71x - 361.5x = 49.
- Simplifying: -290.5x = 49.
- Dividing by -290.5: x = -49 / -290.5 ≈ 0.1686.

9. Since "x" represents a digit, we need to find a whole number between 0 and 9 that is closest to 0.1686, which is 0.

10. Now, substitute the value of "x" into the original number expression (100x + 20x + 0.5x):
- 100(0) + 20(2 * 0) + 0.5(0) = 0 + 0 + 0 = 0.

Therefore, the original 3-digit number is 0.