The value at the tenth place of a three-digit number is twice the value at the unit's place. The value at the hundredth place is 75% of the value at the unit's place. How many such numbers are possible?

To find the possible three-digit numbers, let the digits be H (hundreds), T (tens), and U (units). According to the problem, the tens digit is twice the units digit (T = 2U) and the hundreds digit is 75% of the units digit (H = 0.75U or H = 3/4U). For H to be a non-zero integer (as it is the leading digit of a three-digit number), U must be a multiple of 4. If U = 4, then H = 3 and T = 8, resulting in the number 384. If U = 8, then H = 6 and T = 16; however, T must be a single digit (0-9), so this is impossible. If U = 0, then H = 0, which does not form a three-digit number. Thus, only the number 384 satisfies the conditions. However, the options provided are 0, 1, 2, and 3. Re-evaluating the constraints, if the question implies the hundredth place value (100H) rather than the digit, the logic remains consistent. Since only one such number (384) exists, the correct count is 1.

Sources

1. [1] https://register-kms.ncdd.gov.kh/Resources/E1FI63/316892/Digit%20Problems%20Algebra%20With%20Solutions.pdf