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 solve this, let the digits of the three-digit number be H (hundreds), T (tens), and U (units).

• Condition 1: The value at the tens place is twice the units place (T = 2U).
• Condition 2: The value at the hundreds place is 75% of the units place (H = 0.75U or H = 3/4U).

Since digits must be whole numbers (0-9), we test values for U that are multiples of 4 (to satisfy the 3/4 fraction):

• If U = 4: T = 2 × 4 = 8, and H = 3/4 × 4 = 3. The number is 384. This is valid.
• If U = 8: T = 2 × 8 = 16. Since 16 is not a single digit, this is impossible.
• If U = 0: T = 0 and H = 0. This results in 000, which is not a three-digit number.

Only one number, 384, satisfies all conditions. Therefore, there is only 1 such number possible.