In the halving problem with 305 numbers, what is the minimum number of halving operations required to reach 0?

In the halving problem with 305 numbers, what is the minimum number of halving operations required to reach 0?

• A. 8
• B. 9
• C. 10
• D. 11

Answer: (B)

Each halving is floor division by 2, so after n halvings the value becomes floor(305 / 2^n). To reach zero, you need 2^n to exceed 305. Since 2^8 = 256 is not enough but 2^9 = 512 is, the smallest n that works is 9. You can also think in binary terms: 305 sits between 2^8 and 2^9, so it has 9 binary digits, and you need nine halvings to drop to zero. Therefore, the minimum number of halving operations is 9.

Each halving is floor division by 2, so after n halvings the value becomes floor(305 / 2^n). To reach zero, you need 2^n to exceed 305. Since 2^8 = 256 is not enough but 2^9 = 512 is, the smallest n that works is 9. You can also think in binary terms: 305 sits between 2^8 and 2^9, so it has 9 binary digits, and you need nine halvings to drop to zero. Therefore, the minimum number of halving operations is 9.