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------------------------------------------------ Consider an all-integer problem with 3 variables, each restricted to be between 0 and 4. The number of feasible solutions to this problem CANNOT be:

A. 0
B. 139
C. 30
D. 251
E. Neither 139 nor 251.

The only option that cannot be the number of feasible solutions is Neither 139 nor 251, as it's greater than the maximum possible of 125. Hence, option e is correct.

Here's how we can reason through this problem:

1. **Total possible combinations:** Each variable can take 5 possible values (0, 1, 2, 3, or 4), so the total number of possible combinations is 5 * 5 * 5 = 125.

2. **Feasible solutions:** However, not all combinations might be feasible solutions due to potential constraints in the problem. The number of feasible solutions must be less than or equal to 125.

3. **Analyzing the options:**

- Option a: 0 feasible solutions is possible if all combinations violate constraints.

- Option b: 139 feasible solutions is not possible as it exceeds the maximum of 125.

- Option c: 30 feasible solutions is possible within the range of 0 to 125.

- Option d: 251 feasible solutions is not possible as it exceeds the maximum of 125.

- Option e: This option is true as 139 and 251 are not possible.

Therefore, the only option that cannot be the number of feasible solutions is Neither 139 nor 251, as it's greater than the maximum possible of 125.