The following conjectures were made by students. Are they true or false? Give a counterexample if possible.

Only half of the nonzero even numbers up to 100 are divisible by 4.
My answer is that it is true.

A number is divisible by 8 if it is divisible by 4 and divisible by 2.
My answer is that it is true.

Am I right?

The first is right.

For the second, how about 4, 12, 20...?

To determine whether the conjectures made by the students are true or false, let's analyze them one by one.

1. Only half of the nonzero even numbers up to 100 are divisible by 4.
To test this conjecture, we need to find out how many nonzero even numbers there are up to 100 that are divisible by 4. We can divide all the even numbers up to 100 by 4 and count how many of them have a remainder of 0. If the conjecture is true, we should find that exactly half of the even numbers are divisible by 4.

Nonzero even numbers up to 100: 2, 4, 6, 8, ..., 98, 100

Divisible by 4: 4, 8, 12, ..., 96, 100

Out of these numbers, 4, 8, 12, ..., 96 are divisible by 4.

By counting, we can observe that there are 24 numbers in this list. Since there are 50 nonzero even numbers up to 100, the conjecture is false, as only 24 out of 50 numbers are divisible by 4.

Counterexample: For example, the nonzero even number 16 is divisible by 4, disproving the conjecture.

2. A number is divisible by 8 if it is divisible by 4 and divisible by 2.
To evaluate this conjecture, let's analyze the meaning of divisibility.

For a number to be divisible by 4, it must have a units digit divisible by 4 or be divisible by 2 twice.

For a number to be divisible by 2, it simply needs to have an even units digit.

Therefore, if a number is divisible by 4 and also divisible by 2, it will be divisible by 8.

Hence, the conjecture is true.

In summary:
1. The first conjecture is false, as only 24 out of 50 nonzero even numbers are divisible by 4.
2. The second conjecture is true because a number divisible by 4 and 2 will also be divisible by 8.

Therefore, your answer is incorrect for the first conjecture but correct for the second conjecture.