The easiest way to explain this is by example. 

We can use 16, since 

This makes 4 the correct choice.

The product of a group of nonzero numbers is negative if and only if an odd number of these factors is negative. This occurs in each of these scenarios except for one - all of the numbers (eight) being negative.

Any integer that is divisible by 5 must end in a 0 or a 5, so any integer which, when divided by 5, yields remainder 3 must end in 3 or 8. Since the cubes of 3 and 8 are 27 and 512, the cube of an integer that ends in one of these digits must end in 2 or 7, meaning that, when divided by 5, the remainder will be 2.

Since the numerator is always 1 greater than the denominator, we know that for large enough values of , it's never going to be an integer (one will be even, other odd).  In fact, there are only 2 cases where this can be done.  The first is dividing 0.  Since 0 is divisible by every number, if the numerator is 0, then we will still get an integer.  Thus one answer is

The other answer occurs as a special case as well.  We can divide any number by 1 evenly, so when the denominator is 1 we get an integer:

In every other case, we will have a non-integer.

If a number has a 5 in its base-six representation, then the number divided by 6 yields a remainder of 5. We test each of these five numbers by dividing each by 6 and noting the remainder:

Among the five choices, only 743 yields remainder 5 when divided by 6, so this is the correct choice.

We can divide this group number into two parts. The first part is and the second part is .

The first group is symmetrical, so the sum of this group of numbers is 0. Now for the second part there are only two numbers left according to the question (sum of odd numbers), which is 157 and 159.

Therefore, the answer is .  

There are two prime numbers between 20 and 30; they are 23 and 29. Their product:

The composite numbers that fall in the range 31-40 - those with more than two factors - are 32, 33, 34, 35, 36, 38, 39, and 40. Add them:

We look at the prime factorizations of each of the five choices, and see which one breaks down as the product of more than two primes - this is the correct choice.

None of the above choices are correct, as each can be written as the product of two primes. As for the other choice:

Therefore, it cannot be expressed as the product of two primes. This is the correct choice.

For a number to be divisible by 12, it must be divisible by 3 and 4. 

For a number to be divisible by 4, the last two digits must form a number divisible by 4. This allows us to eliminate 34 and 54.

For a number to be divisible by 3, the digit sum must be divisible by 3. The sum of the first 98 digits, all of which are 1's, is 98; we add this 98 to the sum of each remaining pair of digits to determine which passes this test:

, which is not divisible by 3.

, which is not divisible by 3.

, which is divisble by 3.

64 is the correct choice.