Suppose n is an odd number. Find the cube of n and then divide it by 4. What possible remainders could occur? Check all possible remainders and eliminate any that are impossible.

To find all possible remainders when the cube of an odd number n is divided by 4, we first need to write the cube of n as \( n^3 \) and then divide this by 4.

Since n is an odd number, it can be expressed in the form \( n = 2k + 1 \) where k is an integer. The reason for this is that any odd number can be represented as one more than an even number, and since all even numbers are multiples of 2, an odd number is always 2 times some integer k plus 1.

Now let's find the cube of n, using our expression for the odd number:

\( (2k + 1)^3 = (2k + 1)(2k + 1)(2k + 1) \)

We can expand this using the binomial formula or by multiplying term by term:

\( (2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1 \)

Now, to be divided by 4, we need to break this expression into two parts: the part that gives no remainder (i.e., is perfectly divisible by 4) and the part that determines the remainder.

Notice that \( 8k^3 \) and \( 12k^2 \) are both obviously divisible by 4 since they have 4 as a factor (8 and 12 are multiples of 4).

That leaves us with \( 6k + 1 \). Now, \( 6k \) can be written as \( 4k + 2k \), where again \( 4k \) has no remainder upon division by 4, but \( 2k \) is twice an integer, hence even. So, \( 2k \) divided by 4 will either leave a remainder of 0 (if k is even) or 2 (if k is odd).

Looking at the next term, \( +1 \), this is not divisible by 4 and will always contribute a remainder of 1.

Therefore, the term \( 6k + 1 \) when divided by 4 can leave a remainder of \( 2 + 1 = 3 \) when k is odd, since \( 2k \) will contribute a remainder of 2, and the +1 will add one more to the remainder. If k is even, \( 2k \) leaves no remainder and thus the total remainder is simply that contributed by the +1, which is 1.

In conclusion, the possible remainders when the cube of an odd number (n) is divided by 4 are 1 and 3.

Extra: Understanding this problem requires knowledge of basic algebra and number theory concepts. In particular, it hinges on the properties of odd and even numbers. An even number is any integer that is a multiple of 2 and can be represented as \( 2k \), where k is an integer. An odd number, as explained, is one more than an even number and is written as \( 2k + 1 \).

Cubing an odd number (or any number) involves multiplying that number by itself three times. The cube of an even number retains its even properties, but the cube of an odd number still results in an odd number due to their respective algebraic structures.

When it comes to modular arithmetic, which is the branch of mathematics that studies remainders, we often look for patterns in how numbers behave under division by certain integers. Recognizing that remainders from even numbers divided by 4 are either 0 or 2 greatly simplifies the process of elimination for potential remainders in problems like this. This understanding of number theory is not just academic—it has practical applications in areas such as cryptography, computer science, and solving puzzles.