For any positive integer whose last digit is , the last digit if this integer raised to any power  is the same as the last digit of . So, the last digit of  is the same as the last digit of . Let's try to find a pattern in the last digits of the powers of 3:

For , the last digit is 3

 is 9

 is 7

 is 1

 is 3

So the pattern repeats every 4 consecutive powers. We therefore simply have to divide 47 by 4 to get a remainder of 3. The last digit is given by , or 7. To get the final answer, we simply have to multiply 7 by 8; we get 56, whose last digit is , which is our final answer.

Some trinomials of the form  are factorable as

where , .

Therefore, if the number in the circle makes the polynomial factorable, it can be factored as the product of two whole numbers whose sum is 17.

These numbers are:

Of the five choices, only 56 is not among these numbers, so it makes the polynomial prime. It is the correct choice.

Some trinomials of the form  are factorable as

where , .

Therefore, we are looking for an integer which can be factored as the product of two integers whose difference is 9.

We can start with 1 and 10 and work our way up:

The products will increase, so we can stop, having found our two numbers, 2 and 11. Their product, 22, is the correct choice.

If  is not prime, then, as a quadratic trinomial of the form  it is factorable as 

where  and .

Therefore, we are looking for a whole number that is not the difference of two factors of 60. The integers that are such a difference are

Of the choices, only 23 is not on the list, so it is the correct choice.

The factoring pattern to look for in the second polynomial is the sum of cubes, so the number in the circle must be a perfect cube of a whole number . We can write this polynomial as

which can be factored as 

By the condition of the problem, , so the number that replaces the square is , and the number that replaces the circle is .

There are different ways to approach this problem. We just need to remember three things:

Keeping those in mind, we can simplify the numerator and coefficients:

I'm going to move the negative exponents (number 2 in the list above) in order to make them positive: 

We can now simplify the numerator again with the  exponents (number 3) and the  exponents (number 1): 

The first thing we must do is distribute the exponents outside of the parentheses across each expression (remembering of course that exponents set to another exponent are multiplied).

The last step is to follow the rules of exponent addition/subtraction:

 and 

Therefore:

To answer this question, note that 

and 

.

Therefore, since

it follows that 

and 

.

Rewrite both sides as powers of 2 using the exponent rules as follows:

Since powers of the same base are equal, set the exponents equal to each other:

Rewrite both sides as powers of 3 using the exponent rules as follows:

Since powers of the same base are equal, set the exponents equal to each other: