This is a simple problem that we can solve by rewriting the denominator as a product of its prime numbers. Since it is an odd number, we can start with 3.

3213 has three powers of 3. Let's try to divide the rest, 119, by 7.

119 is the product of 17 and 7, and since both of these factors are prime numbers, we are done calculating the factors of . Now, we can start canceling factors shared by the numerator and the denominator. After canceling shared factors, we are left with , which is the final answer.

Firstly, we must find the value of  for which  is true. You can solve this simply by testing out powers of :

4096 is 2 raised to the power of 12. Now we can easily say that . We can plug  in for  in  and solve:

 is a multiple of 7, so at the very least it includes a 7. Since  and  are both prime numbers, 7 is either  or . To make sure we have 49, the square of 7, into our product, we must take both the squares of  and  or , which is the final answer.

Our first two terms each involve an exponent in parentheses raised to an exponent outside the parentheses, so in this case, for each term, we multiply the exponent inside the parentheses with the exponent outside the parentheses. For the third term, we have the same two numbers raised to a certain power, so we add the exponents. This gives us:

We can then solve the expression:

Recall the rule dealing with raising exponents to a higher power when an exponent in parenthesis is raised to an exponent outside of the parentheses; multiply the exponents together to get the new exponent. Don't be confused by the fractional exponents. Simply multiply across the numerator and across the denominator.

In this case, the expression simplifies down to just .

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 . For our problem, the last digit is then given by . Now, this is pretty complicated to calculate, so we can try to find a pattern in the last digits of the powers of 7.

 is 

 is 

 is 

 is 

 is  again.

 is 9 again.

So, the pattern repeats every 4 numbers. Therefore, if we divide the power of 43 in  by 4, we get a remainder of 3. Therefore, the final answer is the last digit of , or .

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 given by its last digit raised to the power of 56, or .

Let's then try to find a pattern in the last digits of the powers of 9.

 is 

 is 

 is 

We can see that the odd powers of 9 have 9 as their last digits and the even powers of 9 have 1 for their last digits.

Since 56 is an even number, the last digit of  is .

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.