The square of an integer assumes the same parity (even or odd) as the integer itself, so  is odd if and only if  is odd.  is always even, having even factor 4. The sum of any integer and an even integer has the same parity as the first integer. Therefore,  assumes the same parity as , and, subsequently, the same parity as . In other words,  is odd if and only if  is odd.

The product of two or more numbers is 0 if and only if at least one of the factors is 0.

If you know the first statement, you know that neither  nor  is equal to 0, and that either , or both are equal to 0. You have narrowed the answer down to one or two.

If you know the second statement, you know that  and  are nonzero, but you know nothing about . You have narrowed the answer down to none or one.

If you know both statements, though, you know that  is the only one of the four numbers equal to zero, so you have answered the question.

To find the last digit in the base  representation of a number, divide the number by ; the remainder is that digit. If we know the number divided by 8 has remainder 5, we know its last base-eight digit is 5. If we know the number divided by 16 is 13, then we still know that the digit is 5, since the reminder when divided by 8 would be .

Let's first look at what the question is asking for. They want us to determine if C is an integer. Since we know that B is an integer, C will be an integer only if A is an integer. If A not an integer, C will not be an integer. So from statements 1 and 2, we want to see if we can prove if A is definitely an integer. 

First, let's try statement 2, which says that A*B is an integer. Let's see what happens if B is 2. If B is 2, then 2.5*2 = 5 is an integer, but 2*2 = 4 is also an integer. So A in this case could be an integer or a not. So statement 2 alone is not sufficient to get our answer. 

Now, let's go back to statement 1, which says A/B is an integer. Let's name another variable x. Let x = A/B. If x=A/B we can rewrite this as A = x*B. But since we know that x is an integer (given in the statement) and B is a non-zero integer (given in the question), A is therefore an integer! (The product of two integers is ALWAYS an integer.) This statement alone is enough to prove that A is an integer. 

Since, as discussed before, A is definitely an integer, and B is an integer. An integer minus an integer will always be an integer. Therefore C is an integer, and statement 1 is sufficient to answer the question. 

Statement 1 alone provides insufficient information to answer the question; for example, , which is positive, but , which is negative.

Statement 2, however, provides proof that  is positive, since any positive or negative number taken to an even power is positive.

If we know Statement 1, we can rewrite  as follows:

If we know Statement 2, we have a system of linear equations that we can solve to get  and :

 or 

By susbstituting: .

Either statement is sufficient to prove  an integer.

(1) This statement tells us that 2 and 7 are prime factors of . This information is not sufficient.

(2) This statement tells us that 3 and 5 are prime factors of . this information is not sufficient.

Considering both (1) and (2), we have four positive prime factors of  between them.

For statement (1), there are 3 possible two digit positive integers: 30, 12, 21. The remainder when these numbers are divided by 3 is 0. Therefore, statement (1) is sufficient. For statement (2), there are lots of different combinations of integers that have different remainders when divided by 3. For example, the remainder of 30 is 0, but the remainder of 14 is 2. Therefore, statement (2) is insufficient.

Statement (1) tells us that  is an odd number. Since we don’t know if  and  are odd numbers, we cannot decide the sign of .

Statement (2) tells us that  is an even number, since . When we do the multiplication, the product will be even if one or more of the integers are even. So with statement (2), we know for sure that  is even.

There are no perfect square integers between 110 and 120 inclusive:

Also, all perfect squares end in 0, 1, 4, 5, 6, or 9, depending on the last digit of the number being squared. 

From ether statement alone, it therefore follows that the number is not a perfect square integer.