If exactly two of $\displaystyle \left \{ A, B,C, D\right \}$ are odd, then exactly one of the seven expressions being added is odd - namely, the only one that does not have an even factor (for example, if $\displaystyle A$ and $\displaystyle B$ are odd, then the only odd number is $\displaystyle AB$). This makes $\displaystyle AB + BC + CD + AD + AC + BD+ ABCD$ the sum of one odd number and six even number and, subsequently, odd.

If exactly three of $\displaystyle \left \{ A, B,C, D\right \}$ are odd, then exactly three of the seven expressions being added are odd - namely, the three that do not include the even factor (for example, if $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are odd, then the three odd numbers are $\displaystyle AB$, $\displaystyle AC$, and $\displaystyle BC$). This makes $\displaystyle AB + BC + CD + AD + AC + BD+ ABCD$ the sum of three odd numbers and four even numbers and, subsequently, odd.

If all four of $\displaystyle \left \{ A, B,C, D\right \}$ are odd, then all of the seven expressions being added, being the product of only odd numbers, are odd. This makes $\displaystyle AB + BC + CD + AD + AC + BD+ ABCD$ the sum of seven odd numbers, and, subsequently, odd.

The correct choice is that all three scenarios are possible.

Add the ones digits:

$\displaystyle 1+3+5=9$

Since there is no tens digit to carry over, proceed to add the tens digits:

$\displaystyle 1+1+1=3$

The answer is $\displaystyle 39$.

If there are $\displaystyle 257$ students taking Greek, then there are $\displaystyle 257+15$ or $\displaystyle 272$ students taking Latin. However, the question asks how many total students there are in the school, so you must add these two values together to get:

$\displaystyle 257 + 272$ or $\displaystyle 529$ total students.

The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example. 

Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.

even * even = even (2 * 2 = 4)

even * odd = even (2 * 3 = 6)

odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.

even * even * even = even * even = even (2 * 2 * 2 = 8)

odd * odd * even = odd * even = even (1 * 3 * 2 = 6)

When one multiplies two negative numbers (or any even multiple) the result is a positive number. However, when one multiplies three negative numbers (or any odd multiple) the product is negative. If the result of multiplying 8 negatives is odd, the largest number of negative integers will be the largest odd number, in this case 7.

Let's examine the choice $\displaystyle n^2$. We can rewrite $\displaystyle n^2$ as $\displaystyle n \times n$, which would be multiplying an odd number (because $\displaystyle n$ is odd) by an odd number. Multiplying two odd numbers always produces another odd number. So this can't be the correct answer.

Next, let's look at $\displaystyle n^3 + 2$. We can rewrite this as $\displaystyle n^2 \times n + 2$. We already established that $\displaystyle n^2$ must be odd, so then $\displaystyle n^2\times n$ must also be odd. If we then add $\displaystyle 2$ to an odd number, we still get an odd number. So we can eliminate this choice as well.

Now, let's look at the choice $\displaystyle n^2 + 2n$. Let's factor this as $\displaystyle n(n + 2)$. We know that n must be odd, and we know that $\displaystyle n+2$ must be odd. Therefore, $\displaystyle n(n + 2)$ is odd, because multiplying two odd numbers gives us an odd number.

Finally, let's analyze $\displaystyle n^2 + 2n + 1$. We can rewrite this as $\displaystyle (n + 1)(n + 1)$. Since n is odd, $\displaystyle n+1$ must be an even number. When we multiply an even number by an even number, we get an even number, so $\displaystyle (n + 1)(n + 1)$ must be even, and it cannot be odd. 

The answer is $\displaystyle n^2 + 2n + 1$.

Let us analyze I, II, and III one at a time.

Because n and m are both even, if we increase either by 1, the result will be an odd number. Thus, n + 1 and m + 1 are both odd. When two odd numbers are multiplied together, the result is always an odd number. Thus (n + 1)(m + 1) must be an odd number.

Because n and m are even, when we multiply two even numbers together, we always get an even number. Thus nm is even. However, when we then add one to an even number, the result will be an odd number. Thus, nm + 1 is odd.

We just established that nm is even. If we subtract an even number from an even number, the result is always even. Thus, nm – m is an even number.

Only choice I and II will always produce odd numbers.

The answer is I and II only.

The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example. 

Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.

even * even = even (2 * 2 = 4)

even * odd = even (2 * 3 = 6)

odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.

even * even * even = even * even = even (2 * 2 * 2 = 8)

odd * odd * even = odd * even = even (1 * 3 * 2 = 6)

Let us say we have two numbers, $\displaystyle dpi{100} x$ and $\displaystyle dpi{100} y$ whose ones digits are $\displaystyle dpi{100} A$ and B, respectively. If we want to know the ones digits of the product of $\displaystyle dpi{100} x$ and $\displaystyle dpi{100} y$, all we need to do is to look at the ones digit of the product of $\displaystyle dpi{100} A$ and $\displaystyle dpi{100} B$. For example, if we multiply 137 and 219, then the ones digit will be the same as the ones digit of $\displaystyle dpi{100} 7 imes 9=63$. Since the ones digit of 63 is 3, the ones digit of 137 x 219 will also be 3. In short, we really only need to worry about the ones digits of the numbers we multiply when we try to find the ones digit of their product.

We want to find the ones digit of $\displaystyle 2013^{2013}$. An exponent is essentially just a short hand for repeated multiplication. Let us look at the ones digit of the first few exponents of 2013.

 $\displaystyle 2013^1 = 2013$  - the ones digit is 3.

To find the ones digit of the 2013 to the second power, we need to think of it as the product of 2013 and 2013. As discussed previously, if we want the ones digit of two numbers multiplied together, we just need to multiply their ones digits. Thus, if we multiply 2013 by 2013, then the ones digit will be the same as $\displaystyle dpi{100} 3 imes 3=9$.

$\displaystyle 2013^2 = 2013 cdot 2013$  - ones digit is 9.

Next, we want to find the ones digit of 2013 to the third power. In order to do this, we will multiply the square of 2013 by 2013. It does not matter that we do not know exactly what 2013 squared equals, beacuse we only need to worry about the ones digit, which is 9. In other words, 2013 to the third power will have a ones digit that is equal to the ones digit of the product of 9 (which was the ones digit of 2013 squared) and 3 (which is the ones digit of 2013). When we multiply 9 and 3, we get 27, so the ones digit of 2013 to the third power is 7.

$\displaystyle 2013^3 = 2013^2 cdot 2013$  - ones digit of 7.

To find the ones digit of 2013 to the fourth power, we only need to worry about multiplying the ones digit of 2013 to the third power (which is 7) by the ones digit of 2013. When we mulitply 7 and 3, we get 21, which means that the ones digit of 2013 to the fourth power is 1.

$\displaystyle 2013^4 = 2013^3 cdot 2013$  - ones digit of 1.

To find the ones digit of 2013 to the fifth power, we will multiply 1 by 3, which gives us 3.

$\displaystyle 2013^5 = 2013^4 cdot 2013$  - ones digit of 3.

Notice that we are back to a ones digit with 3. If we multiply this by 2013, we will end up with a ones digit of 9. In other words, the ones digits repeat every fourth power.

The value of the ones digits of the powers of 2013 is as follows (starting with 2013 to the first power):

3, 9, 7, 1, 3, 9, 7, 1,....

We essentially want to find the 2013th term of the sequence above. Notice that every fourth term is 1, i.e. the sequence repeats every four terms. If a terms position in the sequence is a multiple of 4, then the term will be 1. In short, the 4th, 8th, 12th, 16th terms, and so on, will be 1. Because 2012 is a multiple of 4, the 2012th term in the sequence will be 1. (We can determine if a number is a multiple of 4 by looking at its last two digits.) This means that that 2013th term will be 3. Thus, 2013 to the power of 2013 has a ones digit of 3.

The answer is 3.

For the product of three integers to be odd, all three integers must themselves be odd.

$\displaystyle 2A$ must be even, so for $\displaystyle C+ 2A$ to be odd, $\displaystyle C$ must be odd. Similarly, for $\displaystyle A+ 2B$ and $\displaystyle B + 2C$ to be odd, respectively, $\displaystyle A$ and $\displaystyle B$ must be odd. 

Therefore, all three of $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ must be odd, and the correct response is III only.