If the largest of the three consecutive odd integers is d, then the three numbers are (in descending order):

d, d – 2, d – 4

This is true because consecutive odd integers always differ by two. Adding the three expressions together, we see that the sum is 3d – 6.

Since \(\displaystyle \dpi{100} p\) and \(\displaystyle \dpi{100} r\) must be positive, eliminate choices with negative numbers or zero. Since they must be distinct (different), eliminate choices where \(\displaystyle \dpi{100} p=r\).  This leaves 4 and 5 (which is the only choice that does not add to 20), and the correct answer, 5 and 15.

Consecutive odd integers differ by 2. If the smallest integer is x, then

x + (x + 2) + (x + 4) = 93

3x + 6 = 93

3x = 87

x = 29

The three numbers are 29, 31, and 33, the largest of which is 33.

The odd/even status of \(\displaystyle B\) is not known, so no information can be determined about that of \(\displaystyle C\).

\(\displaystyle C\) is known to be an integer, so \(\displaystyle 2C\) is an even integer. Added to odd number \(\displaystyle A\), an odd sum is yielded; this is \(\displaystyle D\).

\(\displaystyle D\) is known to be odd, so \(\displaystyle 3D\) is also odd. Added to odd number \(\displaystyle A\), an even sum is yielded; this is \(\displaystyle E\).

\(\displaystyle E\) is known to be even, so \(\displaystyle 4E\) is even. Added to odd number \(\displaystyle A\); an odd sum is yielded; this is \(\displaystyle F\).

The numbers known to be odd are \(\displaystyle D\) and \(\displaystyle F\); the number known to be even is \(\displaystyle E\); nothing is known about \(\displaystyle C\).

A power of an integer takes on the same odd/even status as that integer. Therefore, without knowing the odd/even status of \(\displaystyle B\), we do not know that of \(\displaystyle B^{2}\), and, subsequently, we cannot know that of \(\displaystyle C\). As a result, we cannot know the status of any of the other values of the other three variables in the subsequent statements. Therefore, none of the four choices are correct.

\(\displaystyle 2B\), the product of an even integer and another integer, is even. Therefore, \(\displaystyle C\) is equal to the sum of an odd number \(\displaystyle A\) and an even number \(\displaystyle 2B\), and it is odd.

\(\displaystyle 3C\), the product of odd integers, is odd, so \(\displaystyle D\), the sum of odd integers \(\displaystyle A\) and \(\displaystyle 3C\), is even.

\(\displaystyle 5D\), the product of an odd integer and an even integer, is even, so \(\displaystyle E\), the sum of an odd integer \(\displaystyle A\) and even integer \(\displaystyle 5D\), is odd.

\(\displaystyle 7E\), the product of odd integers, is odd, so \(\displaystyle F\), the sum of odd integers \(\displaystyle A\) and \(\displaystyle 7E\), is even.

The correct response is that \(\displaystyle C\) and \(\displaystyle E\) are odd and that \(\displaystyle D\) and \(\displaystyle F\) are even.

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)