Even / Odd Numbers - GRE Quantitative Reasoning

To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.

For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.

y, y + 2, y + 4, y + 6 . . .

The product of any number and 0 is 0.

Therefore, column B must be greater.

If and are both even whole numbers, then their addition must be an even whole number as well. Although is an even number, it is not a whole number and could therefore not be a solution. This means the only possible solution would be .

Even numbers come in the form 2x, and odd numbers come in the form (2x + 1), where x is an integer. If this is confusing for you, simply plug in numbers such as 1, 2, 3, and 4 to find that:

Any odd number + any even number = odd number

Any odd number + any odd number = even number

Any even number x any number = even number

Any odd number x any odd number = odd number

a(b + c) = odd x (odd + even) = odd x (odd) = odd

When multiplying odd/even numbers, we know that odd * odd = odd, and odd * even = even. We also know that odd + odd = even. We will proceed to evaluate each answer choice, knowing that m, n and p are odd.

(m + 1) * n

m + 1 becomes even. This gives us even * odd = even.

m + n + p + 1

Odd + odd + odd + odd = even + odd + odd = even + even = even.

(m - 2 )* n * p

m - 2 is stil odd. This gives us odd * odd * odd = odd * odd = odd.

m * (n + p)

Odd + odd is even, so here we have odd * even = even.

m * p * (n -1)

n - 1 becomes even so we have odd * odd * even = odd * even = even.

The correct answer is therefore m * p * (n -1).

For this problem, we must recognize under what arithmetic conditions an even or odd number is produced. We do not know what the values for x and y are, but we do know for example that any two numbers added can be either even or odd, and and any number multiplied by two must be even (e.g. 2 * 2 = 4; 3 * 2 = 6), or an odd number multiplied by another odd is always even.

In this situation, we have to ensure that an arithmetic operation (subtraction) must result only in an even number, by further ensuring that the variables themselves are fixed to either odd or even values.

In this situation, we know that due to the "*2 (= 2 * 2)" principle, the 4x component always being an even value. The only situation in which the operation "4x – y2" will be even is if the "y2" term is also even, since an even minus an odd would be odd. The only situation in which the "y2" term is even is if the y itself is even, since an odd number squared always results in an odd value (e.g. 32 = 9; 72 = 25). To ensure that the y itself is even, we must also double it to "2y"(2).

We cannot add 1 to a random variable y as it may still result in an odd y-value. Similarly, since we already stated that a squared value may still be odd, we cannot be sure that squaring the y will also result in an even number.

Doing this by counting the odd numbers will take way too long for the GRE. And if you look at the answer choices, you see five consecutive odd numbers, so one little mistake in counting will give you the wrong answer! Instead, we should use the formula for finding odd numbers: the n_th odd number is 2_n – 1. (The n_th even number is 2_n.) So the 65th odd number is 2 * 65 – 1 = 129.

If there are students taking Greek, then there are or 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:

or total students.

Since , then the final answer will be a number greater than one. The only answer that fits is .

To solve this problem, set up long division for yourself. First, you know that your hundreds digit of the solution will be one, as twelve goes into seventeen one time. Then, you take twelve away from seventeen, and are left with three remaining. Bring down your next digit over from the dividend, and you are left with thirty two, which twelve goes into four times. Already, you know that the answer has to be as none of the other answers fit the correct pattern.

To solve, isolate the variable by dividing both sides of the equation by :

As a check, know that any time you divide an even number by another even number, the result will be even.

To solve, isolate the variable by dividing both sides of your equation by :

As a check, know that any time you divide an even number by another even number, your result will be even.

To solve, first isolate the variable by dividing both sides of the equation by :

As a check, know that any time you divide an even number by another even number, you will get an even result.

To solve, isolate your variable by dividing both sides of the equation by :

As a check, know that any time you divide an even number by another even number, the result will be even.

Set up the equation,

.

Simplify to and solve for x to find .

Therefore the 3 consecutive odd integrers are .

The two requirements for this problem are that both and must be odd, and that . The only answer that fits both of these is . The other answers show either or is an even number.

To solve, divide both sides of the equation by :

As a check, if you divide an odd number by another odd number, your result should be odd.

To solve, isolate your variable by dividing both sides of the equation by :

To solve, first isolate your variable by dividing both sides of the equation by :

As a check, if you divide an odd number by another odd number, your result should be odd.