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.

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

Plugging in the values given we arrive at the total fruit John has:

A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.

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.

Since,must result in a positive whole number. The only answer that fits these requirements of being both positive and whole number is.

First, add the total number of passengers that got off BEFORE the fourth stop. Three plus three is six, so you know that you've lost six total passengers before the fourth stop., so there are ten passengers remaining at the fourth stop, and that's how many get off there.

If it's simpler for you, you can split this problem into two parts: First, takeaway from, and you're left with. Then, take awayfrom(you can even count backwards if necessary), and you'll be left with the final answer,.

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还是很奇怪。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 thereforem * 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”项is also even, since an even minus an odd would be odd. The only situation in which the "y²”项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: then^thodd number is 2n– 1. (Then^theven number is 2n.) So the 65^thodd number is 2 * 65 – 1 = 129.