Even / Odd Numbers - GRE Quantitative Reasoning
Card 0 of 312
If 
is even, and 
is odd. Which of the following must be odd?
If
To solve, pick numbers to represent 
and 
. Let 
and 
. Now try each of the equations given:
.
Only 
works and is thus our answer.
To solve, pick numbers to represent
Only
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The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
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.
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.
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and 
are both even whole numbers.
What is a possible solution for 
?
What is a possible solution for
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 
.
If
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If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
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.
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.
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Choose the answer below which best solves the following problem:
Choose the answer below which best solves the following problem:
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.
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.
Compare your answer with the correct one above
The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
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.
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.
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and are both even whole numbers.
What is a possible solution for ?
What is a possible solution for
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 .
If
Compare your answer with the correct one above
If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
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.
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.
Compare your answer with the correct one above
Choose the answer below which best solves the following problem:
Choose the answer below which best solves the following problem:
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.
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.
Compare your answer with the correct one above
The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
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.
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.
Compare your answer with the correct one above
and are both even whole numbers.
What is a possible solution for ?
What is a possible solution for
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 .
If
Compare your answer with the correct one above
If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
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.
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.
Compare your answer with the correct one above
Choose the answer below which best solves the following problem:
Choose the answer below which best solves the following problem:
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.
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.
Compare your answer with the correct one above
If 
and 
are odd integers, and 
is even, which of the following must be an odd integer?
If
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
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
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If m, n and p are odd integers, which of the following must be an odd integer?
If m, n and p are odd integers, which of the following must be an odd integer?
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).
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).
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The operation ¤ is defined for all integers x and y as: x ¤ y = 4x-y2.
If x and y are positive integers, which of the following cannot produce an odd value?
The operation ¤ is defined for all integers x and y as: x ¤ y = 4x-y2.
If x and y are positive integers, which of the following cannot produce an odd value?
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.
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.
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What is the 65th odd number?
What is the 65th odd 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.
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.
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At a certain high school, everyone must take either Latin or Greek. There are 
more students taking Latin than there are students taking Greek. If there are 
students taking Greek, how many total students are there?
At a certain high school, everyone must take either Latin or Greek. There are
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.
If there are
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If and are odd integers, and is even, which of the following must be an odd integer?
If
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
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
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If m, n and p are odd integers, which of the following must be an odd integer?
If m, n and p are odd integers, which of the following must be an odd integer?
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).
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).
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