Integers - GRE Quantitative Reasoning
Card 0 of 20
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.
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Of a group of 335 graduating high school atheletes, 106 played basketball, 137 ran track and field, and 51 participated in swimming. What is the maximum number of students that did both track and field and swimming upon graduation?
Of a group of 335 graduating high school atheletes, 106 played basketball, 137 ran track and field, and 51 participated in swimming. What is the maximum number of students that did both track and field and swimming upon graduation?
Simply recognize that logically, the participation of either sport is non-exclusive, that is, just because people took track and field does not necessarily mean they did not take swimming as well. As such, those 51 who took swimming could have all potentially done track and field, which means all 51 students.
Simply recognize that logically, the participation of either sport is non-exclusive, that is, just because people took track and field does not necessarily mean they did not take swimming as well. As such, those 51 who took swimming could have all potentially done track and field, which means all 51 students.
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Of 300 students, 120 are enrolled in math club, 150 are enrolled in chess club, and 100 are enrolled in both. How many students are not members of either club?
Of 300 students, 120 are enrolled in math club, 150 are enrolled in chess club, and 100 are enrolled in both. How many students are not members of either club?
There are 120 students in the math club and 150 students in the chess club, for a total membership of 270. However, 100 students are in both clubs, which means they are counted twice. You simply subtract 100 from 270, which will give you a total of 170 different students participating in both clubs. This means that the remaining 130 students do not participate in either club.
There are 120 students in the math club and 150 students in the chess club, for a total membership of 270. However, 100 students are in both clubs, which means they are counted twice. You simply subtract 100 from 270, which will give you a total of 170 different students participating in both clubs. This means that the remaining 130 students do not participate in either club.
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Choose the answer which best solves the following equation:
Choose the answer which best solves the following equation:
When adding integers, one needs to pay close attention to the sign. When you add a negative integer, it's the same thing as subtracting that integer. Therefore:
When adding integers, one needs to pay close attention to the sign. When you add a negative integer, it's the same thing as subtracting that integer. Therefore:
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Choose the answer below which best solves the following equation:
Choose the answer below which best solves the following equation:
The sum of any two negative numbers will be negative. Remember, also, that adding a negative number is like subtracting it. Therefore:
The sum of any two negative numbers will be negative. Remember, also, that adding a negative number is like subtracting it. Therefore:
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Simplify 
Simplify
The answer is 
Make sure to distribute negatives throughout the second half of the equation.
The answer is
Make sure to distribute negatives throughout the second half of the equation.
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Solve for 
:
Solve for
To solve this problem, you need to get your variable isolated on one side of the equation:
Taking this step will elminate the 
on the side with 
:
Now divide by 
to solve for 
:
The important step here is to be able to add the negative numbers. When you add negative numbers, they create lower negative numbers (if you prefer to think about it another way, you can think of adding negative numbers as subtracting one of the negative numbers from the other).
To solve this problem, you need to get your variable isolated on one side of the equation:
Taking this step will elminate the
Now divide by
The important step here is to be able to add the negative numbers. When you add negative numbers, they create lower negative numbers (if you prefer to think about it another way, you can think of adding negative numbers as subtracting one of the negative numbers from the other).
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Solve for 
:
Solve for
To solve this problem, first you must add 
to both sides of the problem. This will yield a result on the right side of the equation of 
, because a negative number added to a negative number will create a lower number (i.e. further away from zero, and still negative). Then you divide both sides by two, and you are left with 
.
To solve this problem, first you must add
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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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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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Assume 
and 
are both even whole numbers and 
.
What is a possible solution of 
?
Assume
What is a possible solution of
Since 
, then the final answer will be a number greater than one. The only answer that fits is 
.
Since
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Choose the answer below which best solves the following equation:
Choose the answer below which best solves the following equation:
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 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
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Solve for 
:
Solve for
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 the equation by
As a check, know that any time you divide an even number by another even number, the result will be even.
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Solve for 
:
Solve for
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, 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.
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