Even / Odd Numbers - SAT Math
Card 0 of 264
The sum of three consecutive odd integers is 93. What is the largest of the integers?
The sum of three consecutive odd integers is 93. What is the largest of the integers?
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.
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.
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Which of the following could represent the sum of 3 consecutive odd integers, given that d is one of the three?
Which of the following could represent the sum of 3 consecutive odd integers, given that d is one of the three?
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 3_d_ – 6.
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 3_d_ – 6.
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, where
and
are distinct positive integers. Which of the following could be values of
and
?
, where
and
are distinct positive integers. Which of the following could be values of
and
?
Since
and
must be positive, eliminate choices with negative numbers or zero. Since they must be distinct (different), eliminate choices where
. This leaves 4 and 5 (which is the only choice that does not add to 20), and the correct answer, 5 and 15.
Since and
must be positive, eliminate choices with negative numbers or zero. Since they must be distinct (different), eliminate choices where
. This leaves 4 and 5 (which is the only choice that does not add to 20), and the correct answer, 5 and 15.
Compare your answer with the correct one above
Solve: 
Solve:
Add the ones digits:

Since there is no tens digit to carry over, proceed to add the tens digits:

The answer is
.
Add the ones digits:
Since there is no tens digit to carry over, proceed to add the tens digits:
The answer is .
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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 more students taking Latin than there are students taking Greek. If there are
students taking Greek, how many total students are there?
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 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.
Compare your answer with the correct one above
Add: 
Add:
Add the ones digit.

Since the there is a tens digit, use that as the carryover to the next term.
Add the tens digit including the carryover.

The hundreds digit is 7.
Combine the ones digit of each calculation in order.
The answer is: 
Add the ones digit.
Since the there is a tens digit, use that as the carryover to the next term.
Add the tens digit including the carryover.
The hundreds digit is 7.
Combine the ones digit of each calculation in order.
The answer is:
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Add: 
Add:
Add the ones digit.

Carry over the one from the tens digit to the next number.
Add the tens digit with the carry over.

Carry over the one from the tens digit to the hundreds digit.
Add the hundreds digit with the carry over.

The thousands digit has no carry over. The second number has no thousands digit. This means that the thousands is one. Combine all the ones digits from each of the previous calculations.
The correct answer is: 
Add the ones digit.
Carry over the one from the tens digit to the next number.
Add the tens digit with the carry over.
Carry over the one from the tens digit to the hundreds digit.
Add the hundreds digit with the carry over.
The thousands digit has no carry over. The second number has no thousands digit. This means that the thousands is one. Combine all the ones digits from each of the previous calculations.
The correct answer is:
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If m and n are both even integers, which of the following must be true?
l. _m_2/_n_2 is even
ll. _m_2/_n_2 is odd
lll. _m_2 + _n_2 is divisible by four
If m and n are both even integers, which of the following must be true?
l. _m_2/_n_2 is even
ll. _m_2/_n_2 is odd
lll. _m_2 + _n_2 is divisible by four
While I & II can be true, examples can be found that show they are not always true (for example, 22/22 is odd and 42/22 is even).
III is always true – a square even number is always divisible by four, and the distributive property tell us that adding two numbers with a common factor gives a sum that also has that factor.
While I & II can be true, examples can be found that show they are not always true (for example, 22/22 is odd and 42/22 is even).
III is always true – a square even number is always divisible by four, and the distributive property tell us that adding two numbers with a common factor gives a sum that also has that factor.
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Let S be a set that consists entirely of even integers, and let T be the set that consists of each of the elements in S increased by two. Which of the following must be even?
I. the mean of T
II. the median of T
III. the range of T
Let S be a set that consists entirely of even integers, and let T be the set that consists of each of the elements in S increased by two. Which of the following must be even?
I. the mean of T
II. the median of T
III. the range of T
S consists of all even integers. If we were to increase each of these even numbers by 2, then we would get another set of even numbers, because adding 2 to an even number yields an even number. In other words, T also consists entirely of even numbers.
In order to find the mean of T, we would need to add up all of the elements in T and then divide by however many numbers are in T. If we were to add up all of the elements of T, we would get an even number, because adding even numbers always gives another even number. However, even though the sum of the elements in T must be even, if the number of elements in T was an even number, it's possible that dividing the sum by the number of elements of T would be an odd number.
For example, let's assume T consists of the numbers 2, 4, 6, and 8. If we were to add up all of the elements of T, we would get 20. We would then divide this by the number of elements in T, which in this case is 4. The mean of T would thus be 20/4 = 5, which is an odd number. Therefore, the mean of T doesn't have to be an even number.
Next, let's analyze the median of T. Again, let's pretend that T consists of an even number of integers. In this case, we would need to find the average of the middle two numbers, which means we would add the two numbers, which gives us an even number, and then we would divide by two, which is another even number. The average of two even numbers doesn't have to be an even number, because dividing an even number by an even number can produce an odd number.
For example, let's pretend T consists of the numbers 2, 4, 6, and 8. The median of T would thus be the average of 4 and 6. The average of 4 and 6 is (4+6)/2 = 5, which is an odd number. Therefore, the median of T doesn't have to be an even number.
Finally, let's examine the range of T. The range is the difference between the smallest and the largest numbers in T, which both must be even. If we subtract an even number from another even number, we will always get an even number. Thus, the range of T must be an even number.
Of choices I, II, and III, only III must be true.
The answer is III only.
S consists of all even integers. If we were to increase each of these even numbers by 2, then we would get another set of even numbers, because adding 2 to an even number yields an even number. In other words, T also consists entirely of even numbers.
In order to find the mean of T, we would need to add up all of the elements in T and then divide by however many numbers are in T. If we were to add up all of the elements of T, we would get an even number, because adding even numbers always gives another even number. However, even though the sum of the elements in T must be even, if the number of elements in T was an even number, it's possible that dividing the sum by the number of elements of T would be an odd number.
For example, let's assume T consists of the numbers 2, 4, 6, and 8. If we were to add up all of the elements of T, we would get 20. We would then divide this by the number of elements in T, which in this case is 4. The mean of T would thus be 20/4 = 5, which is an odd number. Therefore, the mean of T doesn't have to be an even number.
Next, let's analyze the median of T. Again, let's pretend that T consists of an even number of integers. In this case, we would need to find the average of the middle two numbers, which means we would add the two numbers, which gives us an even number, and then we would divide by two, which is another even number. The average of two even numbers doesn't have to be an even number, because dividing an even number by an even number can produce an odd number.
For example, let's pretend T consists of the numbers 2, 4, 6, and 8. The median of T would thus be the average of 4 and 6. The average of 4 and 6 is (4+6)/2 = 5, which is an odd number. Therefore, the median of T doesn't have to be an even number.
Finally, let's examine the range of T. The range is the difference between the smallest and the largest numbers in T, which both must be even. If we subtract an even number from another even number, we will always get an even number. Thus, the range of T must be an even number.
Of choices I, II, and III, only III must be true.
The answer is III only.
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Divide: 
Divide:
Take a known common factor of two and rewrite the fraction.

Dividing the number 143 into 16, the coefficient is 8 since:


There is a remainder of fifteen, which is
.
Combining the coefficient and the remainder as a mixed fraction, this can be rewritten as: 
The answer is: 
Take a known common factor of two and rewrite the fraction.
Dividing the number 143 into 16, the coefficient is 8 since:
There is a remainder of fifteen, which is .
Combining the coefficient and the remainder as a mixed fraction, this can be rewritten as:
The answer is:
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The sum of three consecutive odd integers is 93. What is the largest of the integers?
The sum of three consecutive odd integers is 93. What is the largest of the integers?
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.
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.
Compare your answer with the correct one above
Which of the following could represent the sum of 3 consecutive odd integers, given that d is one of the three?
Which of the following could represent the sum of 3 consecutive odd integers, given that d is one of the three?
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 3_d_ – 6.
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 3_d_ – 6.
Compare your answer with the correct one above
, where
and
are distinct positive integers. Which of the following could be values of
and
?
, where
and
are distinct positive integers. Which of the following could be values of
and
?
Since
and
must be positive, eliminate choices with negative numbers or zero. Since they must be distinct (different), eliminate choices where
. This leaves 4 and 5 (which is the only choice that does not add to 20), and the correct answer, 5 and 15.
Since and
must be positive, eliminate choices with negative numbers or zero. Since they must be distinct (different), eliminate choices where
. This leaves 4 and 5 (which is the only choice that does not add to 20), and the correct answer, 5 and 15.
Compare your answer with the correct one above
Solve: 
Solve:
Add the ones digits:

Since there is no tens digit to carry over, proceed to add the tens digits:

The answer is
.
Add the ones digits:
Since there is no tens digit to carry over, proceed to add the tens digits:
The answer is .
Compare your answer with the correct one above
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 more students taking Latin than there are students taking Greek. If there are
students taking Greek, how many total students are there?
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 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.
Compare your answer with the correct one above
Add: 
Add:
Add the ones digit.

Since the there is a tens digit, use that as the carryover to the next term.
Add the tens digit including the carryover.

The hundreds digit is 7.
Combine the ones digit of each calculation in order.
The answer is: 
Add the ones digit.
Since the there is a tens digit, use that as the carryover to the next term.
Add the tens digit including the carryover.
The hundreds digit is 7.
Combine the ones digit of each calculation in order.
The answer is:
Compare your answer with the correct one above
Add: 
Add:
Add the ones digit.

Carry over the one from the tens digit to the next number.
Add the tens digit with the carry over.

Carry over the one from the tens digit to the hundreds digit.
Add the hundreds digit with the carry over.

The thousands digit has no carry over. The second number has no thousands digit. This means that the thousands is one. Combine all the ones digits from each of the previous calculations.
The correct answer is: 
Add the ones digit.
Carry over the one from the tens digit to the next number.
Add the tens digit with the carry over.
Carry over the one from the tens digit to the hundreds digit.
Add the hundreds digit with the carry over.
The thousands digit has no carry over. The second number has no thousands digit. This means that the thousands is one. Combine all the ones digits from each of the previous calculations.
The correct answer is:
Compare your answer with the correct one above
If
is an odd integer and
is an even integer, which of the following must true of
?
If is an odd integer and
is an even integer, which of the following must true of
?
An even number subtracted from an odd number will always produce an odd result.
None of the other answer choices are correct.
An even number subtracted from an odd number will always produce an odd result.
None of the other answer choices are correct.
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Subtract: 
Subtract:
When subtracting even numbers, we will know that the ones digit will also be an even number.
Borrow a one from the tens digit of 64, since we cannot subtract the ones digits.
After borrowing, we can then subtract.

Subtract the tens place. The tens place of 64 becomes a 5.

Combine the two numbers.
The answer is
.
When subtracting even numbers, we will know that the ones digit will also be an even number.
Borrow a one from the tens digit of 64, since we cannot subtract the ones digits.
After borrowing, we can then subtract.
Subtract the tens place. The tens place of 64 becomes a 5.
Combine the two numbers.
The answer is .
Compare your answer with the correct one above
If m and n are both even integers, which of the following must be true?
l. _m_2/_n_2 is even
ll. _m_2/_n_2 is odd
lll. _m_2 + _n_2 is divisible by four
If m and n are both even integers, which of the following must be true?
l. _m_2/_n_2 is even
ll. _m_2/_n_2 is odd
lll. _m_2 + _n_2 is divisible by four
While I & II can be true, examples can be found that show they are not always true (for example, 22/22 is odd and 42/22 is even).
III is always true – a square even number is always divisible by four, and the distributive property tell us that adding two numbers with a common factor gives a sum that also has that factor.
While I & II can be true, examples can be found that show they are not always true (for example, 22/22 is odd and 42/22 is even).
III is always true – a square even number is always divisible by four, and the distributive property tell us that adding two numbers with a common factor gives a sum that also has that factor.
Compare your answer with the correct one above