Even / Odd Numbers - PSAT Math
Card 1 of 231
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
Tap to reveal answer
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
Tap to reveal answer
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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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?
Tap to reveal answer
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?
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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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p+r=20, where p and r are distinct positive integers. Which of the following could be values of p and r?
p+r=20, where p and r are distinct positive integers. Which of the following could be values of p and r?
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Since p and r must be positive, eliminate choices with negative numbers or zero. Since they must be distinct (different), eliminate choices where p=r. This leaves 4 and 5 (which is the only choice that does not add to 20), and the correct answer, 5 and 15.
Since p and r must be positive, eliminate choices with negative numbers or zero. Since they must be distinct (different), eliminate choices where p=r. This leaves 4 and 5 (which is the only choice that does not add to 20), and the correct answer, 5 and 15.
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You are given that 
are all positive integers. Also, you are given that:
is an odd number. 
can be even or odd. What is known about the odd/even status of the other four numbers?
You are given that
Tap to reveal answer
The odd/even status of 
is not known, so no information can be determined about that of 
.
is known to be an integer, so 
is an even integer. Added to odd number 
, an odd sum is yielded; this is 
.
is known to be odd, so 
is also odd. Added to odd number 
, an even sum is yielded; this is 
.
is known to be even, so 
is even. Added to odd number 
; an odd sum is yielded; this is 
.
The numbers known to be odd are 
and 
; the number known to be even is 
; nothing is known about 
.
The odd/even status of
The numbers known to be odd are
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You are given that 
are all positive integers. Also, you are given that:
is an odd number. 
can be even or odd. What is known about the odd/even status of the other four numbers?
You are given that
Tap to reveal answer
A power of an integer takes on the same odd/even status as that integer. Therefore, without knowing the odd/even status of 
, we do not know that of 
, and, subsequently, we cannot know that of 
. As a result, we cannot know the status of any of the other values of the other three variables in the subsequent statements. Therefore, none of the four choices are correct.
A power of an integer takes on the same odd/even status as that integer. Therefore, without knowing the odd/even status of
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You are given that are all positive integers. Also, you are given that:
You are given that 
is odd, but you are not told whether 
is even or odd. What can you tell about whether the values of the other four variables are even or odd?
You are given that
You are given that
Tap to reveal answer
, the product of an even integer and another integer, is even. Therefore, 
is equal to the sum of an odd number 
and an even number 
, and it is odd.
, the product of odd integers, is odd, so 
, the sum of odd integers 
and 
, is even.
, the product of an odd integer and an even integer, is even, so 
, the sum of an odd integer 
and even integer 
, is odd.
, the product of odd integers, is odd, so 
, the sum of odd integers 
and 
, is even.
The correct response is that 
and 
are odd and that 
and 
are even.
The correct response is that
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Solve: 
Solve:
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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
Tap to reveal answer
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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, 
, 
, and 
are positive integers.
is odd.
Which of the following is possible?
I) Exactly two of 
are odd.
II) Exactly three of 
are odd.
III) All four of 
are odd.
Which of the following is possible?
I) Exactly two of
II) Exactly three of
III) All four of
Tap to reveal answer
If exactly two of 
are odd, then exactly one of the seven expressions being added is odd - namely, the only one that does not have an even factor (for example, if 
and 
are odd, then the only odd number is 
). This makes 
the sum of one odd number and six even number and, subsequently, odd.
If exactly three of 
are odd, then exactly three of the seven expressions being added are odd - namely, the three that do not include the even factor (for example, if 
, 
, and 
are odd, then the three odd numbers are 
, 
, and 
). This makes 
the sum of three odd numbers and four even numbers and, subsequently, odd.
If all four of 
are odd, then all of the seven expressions being added, being the product of only odd numbers, are odd. This makes 
the sum of seven odd numbers, and, subsequently, odd.
The correct choice is that all three scenarios are possible.
If exactly two of
If exactly three of
If all four of
The correct choice is that all three scenarios are possible.
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When 8 integers are multiplied their product is negative, then at most how many of the integers can be negative?
When 8 integers are multiplied their product is negative, then at most how many of the integers can be negative?
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When one multiplies two negative numbers (or any even multiple) the result is a positive number. However, when one multiplies three negative numbers (or any odd multiple) the product is negative. If the result of multiplying 8 negatives is odd, the largest number of negative integers will be the largest odd number, in this case 7.
When one multiplies two negative numbers (or any even multiple) the result is a positive number. However, when one multiplies three negative numbers (or any odd multiple) the product is negative. If the result of multiplying 8 negatives is odd, the largest number of negative integers will be the largest odd number, in this case 7.
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If 
is an odd integer, all of the following must be odd integers EXCEPT:
If 
Tap to reveal answer
Let's examine the choice 
. We can rewrite 
as 
, which would be multiplying an odd number (because 
is odd) by an odd number. Multiplying two odd numbers always produces another odd number. So this can't be the correct answer.
Next, let's look at 
. We can rewrite this as 
. We already established that 
must be odd, so then 
must also be odd. If we then add 
to an odd number, we still get an odd number. So we can eliminate this choice as well.
Now, let's look at the choice 
. Let's factor this as 
. We know that n must be odd, and we know that 
must be odd. Therefore, 
is odd, because multiplying two odd numbers gives us an odd number.
Finally, let's analyze 
. We can rewrite this as 
. Since n is odd, 
must be an even number. When we multiply an even number by an even number, we get an even number, so 
must be even, and it cannot be odd.
The answer is 
.
Let's examine the choice 
Next, let's look at 
Now, let's look at the choice 
Finally, let's analyze 
The answer is 
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If n and m are both positive even integers, which of the following must be odd?
I. (n + 1)(m + 1)
II. nm + 1
III. nm + m
If n and m are both positive even integers, which of the following must be odd?
I. (n + 1)(m + 1)
II. nm + 1
III. nm + m
Tap to reveal answer
Let us analyze I, II, and III one at a time.
Because n and m are both even, if we increase either by 1, the result will be an odd number. Thus, n + 1 and m + 1 are both odd. When two odd numbers are multiplied together, the result is always an odd number. Thus (n + 1)(m + 1) must be an odd number.
Because n and m are even, when we multiply two even numbers together, we always get an even number. Thus nm is even. However, when we then add one to an even number, the result will be an odd number. Thus, nm + 1 is odd.
We just established that nm is even. If we subtract an even number from an even number, the result is always even. Thus, nm – m is an even number.
Only choice I and II will always produce odd numbers.
The answer is I and II only.
Let us analyze I, II, and III one at a time.
Because n and m are both even, if we increase either by 1, the result will be an odd number. Thus, n + 1 and m + 1 are both odd. When two odd numbers are multiplied together, the result is always an odd number. Thus (n + 1)(m + 1) must be an odd number.
Because n and m are even, when we multiply two even numbers together, we always get an even number. Thus nm is even. However, when we then add one to an even number, the result will be an odd number. Thus, nm + 1 is odd.
We just established that nm is even. If we subtract an even number from an even number, the result is always even. Thus, nm – m is an even number.
Only choice I and II will always produce odd numbers.
The answer is I and II only.
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odd * odd * odd =
odd * odd * odd =
Tap to reveal answer
The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example.
Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.
even * even = even (2 * 2 = 4)
even * odd = even (2 * 3 = 6)
odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.
even * even * even = even * even = even (2 * 2 * 2 = 8)
odd * odd * even = odd * even = even (1 * 3 * 2 = 6)
The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example.
Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.
even * even = even (2 * 2 = 4)
even * odd = even (2 * 3 = 6)
odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.
even * even * even = even * even = even (2 * 2 * 2 = 8)
odd * odd * even = odd * even = even (1 * 3 * 2 = 6)
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When $2013^{2013}$ is evaluated, what number is in the ones digit?
When $2013^{2013}$ is evaluated, what number is in the ones digit?
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Let us say we have two numbers, dpi{100} x and dpi{100} y whose ones digits are dpi{100} A and B, respectively. If we want to know the ones digits of the product of dpi{100} x and dpi{100} y, all we need to do is to look at the ones digit of the product of dpi{100} A and dpi{100} B. For example, if we multiply 137 and 219, then the ones digit will be the same as the ones digit of dpi{100} 7 imes 9=63. Since the ones digit of 63 is 3, the ones digit of 137 x 219 will also be 3. In short, we really only need to worry about the ones digits of the numbers we multiply when we try to find the ones digit of their product.
We want to find the ones digit of $2013^{2013}$. An exponent is essentially just a short hand for repeated multiplication. Let us look at the ones digit of the first few exponents of 2013.
$2013^1$ = 2013 - the ones digit is 3.
To find the ones digit of the 2013 to the second power, we need to think of it as the product of 2013 and 2013. As discussed previously, if we want the ones digit of two numbers multiplied together, we just need to multiply their ones digits. Thus, if we multiply 2013 by 2013, then the ones digit will be the same as dpi{100} 3 imes 3=9.
$2013^2$ = 2013 cdot 2013 - ones digit is 9.
Next, we want to find the ones digit of 2013 to the third power. In order to do this, we will multiply the square of 2013 by 2013. It does not matter that we do not know exactly what 2013 squared equals, beacuse we only need to worry about the ones digit, which is 9. In other words, 2013 to the third power will have a ones digit that is equal to the ones digit of the product of 9 (which was the ones digit of 2013 squared) and 3 (which is the ones digit of 2013). When we multiply 9 and 3, we get 27, so the ones digit of 2013 to the third power is 7.
$2013^3$ = $2013^2$ cdot 2013 - ones digit of 7.
To find the ones digit of 2013 to the fourth power, we only need to worry about multiplying the ones digit of 2013 to the third power (which is 7) by the ones digit of 2013. When we mulitply 7 and 3, we get 21, which means that the ones digit of 2013 to the fourth power is 1.
$2013^4$ = $2013^3$ cdot 2013 - ones digit of 1.
To find the ones digit of 2013 to the fifth power, we will multiply 1 by 3, which gives us 3.
$2013^5$ = $2013^4$ cdot 2013 - ones digit of 3.
Notice that we are back to a ones digit with 3. If we multiply this by 2013, we will end up with a ones digit of 9. In other words, the ones digits repeat every fourth power.
The value of the ones digits of the powers of 2013 is as follows (starting with 2013 to the first power):
3, 9, 7, 1, 3, 9, 7, 1,....
We essentially want to find the 2013th term of the sequence above. Notice that every fourth term is 1, i.e. the sequence repeats every four terms. If a terms position in the sequence is a multiple of 4, then the term will be 1. In short, the 4th, 8th, 12th, 16th terms, and so on, will be 1. Because 2012 is a multiple of 4, the 2012th term in the sequence will be 1. (We can determine if a number is a multiple of 4 by looking at its last two digits.) This means that that 2013th term will be 3. Thus, 2013 to the power of 2013 has a ones digit of 3.
The answer is 3.
Let us say we have two numbers, dpi{100} x and dpi{100} y whose ones digits are dpi{100} A and B, respectively. If we want to know the ones digits of the product of dpi{100} x and dpi{100} y, all we need to do is to look at the ones digit of the product of dpi{100} A and dpi{100} B. For example, if we multiply 137 and 219, then the ones digit will be the same as the ones digit of dpi{100} 7 imes 9=63. Since the ones digit of 63 is 3, the ones digit of 137 x 219 will also be 3. In short, we really only need to worry about the ones digits of the numbers we multiply when we try to find the ones digit of their product.
We want to find the ones digit of $2013^{2013}$. An exponent is essentially just a short hand for repeated multiplication. Let us look at the ones digit of the first few exponents of 2013.
$2013^1$ = 2013 - the ones digit is 3.
To find the ones digit of the 2013 to the second power, we need to think of it as the product of 2013 and 2013. As discussed previously, if we want the ones digit of two numbers multiplied together, we just need to multiply their ones digits. Thus, if we multiply 2013 by 2013, then the ones digit will be the same as dpi{100} 3 imes 3=9.
$2013^2$ = 2013 cdot 2013 - ones digit is 9.
Next, we want to find the ones digit of 2013 to the third power. In order to do this, we will multiply the square of 2013 by 2013. It does not matter that we do not know exactly what 2013 squared equals, beacuse we only need to worry about the ones digit, which is 9. In other words, 2013 to the third power will have a ones digit that is equal to the ones digit of the product of 9 (which was the ones digit of 2013 squared) and 3 (which is the ones digit of 2013). When we multiply 9 and 3, we get 27, so the ones digit of 2013 to the third power is 7.
$2013^3$ = $2013^2$ cdot 2013 - ones digit of 7.
To find the ones digit of 2013 to the fourth power, we only need to worry about multiplying the ones digit of 2013 to the third power (which is 7) by the ones digit of 2013. When we mulitply 7 and 3, we get 21, which means that the ones digit of 2013 to the fourth power is 1.
$2013^4$ = $2013^3$ cdot 2013 - ones digit of 1.
To find the ones digit of 2013 to the fifth power, we will multiply 1 by 3, which gives us 3.
$2013^5$ = $2013^4$ cdot 2013 - ones digit of 3.
Notice that we are back to a ones digit with 3. If we multiply this by 2013, we will end up with a ones digit of 9. In other words, the ones digits repeat every fourth power.
The value of the ones digits of the powers of 2013 is as follows (starting with 2013 to the first power):
3, 9, 7, 1, 3, 9, 7, 1,....
We essentially want to find the 2013th term of the sequence above. Notice that every fourth term is 1, i.e. the sequence repeats every four terms. If a terms position in the sequence is a multiple of 4, then the term will be 1. In short, the 4th, 8th, 12th, 16th terms, and so on, will be 1. Because 2012 is a multiple of 4, the 2012th term in the sequence will be 1. (We can determine if a number is a multiple of 4 by looking at its last two digits.) This means that that 2013th term will be 3. Thus, 2013 to the power of 2013 has a ones digit of 3.
The answer is 3.
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odd * odd * odd =
odd * odd * odd =
Tap to reveal answer
The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example.
Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.
even * even = even (2 * 2 = 4)
even * odd = even (2 * 3 = 6)
odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.
even * even * even = even * even = even (2 * 2 * 2 = 8)
odd * odd * even = odd * even = even (1 * 3 * 2 = 6)
The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example.
Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.
even * even = even (2 * 2 = 4)
even * odd = even (2 * 3 = 6)
odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.
even * even * even = even * even = even (2 * 2 * 2 = 8)
odd * odd * even = odd * even = even (1 * 3 * 2 = 6)
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If 
and 
are both odd integers, which of the following is not necessarily odd?
If
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With many questions like this, it might be easier to plug in numbers rather than dealing with theoretical variables. However, given that this question asks for the expression that is not always even or odd but only not necessarily odd, the theoretical route might be our only choice.
Therefore, our best approach is to simply analyze each answer choice.
: Since 
is odd, 
is also odd, since and odd number multiplied by an odd number yields an odd product. Since 
is also odd, multiplying it by 
will again yield an odd product, so this expression is always odd.
: Since 
is odd, multiplying it by 2 will yield an even number. Subtracting this number from 
will also give an odd result, since an odd number minus an even number gives an odd number. Therefore, this answer is also always odd.
: Since both numbers are odd, their product will also always be odd.
: Since 
is odd, multiplying it by 2 will give an even number. Since 
is odd, subtracting it from our even number will give an odd number, since an even number minus and odd number is always odd. Therefore, this answer will always be odd.
: Since both numbers are odd, there sum will be even. However, dividing an even number by another even number (2 in our case) does not always produce an even or an odd number. For example, 5 and 7 are both odd. Their sum, 12, is even. Dividing by 2 gives 6, an even number. However, 5 and 9 are also both odd. Their sum, 14, is even, but dividing by 2 gives 7, an odd number. Therefore, this expression isn't necessarily always odd or always even, and is therefore our answer.
With many questions like this, it might be easier to plug in numbers rather than dealing with theoretical variables. However, given that this question asks for the expression that is not always even or odd but only not necessarily odd, the theoretical route might be our only choice.
Therefore, our best approach is to simply analyze each answer choice.
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Theodore has 
jelly beans. Portia has three times that amount. Harvey has five times as many as she does. What is the total count of jelly beans in the whole group?
Theodore has
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To find the answer to this question, calculate the total jelly beans for each person:
Portia: 
* <Theodore's count of jelly beans>, which is 
or 
Harvey: 
* <Portia's count of jelly beans>, which is 
or 
So, the total is:
(Do not forget that you need those original 
for Theodore!)
To find the answer to this question, calculate the total jelly beans for each person:
Portia:
Harvey:
So, the total is:
(Do not forget that you need those original
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You are given that 
, 
, and 
are positive integers, and
In which of the following cases is 
odd?
I) Exactly one of 
is odd.
II) Exactly two of 
are odd.
III) Exactly three of 
are odd.
You are given that
In which of the following cases is
I) Exactly one of
II) Exactly two of
III) Exactly three of
Tap to reveal answer
For the product of three integers to be odd, all three integers must themselves be odd.
At least two of 
must have the same odd/even status. The sum of those two numbers must be even, and since it is a factor of 
, then 
itself must be even.
For the product of three integers to be odd, all three integers must themselves be odd.
At least two of
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