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Re: A set of 6 consecutive integers has a product Q [#permalink]
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Let's say that the 6 consecutive integers are x, x + 1, x + 2, x + 3, x + 4, and x + 5. We are told that their product is equal to Q. So,
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = Q

We can check each option individually to find out which ones could be true:

Option A):
The product of these 6 consecutive integers can be greater than 0 when all of the integers are also greater than 0. If x > 0, then:
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = Q > 0

Therefore, Option A) could be true.

Option B):
Similarly, the product of these 6 consecutive integers can be equal to 0 when any one of the consecutive integers is equal to 0. If x = 0, then:
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = Q = 0

Therefore, Option B) could be true.

Option C):
When some of the consecutive integers are less than 0, then the product of them can not be negative. For example, if x = -3, then the consecutive integers will be -3, -2, -1, 0, 1, and 2. Their product will be:
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = (-3) * (-2) * (-1) * (0) * (1) * (2)

(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = 0

That's why, the Option C) can not be true.

Option D):
The mean of the set of these 6 consecutive integers x, x + 1, x + 2, x + 3, x + 4, and x + 5 will be:
Mean = \(\frac{(x) + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5)}{6}\)

Mean = \(\frac{(x + x + x + x + x + x) + (1 + 2 + 3 + 4 + 5)}{6}\)

Mean = \(\frac{(6x) + (15)}{6}\)

Mean = \(\frac{3 * (2x + 5)}{2 * 3}\)

Mean = \(\frac{(2x + 5)}{2}\)

The median of the set of these 6 consecutive integers x, x + 1, x + 2, x + 3, x + 4, and x + 5 will be:
Median = \(\frac{(x + 2) + (x + 3)}{2}\)

Median = \(\frac{(2x + 5)}{2}\)

Mean = Median

It means that the Option D) is true as well.

Option E):
Finally, we can see that the mean will be less than 0 when most of the consecutive integers are also less than 0. For example, if x = -3, then:
Mean} = \(\frac{(2 * (-3) + 5)}{2}\)

Mean = \(\frac{(-6 + 5)}{2}\)

Mean = \(\frac{(-1)}{2}\)

That's why, the Option E) can not be true.

Hence, the correct answers are Option A), Option B), and Option D).
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Re: A set of 6 consecutive integers has a product Q [#permalink]
RA911 wrote:
Let's say that the 6 consecutive integers are x, x + 1, x + 2, x + 3, x + 4, and x + 5. We are told that their product is equal to Q. So,
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = Q

We can check each option individually to find out which ones could be true:

Option A):
The product of these 6 consecutive integers can be greater than 0 when all of the integers are also greater than 0. If x > 0, then:
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = Q > 0

Therefore, Option A) could be true.

Option B):
Similarly, the product of these 6 consecutive integers can be equal to 0 when any one of the consecutive integers is equal to 0. If x = 0, then:
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = Q = 0

Therefore, Option B) could be true.

Option C):
When some of the consecutive integers are less than 0, then the product of them can not be negative. For example, if x = -3, then the consecutive integers will be -3, -2, -1, 0, 1, and 2. Their product will be:
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = (-3) * (-2) * (-1) * (0) * (1) * (2)

(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = 0

That's why, the Option C) can not be true.

Option D):
The mean of the set of these 6 consecutive integers x, x + 1, x + 2, x + 3, x + 4, and x + 5 will be:
Mean = \(\frac{(x) + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5)}{6}\)

Mean = \(\frac{(x + x + x + x + x + x) + (1 + 2 + 3 + 4 + 5)}{6}\)

Mean = \(\frac{(6x) + (15)}{6}\)

Mean = \(\frac{3 * (2x + 5)}{2 * 3}\)

Mean = \(\frac{(2x + 5)}{2}\)

The median of the set of these 6 consecutive integers x, x + 1, x + 2, x + 3, x + 4, and x + 5 will be:
Median = \(\frac{(x + 2) + (x + 3)}{2}\)

Median = \(\frac{(2x + 5)}{2}\)

Mean = Median

It means that the Option D) is true as well.

Option E):
Finally, we can see that the mean will be less than 0 when most of the consecutive integers are also less than 0. For example, if x = -3, then:
Mean} = \(\frac{(2 * (-3) + 5)}{2}\)

Mean = \(\frac{(-6 + 5)}{2}\)

Mean = \(\frac{(-1)}{2}\)

That's why, the Option E) can not be true.

Hence, the correct answers are Option A), Option B), and Option D).


Hi, cannot option E be true ? for example 1,2,3,4,5,6 have a mean of 3.5 which is positive.
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