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In the series of all consecutive integers from 1 to n inclusive-
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13 Mar 2021, 20:20
13 Mar 2021, 20:20
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93% (01:31) correct
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In the series of all consecutive integers from 1 to n inclusive, the product of all the numbers in the series is divisible by 990.
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
n |
9 |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
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Re: In the series of all consecutive integers from 1 to n inclusive-
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13 Mar 2021, 21:51
13 Mar 2021, 21:51
1
swiikiller wrote:
In the series of all consecutive integers from 1 to n inclusive, the product of all the numbers in the series is divisible by 990.
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
n |
9 |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
\(990 = (9)(11)(10) = (2)(3^2)(5)(11)\)
Since \(n!\) is divisible by \(990\), it must have all the factors of \(990\)
So, \(n\) must have \(11\)
Col. A: 11
Col. B: 9
Hence, option A
In the series of all consecutive integers from 1 to n inclusive-
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15 Mar 2021, 13:43
15 Mar 2021, 13:43
1
990 is divisible by 11, a prime number. Thus, any series that is divisible by 990 must include 11, so n>9. Alternatively, you can multiply the numbers in the series and then divide by 990. Since the resulting number is a fraction, you can conclude that n>9.
A is the correct answer
A is the correct answer
