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Updated on: 07 Jul 2019, 23:23
Originally posted by nailgmattoefl on 28 Dec 2009, 11:47.
Last edited by Bunuel on 07 Jul 2019, 23:23, edited 4 times in total.
Last edited by Bunuel on 07 Jul 2019, 23:23, edited 4 times in total.
Edited the question and added the OA
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
82% (01:17) correct
18%
(01:29)
wrong
based on 7138
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If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A. 10
B. 11
C. 12
D. 13
E. 14
A. 10
B. 11
C. 12
D. 13
E. 14
Show SpoilerThis question has different answer choices in OG 2020
A. 8
B. 9
C. 10
D. 11
E. 12
B. 9
C. 10
D. 11
E. 12
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If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A. 10
B. 11
C. 12
D. 13
E. 14
Given:
\(n!=990*k=2*5*3^2*11*k\)
\(n!=2*5*3^2*11*k\)
This means that \(n!\) must have all factors of 990 to be a multiple of 990, hence it must also have 11. So the least value of \(n\) is 11 (notice that 11! will include all the other factors of 990 as well, otherwise the least value of \(n\) would have been larger).
Answer: B.
Hope it's clear.
A. 10
B. 11
C. 12
D. 13
E. 14
Given:
\(n!=990*k=2*5*3^2*11*k\)
\(n!=2*5*3^2*11*k\)
This means that \(n!\) must have all factors of 990 to be a multiple of 990, hence it must also have 11. So the least value of \(n\) is 11 (notice that 11! will include all the other factors of 990 as well, otherwise the least value of \(n\) would have been larger).
Answer: B.
Hope it's clear.
18 Sep 2010, 15:18
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This question tests the concept of prime factorization. In order to figure out if 990 can be a divisor of the unknown product you need to determine all of the prime factors that combine to successfully divide 990 without leaving a remainder.
The prime factors of 990 are 2, 3, 3, 5, 11
Since 11 is a prime factor of 990, you need to have the number present somewhere in the unknown product. Answer A is the product of integers from 1 to 10 and therefore does not meet the necessary criteria. Answer B includes 11 and therefore is the answer.
The prime factors of 990 are 2, 3, 3, 5, 11
Since 11 is a prime factor of 990, you need to have the number present somewhere in the unknown product. Answer A is the product of integers from 1 to 10 and therefore does not meet the necessary criteria. Answer B includes 11 and therefore is the answer.
