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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1053
GPA: 3.39
If n is a positive integer and the product of all integers from 1 to n
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24 Feb 2023, 08:36
24 Feb 2023, 08:36
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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
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1053
GPA: 3.39
Re: If n is a positive integer and the product of all integers from 1 to n
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27 Feb 2023, 02:15
27 Feb 2023, 02:15
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Expert Reply
To solve this problem, we need to consider the prime factorization of 990, which is 2 × 3^2 × 5 × 11. The product of all integers from 1 to n can be written as n!, the factorial of n. To ensure that n! is a multiple of 990, n! must have at least one factor of 2, two factors of 3, one factor of 5, and one factor of 11.
The smallest integer that satisfies these conditions is 11, because:
The factors of 2 in 11! are: 2, 4, 6, 8, and 10.
The factors of 3 in 11! are: 3, 6, and 9.
The factors of 5 in 11! are: 5 and 10.
The factors of 11 in 11! is: 11.
Thus, the least possible value of n is 11, and the answer is (B).
The smallest integer that satisfies these conditions is 11, because:
The factors of 2 in 11! are: 2, 4, 6, 8, and 10.
The factors of 3 in 11! are: 3, 6, and 9.
The factors of 5 in 11! are: 5 and 10.
The factors of 11 in 11! is: 11.
Thus, the least possible value of n is 11, and the answer is (B).
gmatclubot
Re: If n is a positive integer and the product of all integers from 1 to n [#permalink]
27 Feb 2023, 02:15
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