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If n is a positive integer, and the product of all integers

This topic has 3 expert replies and 1 member reply
lucas211 Senior | Next Rank: 100 Posts Default Avatar
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If n is a positive integer, and the product of all integers

Post Wed May 25, 2016 9:51 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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

    Thanks in advance Smile

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    Post Wed May 25, 2016 10:41 am
    Quote:
    If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990. What is least possible value of n?
    a) 10
    b) 11
    c) 12
    d) 13
    e) 14
    A lot of integer property questions can be solved using prime factorization.
    For questions involving divisibility, divisors, factors and multiples, we can say:
    If N is divisible by k, then k is "hiding" within the prime factorization of N
    Similarly, we can say:
    If N is is a multiple of k, then k is "hiding" within the prime factorization of N

    Examples:
    24 is divisible by 3 <--> 24 = 2x2x2x3
    70 is divisible by 5 <--> 70 = 2x5x7
    330 is divisible by 6 <--> 330 = 2x3x5x11
    56 is divisible by 8 <--> 56 = 2x2x2x7

    So, if if some number is a multiple of 990, then 990 is hiding in the prime factorization of that number.

    Since 990 = (2)(3)(3)(5)(11), we know that one 2, two 3s, one 5 and one 11 must be hiding in the prime factorization of our number.

    For 11 to appear in the product of all the integers from 1 to n, n must equal 11 or more.
    So, the answer is B

    Cheers,
    Brent

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    Thanked by: lucas211
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    Post Wed May 25, 2016 6:14 pm
    Hi lucas211,

    For a number to be a multiple of 990, that number must have the exact same prime factors (including duplicates) as 990 (but may have "extra" prime factors as well).

    This prompt adds the extra stipulation that N has to be as SMALL as possible, so after factoring 990, we need to find the smallest product that "holds" the 2, 5, 11, and two 3s that make up 990.

    1(2)(3)(4)....(11) is the smallest product that does that, so N = 11.

    Final Answer: B

    GMAT assassins aren't born, they're made,
    Rich

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    Thanked by: lucas211

    GMAT/MBA Expert

    Post Thu May 26, 2016 2:44 pm
    If we've got n! = 9 * 10 * 11, we need n! to contain 9, 10, and 11 somewhere in its prime factorization.

    Since 11 can only appear in multiples of 11, we need a multiple of 11. The smallest positive one is 11 itself, so n ≥ 11, and the least possible n is 11 itself.

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    shwetagoyal121 Newbie | Next Rank: 10 Posts
    Joined
    09 May 2016
    Posted:
    3 messages
    Post Sun May 29, 2016 11:27 pm
    Brent@GMATPrepNow wrote:
    Quote:
    If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990. What is least possible value of n?
    a) 10
    b) 11
    c) 12
    d) 13
    e) 14
    A lot of integer property questions can be solved using prime factorization.
    For questions involving divisibility, divisors, factors and multiples, we can say:
    If N is divisible by k, then k is "hiding" within the prime factorization of N
    Similarly, we can say:
    If N is is a multiple of k, then k is "hiding" within the prime factorization of N

    Examples:
    24 is divisible by 3 <--> 24 = 2x2x2x3
    70 is divisible by 5 <--> 70 = 2x5x7
    330 is divisible by 6 <--> 330 = 2x3x5x11
    56 is divisible by 8 <--> 56 = 2x2x2x7

    So, if if some number is a multiple of 990, then 990 is hiding in the prime factorization of that number.

    Since 990 = (2)(3)(3)(5)(11), we know that one 2, two 3s, one 5 and one 11 must be hiding in the prime factorization of our number.

    For 11 to appear in the product of all the integers from 1 to n, n must equal 11 or more.
    So, the answer is B

    Cheers,
    Brent
    The answer would be 11.

    Look at this question from the prospective options angle as well.

    The first probable answer is 10 ; (because that is the smallest value)

    Product of all the integers till n= 990 x (Some other number say Y)

    Case I:

    Now if n=10, whether the product is a multiple of 990 will depend on whether Y has 11 as a factor or not.

    case II:

    now if n=11 (next smallest value)

    the product has all the factors to make it a multiple of 990 irrespective of the value of Y.

    Hence 11 is the right answer.

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