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If n is a positive integer and the product of all integers
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Updated on: 27 Feb 2014, 04:56
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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 I got the right answer 11 by stating: > 990 can be divided by 11
Is is the right way to solve that?
Thanks in advance.
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Originally posted by nailgmattoefl on 28 Dec 2009, 11:47.
Last edited by Bunuel on 27 Feb 2014, 04:56, edited 2 times in total.
Edited the question and added the OA




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Re: Math Questions
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28 Dec 2009, 12:22
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 We are told that \(n!=990*k=2*5*3^2*11*k\) > \(n!=2*5*3^2*11*k\) which means that \(n!\) must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of \(n\) is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger).. Answer: B. Hope it's clear.
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Re: Multiple of 990
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18 Sep 2010, 15:18
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.




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Re: GMAT Prep q#11 need help
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27 Feb 2010, 21:50
divanshuj wrote: 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 for n?
a) 10 b) 11 c) 12 d) 13 e) 14 990 = 2 * 5 * 3 * 3* 11 so product from 1 to n should have all the above numbers. n = 11 will have all the above numbers as factors of 1 * 2* ..... * 11 B



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Re: n is a positive int and the product of all ints from 1 to n
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02 Aug 2010, 00:00
Hi,
The question says that 990 is a factor of n!. 990 = 9*110 = 9*11*10 For 990 to be a factor of n!, n should at least be 11 so that it includes the common factor 11 that is found in 990.
regards, Jack



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Re: n is a positive int and the product of all ints from 1 to n
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02 Aug 2010, 07:18
990= 2x3x3x5x11 So 11 has to be one of the prime factor So 11!
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Re: Multiple of 990
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18 Sep 2010, 15:59
The least possible value of such a n will be at least the highest prime factor of 990, which is 11 The only other thing to check is if the rest of the factors of 990, can be obtained by multiplication of numbers less than 11. Otherwise n may be greater than 11 But that is easy to see since 990=9*10*11. So taking all the numbers below 11 will suffice.
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If n is a positive integer and the product of all integers
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13 Feb 2011, 18:53



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Re: If n is a positive integer and the product of all the integers from 1
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24 Aug 2011, 08:13
You need to primarily find the primes of 990: 2,3,3,5,11 The integer must contain all these primes By knowing 11 is an prime, we know the answer can be B,C,D or E as these all contain 11. 11=1,2,3,4,5,6,7,8,9,10,11 11 contains the 2,3,5 and 11 explicitly and the second 3 comes from the 6 (2*3), therefore the answer is 11. This is because we know a number that contains all the primes of 990 will be a multiple of 990. Hope this helps
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Re: If n is a positive integer and the product of all the integers from 1
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24 Aug 2011, 09:01
Here's my understanding of the term of multiple. If a number Y can be divided into another number X without a remainder resulting it is considered a factor of that number. Likewise, if a number X can be divided by another number Y cleanly then X is a considered a multiple of Y. Now this is what I don't understand. The question reads: 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 the least possible value of n? So when I first saw this I interpreted it as (1*2*3........*n)/990 should end up without a remainder. I interpreted it this way because the question read "the product of all integers from 1 to n, inclusive." Then the question asked for what the smallest value of n could be, which confused me. You wrote: Quote: You need to primarily find the primes of 990: 2,3,3,5,11 The integer must contain all these primes I follow this. Quote: By knowing 11 is an prime, we know the answer can be B,C,D or E as these all contain 990.
11=1,2,3,4,5,6,7,8,9,10,11
11 contains the 2,3,5 and 11 explicitly and the second 3 comes from the 6 (2*3), therefore the answer is 11.
This is because we know a number that contains all the primes of 990 will be a multiple of 990. This is where I get lost. I understand that the series of 1 to 11 contains all the prime numbers that are factors of 990. But how did you reach the conclusion that the answer is 11? I guess what I don't understand is how you were able to interpret the question the way you did? Is it just from experience of lots of practice problems? Or in the worst case scenario where I don't understand the question and just memorize the question format, I have two follow questions: Lets say the question read: Quote: If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 2890, what is the least possible value of n? Would the answer in this situation be 17 in this case (assuming it was an option) Or alternatively if the question read: Quote: If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 1960, what is the least possible value of n? How would you solve that?



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Re: If n is a positive integer and the product of all the integers from 1
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24 Aug 2011, 09:29
The answer would be the number which contains all the factors of the figure. The series of 1 to 11 contained all the factors of 990. 2,3,3,5 and 11. Because 990 is made up from these 5 numbers such that 2*3*3*5*11 = 990 If you times the other numbers (1,4,6,7,8,9,10) in the series you will get a multiple of 990 from this. I picked 11 because it asked for the least possible value of n. Picking 12,13,14 would result in a multiple of 990 but the number is more than the least possible value of 11. The best way to interpret problems is through practice, you will come to understand questions better over time. Always remember GMAT questions are simple questions made to look complicated.
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Re: If n is a positive integer and the product of all the integers from 1
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24 Aug 2011, 09:50
just entered this in google and you're right! (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11) / 990 = 40 320
meanwhile (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10) / 990 = 3 665.45455 wouldn't work
It makes sense because the highest prime factor of 990 must also be a factor of any multiple of 990.
I guess I need to do more practice problems involving multiples and number properties.



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Re: If n is a positive integer and the product of all the integers from 1
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24 Aug 2011, 11:23
990= 3*11*5*6 N must be 11 to satisfy the condition. Hence B is the answer.
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Re: If n is a positive integer and the product of all the integers from 1
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02 Sep 2011, 00:00
Using trial and error method clearly we need to see how we can calculate 990 by using consecutive integers. Clearly for 990 a prime factor is 11 so we must need it. 9*10*11=990 and multiplying it with other +ve integers will result in a # which is always a multiple of 11.
So 11 is the answer.In this way the problem can be solved quickly. Let me know if this helps



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Re: If n is a positive integer and the product of all the integers from 1
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02 Sep 2011, 00:57
prime factorization of 990=3*3*11*2*5
As the product of 1 through n inclusive is a multiple of 990, the least number that n could be is 11.



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Re: If n is a positive integer and the product of all the integers from 1
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02 Sep 2011, 09:24
990 = 2*3^2*5*11
1*2*.....n = multiple of 990
=> n*n1*.....1 = multiple of 990
=> n! = multiple of 990
= 2*3^2*5*11 * p (some multiple p)
and we have to find a minimum value for n that satisfies the above condition.
which tells us n! needs to have all prime factors of 990 .i.e 2,3,5,11 prime factors.
=> minimum value of n is 11.



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Re: If n is a positive integer and the product of all the integers from 1
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03 Sep 2011, 11:11
Since we need to find the minimum value of n, So the greatest prime factor of 990 will be the answer. 990 = 2 * 3^2 * 5 * 11. Hence the minimum value of n such that n! ( 1*2* ... *n) is divisible by 990 is 11.
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Re: If n is a positive integer and the product of all the integers from 1
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06 Sep 2011, 13:31
SunnyDelight wrote: Or alternatively if the question read: Quote: If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 1960, what is the least possible value of n? How would you solve that? 1960 is divisible by 10 ( or 2*5) = 196 196 is divisible by 4 = > 49 49 = 7 *7 hence 1960 is divisible by 2,4,5, 7 ,7 if there were only one 7, i.e if the problem were 280 then the least possible value that is a multiple would be 7 ( since the value is 1*2*3*4*5*6*7) i.e 280 * 3*6 however since we have 2 7's the second 7 can only be used in the digit 14.. hence the minimum value of n required will be 14 hope it helps



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Re: Math Questions
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11 Apr 2012, 02:46
siddhans wrote: Bunuel wrote: 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
We are told that \(n!=990*k=2*5*3^2*11*k\) > \(n!=2*5*3^2*11*k\) which means that \(n!\) must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of \(n\) is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger)..
Answer: B.
Hope it's clear. Why is this k introduced? Whats the concept of introducing k ?? n ! = 990 * k ?? I am confused... We are told that n! is a multiple of 990, which can be expressed as n!=990*k, for some positive integer multiple k.
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Re: Math Questions
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27 Oct 2012, 12:04
Bunuel wrote: 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
We are told that \(n!=990*k=2*5*3^2*11*k\) > \(n!=2*5*3^2*11*k\) which means that \(n!\) must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of \(n\) is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger)..
Answer: B.
Hope it's clear. out of curiousness, i would like to clarify something here. In case if we have a option like (i know option are pathetic) A. 2 B. 5 C. 11 D. 13 E. any option.. then what would be the answer.. BTW when i try to solve this question i was only having the question(without option) , I took the prime factor stuff & chosen the least one from that.. please help me here to understand more.




Re: Math Questions
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