I'd like to check if I understand how to solve below problem in bold. I believe you need to factor 990 which gives you:
990
10 * 99
2 * 5 * 11 * 9
2 * 5 * 11 * 3 * 3
So the answer must be 11. Is this correct?
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?
Answers:
10
11
12
13
14
Multiples Problem
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Hi melanie.espeland,
Yes, you are correct! 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
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Rich
Yes, you are correct! 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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A lot of integer property questions can be solved using prime factorization.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
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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Given the below methods, I wouldn't rehash and would only suggest thatmelanie.espeland wrote:I'd like to check if I understand how to solve below problem in bold. I believe you need to factor 990 which gives you:
990
10 * 99
2 * 5 * 11 * 9
2 * 5 * 11 * 3 * 3
So the answer must be 11. Is this correct?
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?
Answers:
10
11
12
13
14
1) Every such question requires the prime factorization (preferrably to identify the biggest prime available among factors of the number)
2) Such questions are easier to attempt with options
3) One should start with the smallest option
4) The combination of Factorization and checking option would be ideal way to attempt such question
e.g.
Step-1: 990 = 2 x 5 x 9 x 11 (Hence biggest Prime is 11)
Step-2: Look at options and realize that answer must have 11 in the range
So straight Option B
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Hi All ,
In a question we are asked what is the least possible value of n?
so how come 11 is Answer?
Please advise.
Thanks ,
Shreyans
In a question we are asked what is the least possible value of n?
so how come 11 is Answer?
Please advise.
Thanks ,
Shreyans
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The product of all of the possible integers between 1 and n, inclusive, is n!.j_shreyans wrote:Hi All ,
In a question we are asked what is the least possible value of n?
so how come 11 is Answer?
Please advise.
Thanks ,
Shreyans
We can PLUG IN THE ANSWERS, which represent the least possible value of n such that n! is divisible by 990.
Since the question stem asks for the LEAST POSSIBLE VALUE of n, start with the smallest answer choice.
When the correct answer choice is plugged in, n!/990 = integer.
Answer choice A: 10
10!/990 = (10*9*8*7*6*5*4*3*2)/(9*10*11) = (8*7*6*5*4*3*2)/11.
Doesn't work.
Eliminate A.
Answer choice B: 11
11!/990 = (11*10*9*8*7*6*5*4*3*2)/(9*10*11) = 8*7*6*5*4*3*2.
Success!
The correct answer is B.
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As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Hi j_shreyans,
This series of posts provides a nice explanation of the "math" involved and the steps necessary to answer the question. Was there anything about it that you specifically want to know more about?
GMAT assassins aren't born, they're made,
Rich
This series of posts provides a nice explanation of the "math" involved and the steps necessary to answer the question. Was there anything about it that you specifically want to know more about?
GMAT assassins aren't born, they're made,
Rich
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