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Multiples Problem

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Multiples Problem

by me_1234 » Tue Dec 09, 2014 3:04 pm
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
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by [email protected] » Tue Dec 09, 2014 4:39 pm
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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by me_1234 » Tue Dec 09, 2014 5:06 pm
Thank you
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by Brent@GMATPrepNow » Tue Dec 09, 2014 10:17 pm
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

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by GMATinsight » Wed Dec 10, 2014 2:03 am
melanie.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
Given the below methods, I wouldn't rehash and would only suggest that

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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by me_1234 » Wed Dec 10, 2014 9:55 am
Brent@GMATPrepNow and GMATinsight, thank you both.
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by j_shreyans » Sat Feb 21, 2015 3:08 am
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
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by GMATGuruNY » Sat Feb 21, 2015 3:29 am
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
The product of all of the possible integers between 1 and n, inclusive, is n!.

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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by [email protected] » Sat Feb 21, 2015 11:03 am
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?

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by j_shreyans » Sun Feb 22, 2015 1:59 am
Hi All ,

Thanks , if anything will get back to you.

Thanks ,

Shreyans
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