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Product of consequtive integers problem

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Product of consequtive integers problem

by chuang2 » Sun Oct 19, 2008 10:50 am
Question:

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

I know this should be a easy question but could someone walk me through the work process to figuring this out?

Any help would be greatly appreciated, thanks!
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Re: Product of consequtive integers problem

by rippersid » Mon Oct 20, 2008 12:17 am
Is the answer 11?
chuang2 wrote:Question:

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

I know this should be a easy question but could someone walk me through the work process to figuring this out?

Any help would be greatly appreciated, thanks!
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by chuang2 » Mon Oct 20, 2008 9:16 am
yup that is the correct answer. How did you get to that?
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by rippersid » Mon Oct 20, 2008 12:46 pm
I will try my best to explain in the simplest manner possible. So if you don't get something, just hit me back and I will make things simpler. So here is how you do this problem.
This is a problem based on the definition of a factorial.
A factorial is defined as N!= N*(N-1)*(N-2)*(N-3)..............3*2*1
This is what has been given to you when they say N is a positive integer and a product of all integers from 1 to N.
Now this N is also a multiple of 990. So if you divide this N by 990, you will get an integer.
The trick in this question lies in the fact that 990 can be written as 11*10*9.
Thus, you know, by the definition for factorial above, that N has to be atleast 11 so that N!/990 becomes an integer.
Case in point, if N were only 10, you'd still have a 11( 11*10*9, where the 10 and 9 would be common in both the numerator and denominator) in the denominator, and it would not factor out completely, and thus would not be an integer.
So the minimum value of N you need is 11.

Let me know if you need any further clarifications.
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by GMATCHPOINT » Mon Oct 20, 2008 3:54 pm
rippersid,
Would you rephrase this part pls? I understood the rest, but not from this point. (see below)
"....Thus, you know, by the definition for factorial above, that N has to be atleast 11 so that N!/990 becomes an integer.
Case in point, if N were only 10, you'd still have a 11( 11*10*9, where the 10 and 9 would be common in both the numerator and denominator) in the denominator, and it would not factor out completely, and thus would not be an integer.
So the minimum value of N you need is 11...."

thank you!
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by rippersid » Mon Oct 20, 2008 7:39 pm
Here goes.
N! = N*(N-1)*(N-2)..........3*2*1
This N, as given by the problem is a multiple of 990.
Say N=12.
Then 12! = 12*11*10*9*8*7*6*5*4*3*2*1. As you can see, this is also a multiple of 990 because 990 = 11*10*9.
Now let N = 11.
11! = 11*10*9*8*7*6*5*4*3*2*1 which is the LEAST possible value of N such that it is a multiple of 990 (11*10*9)
Now say N = 10
10! = 10*9*8*7*6*5*4*3*2*1
Now this is not a multiple of 990 because you are missing a prime factor, namely 11.
In other words, anything less than N=11, will not be a multiple of 990 because you will not have 11 as one of the factors for the number. (Try factoring 990, and you will realize that 11, is one of the prime factors required).

Since the questions asks for the least value of N, N = 11 satisfies this. All values of N! for N>11 will always be a multiple of 990.

I apologize if my previous post did not make sense.

I hope that this clears things up.
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by GMATCHPOINT » Tue Oct 21, 2008 1:20 pm
now is perfect! tks!
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by rippersid » Tue Oct 21, 2008 11:42 pm
The pleasure is mine.
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quicker way....

by MrGreedy » Wed Dec 10, 2008 11:14 am
I went about this question another way....

I broke down 990 into it's prime factors

3 - 990
3 - 330
11- 110
5 - 10
2 - 2


990 = 2 * 3 * 3* 5 * 11

Therefore, the least possible value for N would be 11 i.e. a multiple of 11 from the list
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by niraj_a » Wed Dec 10, 2008 11:18 am
MrGreedy,

small flaw in your statement. by your method, 5 * 2 = 10 would also work, right?
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by MrGreedy » Wed Dec 10, 2008 11:48 am
I see what you mean...but in my statement I assumed the least possible multiple had to contain the greatest prime of 990 such as 11....10 wouldn't work.

!10 = 3 628 800

3 628 800 / 990 = 36653.45454545 recurring

!11 = 39 916 800

39 916 800 / 990 = 40320

But if I'm going about this the wrong way, please let me know..just started getting back into it for the GMAT.
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by MrGreedy » Wed Dec 10, 2008 12:09 pm
I'm also using this question as reference:

https://www.beatthegmat.com/multiples-gm ... 12490.html
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by alexdallas » Sat Jul 18, 2009 9:00 am
tks ripper.
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by uptowngirl92 » Mon Sep 28, 2009 6:48 pm
I am not too sure i got this one..
990=11 x 3 x 3 x 5 x 2..ok..
So since we want the MINIMUM value of n we didide this by the MAXIMUM prime factor which is 11?
Am i understanding it correctly??
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factor first

by artistocrat » Sat Oct 17, 2009 3:37 pm
In 990 factored you have 3*3*2*2*5*5*11, so you need 2 3's, 2 2's, 2 5's and an 11, all of which are contained in N=11 or =11*(2*5)*(3*3)*(2*2*2)*7*(2*3)*5 and so on.
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