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Math question [#permalink] New post 08 Nov 2006, 10:09
I could figure this out by trying to divide each number by 990, but I am wondering if you guys know a shortcut.

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?

A) 10
B) 11
C) 12
D) 13
E) 14

The answer is B)11

Anyone know a quick method for figuring this out?

Thanks,

Andrew
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 [#permalink] New post 08 Nov 2006, 10:13
Basically, this is n!

990 = 2x3x3x5x11

n! has to include all these factors so the smallest is 11

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 [#permalink] New post 08 Nov 2006, 10:27
Where do you pull 2X3X3X5X11 out of? Obviously I see that equals 990, but where do you come up with that? Why not 5X9X22? Basically whats the process? can you explain some of the concepts for a guy who hasn't had math in years. If I get a problem like this I should..... and then give me a step by step. Pretend you are explaining it to a kid with a learning disability instead of your buddy from matrix algebra.

Thanks!
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 [#permalink] New post 08 Nov 2006, 10:33
Sorry about that.

What you need to do is break it down to its prime factors - basically all the factors of 990 that are prime numbers.

The best way to do this is break it down in steps untill all the factors are prime numbers.

990 = 99 x 10 = 9x11x10=3x3x11x2x5. This way you know that if n does not contain any of these factors, it will not divide into 990.

Now 22! may be a valid answer, but its not the smallest. You could choose 10! bu that doesn't include 11. I hope that makes sense.

PS. I lost all my friends from Matrix Algebra. They joined the chess club.
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thanks! [#permalink] New post 08 Nov 2006, 10:41
haha, yeah, that makes sense. Thanks. The other thing I'm still kind of wondering is that just because something is factor of both 990 and n! (11! for example) that doesn't necessarily mean that n! is a multiple of 990 right?

For example 12 is a factor of both 24 and 60, 60 is not a multiple of 24. Do I have make sure after figuring out that 11 is a factor of 990 that 11! is indeed a multiple of 990? I feel like I am missing another principle her.?
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 [#permalink] New post 08 Nov 2006, 10:41
Andrew -

That 990 = 2x3x3x5x11 is the "prime factorization" of 990. Each of those numbers is prime, meaning that they cannot be divided by any number except themselves and 1. The "Fundamental Theorem of Arithmetic" says that each number is uniquely factorable into the product of primes. As the name suggests, this theorem is very useful (as hosam nicely showed in his solution).

Your factorization includes 9 and 22, neither of which is prime so it is not the "prime factorization". A really important use of the theorem for the GMAT is that if a number N is divisible by a number k that means that all the prime factors of k must also be in N. For example, suppose the question says find the smallest number k so that 270k is divisible by 33. This looks like a nightmare problem on the calculator, but using the FT it's easy.

270 = 3x3x3x2x5
33 = 3x11

so the answer to the question is that the product 270k needs to include an 11 and a 3. It does include a 3 already, so the answer is 11.

Hope this helps.
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one more thing [#permalink] New post 08 Nov 2006, 10:45
Thanks guys, this is really helping me get a grasp on this. One last question, how do you break things down so quick into their prime factors? Like 270 = 3x3x3x2x5. Is there a trick to this or are you just figuring out the prime factors of 27 and then adding 2X5?
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 [#permalink] New post 08 Nov 2006, 10:47
Hi again

It doesn't matter how you do it. You can find any two factors like you just did 27 and 10 and break them down into their prime factors.

27 = 3x3x3
10 = 2x5

Or you could choose 30 x 9
30 = 3x2x5
9 = 3x3

Both ways you end up with 2x3x3x3x5
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 [#permalink] New post 08 Nov 2006, 11:10
I completely agree with hosam's answer. The only thing I would add is that there is no general way to get the prime factorization except coming up with a prime that divides the number and reducing the problem. So for the 270 you need to think of something that divides 270 and go from there. If the number is 15877, I don't have a solution either except to start trying numbers (and it might be prime). The GMAT would only have easily factored numbers like 270 for this kind of problem.
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 [#permalink] New post 08 Nov 2006, 17:11
yes. since 11 must be a factor, then at least n! must contain 11. so 11 is the smallest possible.
  [#permalink] 08 Nov 2006, 17:11
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