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The least possible value for n

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by nickolhomero » Sun May 03, 2009 2:48 pm
Hi everybody,

I ran into this problem that I can't figure out. Can someone please explain why is that the correct answer?

Thank you :)
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by Feep » Mon May 04, 2009 2:45 am
The trick here is to break down 990 into its prime factorization, which is 2 * 3 * 3 * 5 * 11. We absolutely NEED an 11 somewhere in the product to achieve 990, and then we just need to make sure the other prime factors are satisfied in the numbers 1-10. 2? Check. 3? Check. 5? Check. We need one more 3, but that is satisfied by either 6 (2 * 3) or 9 (3 * 3). So, we've got everything we need.

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I tutor GMAT/GRE level mathematics privately in the Los Angeles region, as well as via Skype for a discounted rate. Send me a message if you're interested.
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