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Goldfinger2001
- Senior | Next Rank: 100 Posts
- Posts: 50
- Joined: Mon Jan 03, 2011 9:22 am
Hey Goldfinger,
Really good question here - thanks for sharing! To answer this one, think about what we know has to be true of n:
n! (the product of all integers from 1 to n) MUST BE a multiple of 990.
What does it take to be a multiple of 990? When you're dealing with factors and multiples, you'll typically want to break down numbers into prime factors - prime factors tell you the essential components of a number like 990, almost like breaking a substance down into its atomic components or something like that (water = H20 which means you need 2 hydrogen atoms for every one oxygen atom - the process of prime factorization is similar...it breaks numbers down to their fundamental components).
990 factors out to 99*10, which gets us the prime factors:
99 = 3*3*11
10 = 2*5
So for our number to be a multiple of 990, it MUST have the following prime factors: 2*3*3*5*11
In order to get that 11, the absolute minimum value of n has to be 11, because as a prime number 11 isn't going to be constructed by multiplying any two other integers together. So if n is 11, then n! will contain:
1*2*3*4*5*6*7*8*(3*3)*10*11
(note that we can break that 9 down into 3*3 to ensure that we have that second 3 covered)
n = 11 satisfies all of our requirements to be a multiple of 990, and we know that it's the lowest such number because of that necessary 11. Therefore, the answer is 11.
