woohoo921 wrote:
KarishmaB wrote:
flood wrote:
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
I got this one right but I'm not too clear on the logic behind it.
Can you multiply two numbers (other than 11) and get an 11? No because 11 is prime. If 11 is a part of the product of 3 numbers, say a*b*c, it must have been one of the numbers.
Given that the product is 990x (where x is some integer). When we start breaking down 990 into factors, we start with 9*11*10. We see that 9 and 10 will be further broken down into smaller factors but 11 will not be. It is prime. So one of the numbers which multiplied to give 990x must have been 11.
Since consecutive integers 1 to n were multiplied, n is either 11 or greater than 11 to include 11 in the numbers multiplied.
Least possible value of n is 11.
KarishmaBWould you be able to explain why we only look at the prime numbers? Thank you!
Consider this first: If n! is a multiple of 6, what is the smallest value of n?
Is it 6? Can we say that 1*2*3*4*5*6 = n! and this is a multiple of 6 so n MUST be at least 6? No.
How about 3!
3! = 1*2*3 = 6
This is a multiple of 6 so n = 3 is the smallest value of n.
Now consider this:
If n! is a multiple of 5, what is the smallest value of n?
Will n = 2 or 3 or 4 work? Can I multiply any of these to give 5? No because 5 is prime. I must have a 5 in the product.
So n must be at leats 5.
1*2*3*4*5 is the smallest n! that is a multiple of 5. Here the smallest value of n is 5.
The point is we can make composite numbers by using smaller numbers (their factors) but we cannot make prime numbers.
SO to have 990 as a factor, I need 2*3*3*5*11
If n! is 1*2*3*4*5*6, I have already got 2, 3, 3, 5 but I don't have 11 yet. Will 7! gives me 11? No. Neither will 8!, 9! or 10!.
I need 11! because that is when I will get my first 11. That is why we focus on prime numbers.
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