If n is a positive integer and the product of all integers from 1 to n

We are given that n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990. Thus:

n! / 990 = integer

To determine the minimum value of n, we need to break 990 into its prime factors.

990 = 10 x 99 = 5 x 2 x 3^2 x 11

Thus:

n! / (5 x 2 x 3^2 x 11) = integer

Since n! must be divisible by 2, 5, 9, and 11, the minimum value of n must be 11. Recall that 11! is divisible by 11 and by any positive integer less than 11. In other words, 11! / 990 = integer.

Answer: B

Solution:

Given- Product of all integers from 1 to n, inclusive, is a multiple of 990

=> 1x2x3x4x..n is a multiple of 990 = 990 m

=> n! = 990 m

=> n! = 2 x 5 x 3^2 x 11 x m

Out of the options 11! will have all the factors of 990 without missing any and hence the smallest value of n would be 11.

(option b)

Devmitra Sen
GMAT SME

Hi Bunuel,

I solved it similar toh this one..but came to the conclusion that 10 must be the least number. I have highlighted why I say that. why is that reasoning incorrect?

Hi ueh55406,

One of the 'restrictions' in the prompt is that the product has to be a MULTIPLE OF 990. If you multiply the integers from 1 to 10, inclusive, then you will NOT get a multiple of 990. Since 990 is a multiple of 11, any multiple of 990 (re: 990, 1980, 2970, etc.) will ALSO be a multiple of 11.

When we prime-factor that large number (whatever it is), the number 11 must be in that group of primes - but it does not occur anywhere in the first 10 integers, meaning that N cannot be 10.

GMAT assassins aren't born, they're made,
Rich

Answer: Option B

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KarishmaB
Would 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.

what is the difficulty level of this question ?

You can check difficulty level of a question along with the stats on it in the first post. For this question Difficulty = Sub-600 Level. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question.

Hi BrentBrentGMATPrepNow, Could you elaborate a bit more on this "For 11 to appear in the product of all the integers from 1 to n, n must equal 11 or more. " ? why do we need 11 to appear in it? Question asked the least possible value of n? but 11 is the highest no in the prime factorization?

First recognize that "the product of all the integers from 1 to n, inclusive" is the same as n!

Let's take the prime factorization of 4!
4! = (4)(3)(2)(1) = (2)(2)(3)(2)(1)
Since there is no 5 hiding in the prime factorization of 4!, we know that 4! Is not divisible by 5.

Similarly, if we were to take the prime factorization of 10!, we would see that there is no 11 hiding in the prime factorization of 10!, which means 10! is not divisible by 11.

However, if we were to take the prime factorization of 11!, we would see that there IS an 11 hiding in the prime factorization of 11!, which means 11! IS divisible by 11.

Great explanation thanks BrentGMATPrepNow. So it's the biggest number of x in the prime factorisation of x! in order to be divisible by x.

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