I'm not sure I understand the question correctly.

The product of all the numbers from 1 to 6 inclusive is 720, and the product of all the numbers from 1 to 7 inclusive is 5040. So, it does not seem that there exists an n such that the product of the numbers from 1 to n equals 990.

Is there an "answer does not exist" answer choice?

Tatiana

DUDE you have made the same mistake while re posting the question.

Here is the actual question.

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 ?


If I take you answer which is 10

Then, what the question stem means is 10! is a multiple of 990.

10! = 10*9*8*7*6*5*4*3*2*1=3628800

990 = 9*10*11

I'll rephrase the question for you.

Is 3628800 a multiple of 990?

The answer is no if n = 10, simply because n! which is 10! does not have 11

Now lets look at 11

11! = 11*10*9*8*7*6*5*4*3*2*1
990 =9*10*11

In this case answer is yes 11! is a multiple of 990.

In the test all you have to do is look at the prime factors.

if x is a multiple of y, then x should have all the prime factors of y, but it is not necessary for y to have all the prime factors of x.

I hope this clears everything. Let me know if you still have any doubts.

All the Best.

Prime factor 990.

990 = 2 X 5 X 3^2 X 11

Since the largest prime factor of 990 is 11, choose 11 it must be the least possible value for n.

gmatutor wrote:Prime factor 990.

990 = 2 X 5 X 3^2 X 11

Since the largest prime factor of 990 is 11, choose 11 it must be the least possible value for n.

While, that's right in this case, it isn't universally correct...

For example, if we were doing the same question but for the number 1920:
1920 = 2^7*3*5

In this case, 5 is the largest prime factor, but 5! would not be the appropriate answer.

We'd need to make sure that all of the prime factors are represented so:
2*3*4*5*6*7*8 or 8! would be the right answer.

Absolutely!

Forgive my imprecise language.

You're forgiven!