If n is a positive integer and the product of all 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

This question tests the concept of prime factorization. In order to figure out if 990 can be a divisor of the unknown product you need to determine all of the prime factors that combine to successfully divide 990 without leaving a remainder.

The prime factors of 990 are 2, 3, 3, 5, 11

Since 11 is a prime factor of 990, you need to have the number present somewhere in the unknown product. Answer A is the product of integers from 1 to 10 and therefore does not meet the necessary criteria. Answer B includes 11 and therefore is the answer.

Hi,

The question says that 990 is a factor of n!.
990 = 9*110 = 9*11*10
For 990 to be a factor of n!, n should at least be 11 so that it includes the common factor 11 that is found in 990.


regards,
Jack

The least possible value of such a n will be at least the highest prime factor of 990, which is 11

The only other thing to check is if the rest of the factors of 990, can be obtained by multiplication of numbers less than 11. Otherwise n may be greater than 11

But that is easy to see since 990=9*10*11. So taking all the numbers below 11 will suffice.
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You need to primarily find the primes of 990:
2,3,3,5,11
The integer must contain all these primes

By knowing 11 is an prime, we know the answer can be B,C,D or E as these all contain 11.
11=1,2,3,4,5,6,7,8,9,10,11

11 contains the 2,3,5 and 11 explicitly and the second 3 comes from the 6 (2*3), therefore the answer is 11.

This is because we know a number that contains all the primes of 990 will be a multiple of 990.

Hope this helps

just entered this in google and you're right! (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11) / 990 = 40 320

meanwhile (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10) / 990 = 3 665.45455 wouldn't work

It makes sense because the highest prime factor of 990 must also be a factor of any multiple of 990.

I guess I need to do more practice problems involving multiples and number properties.

to cement my understanding, maximum value of n can be anything, right?
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Hi Bunuel,

I am not sure why have excluded 10, which seems to be one of the factors (5*2).

Here's a more elaborated thinking response to what you all have summarized above -- but for those who need the extra steps, try this.

990* multiple = product of 1 to n
990 * multiple = 1 * 2 * 3 * 4 * 5 * ... * n
11 * 3 * 3 * multiple = 1 * 2 * 3 * 4 * 5 * ... * n

We need to find the smallest value for n. We know in the factor tree, any number can be broken down to its lowest level in the factor tree. And that factor tree will comprise of all prime factors.

The factors on the left side are:
11 and 3 and 'multiple' -- the multiple can be anything.

We know that we want 990 to go into the right side (we don't care how many times).

The 990 consists of 11 and 3^2. So essentially it's prime factors are 11 and 3. These lowest level prime factors must go into the right side -- so the right side at the very least must contain the lowest prime factors from the left side.

So, we see 1 * 2 * 3 * 4 * 5 * ... * n

contains a 3.

But it does not contain an '11'.
11 is the answer.

Hope that helps.

product of all the integers from 1 to n = n!
990 = 9*10*11
i.e. n! must be divisible by 9, 10 and 11 all for n! to be divisible by 990
i.e. minimum value of n must be 11 for n! to be divisible by 11 and 10 and 9 both

P.S. 11!= 11*10*9*8*7*6*5*4*3*2*1

n = 11

Answer: option D
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