Xbond 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

A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N
Similarly, we can say:
If N is is a multiple of k, then k is "hiding" within the prime factorization of N

Examples:
24 is divisible by 3 <--> 24 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7

So, if if some number is a multiple of 990, then 990 is hiding in the prime factorization of that number.

Since 990 = (2)(3)(3)(5)(11), we know that one 2, two 3s, one 5 and one 11 must be hiding in the prime factorization of our number.

For 11 to appear in the product of all the integers from 1 to n, n must equal 11 or more.
So, the answer is B

Cheers,
Brent

Xbond wrote:Hi there,

Could you give me your ELEGANT, QUICK, and EFFICIENT method to resolve (with these concepts and with finding the solution).


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

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