Gmat_mission wrote: ↑

Fri Jan 08, 2021 4:30 am

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

Answer: B

Source: Official Guide

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

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)

For more on this concept, see the 2nd half of this free video: https://www.gmatprepnow.com/module/gmat- ... /video/825

So, 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

Gmat_mission wrote: ↑

Fri Jan 08, 2021 4:30 am

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

Answer: B

Source: Official Guide

Solution:

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