\(n\) is a positive integer, and \(k\) is the product of all integers from \(1\) to \(n\) inclusive. If \(k\) is a multi

[M7MBA]

\(n\) is a positive integer, and \(k\) is the product of all integers from \(1\) to \(n\) inclusive. If \(k\) is a multiple of \(1440,\) then the smallest possible value of \(n\) is

A. 8
B. 12
C. 16
D. 18
E. 24

Answer: A

Source: Magoosh

[Scott@TargetTestPrep]

\(n\) is a positive integer, and \(k\) is the product of all integers from \(1\) to \(n\) inclusive. If \(k\) is a multiple of \(1440,\) then the smallest possible value of \(n\) is

A. 8
B. 12
C. 16
D. 18
E. 24

Answer: A

Solution:

We see that k = n!. Now, let’s factor 1440:

1440 = 12^2 x 10 = (2^2 x 3)^2 x 2 x 5 = 2^5 x 3^2 x 5

Since 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 2^3 x 7 x 2 x 3 x 5 x 2^2 x 3 x 2 = 2^7 x 3^2 x 5 x 7 is divisible by 1440 (notice the exponents of 2, 3, and 5 in 8! are greater than or equal to the respective exponents in 1440), the smallest possible value of n is 8.

Answer: A