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Let's factorize 24. We see that 24 = 2^3*3; thus, for n^3 to be divisible by 24 = 2^3*3, n must be a factor of 2 and 3. Orn must be at least 2*3 = 6. So with n =6, we have n^3 = 6^3 = (2*3)^3 = 2^3*3^3, divisible by 24.BTGmoderatorDC wrote:If n is an integer and n^3 is divisible by 24, what is the largest number that must be a factor of n?
(A) 1
(B) 2
(C) 6
(D) 8
(E) 12
OA C
Source: Manhattan Prep
Since a perfect cube has unique prime factors in multiples of 3 and since 24 = 2^3 x 3^1, we see that the smallest value of n^3 is 2^3 x 2^3, and thus the smallest value of n is 6. Since the largest factor of 6 is 6, then the largest number that must be a factor of n is 6.BTGmoderatorDC wrote: ↑Thu Sep 19, 2019 4:46 pmIf n is an integer and n^3 is divisible by 24, what is the largest number that must be a factor of n?
(A) 1
(B) 2
(C) 6
(D) 8
(E) 12
OA C
Source: Manhattan Prep
