weilun85 Posted May 3, 2015 Share Posted May 3, 2015 Hi all, I encountered the following problem in pg 534 of the Manhatten GRE 5lb. book and is unable to understand the given explanation. Appreciate if anyone could shed some light on this. If n is an integer and n3 is divisible by 24, what is the largest number that must be a factor of n? (A) 1 (B) 2 © 6 (D) 8 (E) 12 Quote Link to comment Share on other sites More sharing options...
Brent Hanneson Posted May 20, 2015 Share Posted May 20, 2015 If n is an integer and n3 is divisible by 24, what is the largest number that must be a factor of n? (A) 1 (B) 2 © 6 (D) 8 (E) 12 First, if n is an integer and n3 is divisible by 24, then we can say that n3 = 24k for some integer k. We can also rewrite this as: (n)(n)(n) = 24k for some integer k Now let's rewrite 24 as a product of primes to get: (n)(n)(n) = (2)(2)(2)(3)k This tells us that there is at least one 2 "hiding" in the prime factorization of n It also tells us that there is at least one 3 "hiding" in the prime factorization of n. So, if we're certain that there is at least one 2 and one 3 "hiding" in the prime factorization of n, we can be certain that n is divisible by 2 and by 3. In other words, we can be certain that n is divisible by 6. Of course it COULD be divisible by a number greater than 6, but the question asks us to determine the largest number that must be a factor of n. Answer: C Cheers, Brent Quote Link to comment Share on other sites More sharing options...
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