jayoptimist wrote:If "p" is a positive integer and "p^3" is divisible by 144, then the largest positive integer that must divide "p" is
(A) 2
(B) 3
(C) 4
(D) 6
(E) 12
Ans is E
Hi!
Since p is a positive integer, p^3 is a perfect cube. Here's an important rule about prime factoring of perfect squares, cubes, etc...:
perfect squares contain prime factors in pairs;
perfect cubes contain prime factors in triplets;
and so on...
For example:
36 is a perfect square; the prime factorization of 36 is 2*2*3*3;
49 is a perfect square; the prime factorization of 49 is 7*7;
27 is a perfect cube; the prime factorization of 27 is 3*3*3; and
216 is a perfect cube; the prime factorization of 216 is 2*2*2*3*3*3.
So, we know that p^3 must have triplets of prime factors. We also know that p^3 is divisible by 144, i.e. p^3/144 = integer.
Let's break down 144 into prime factors:
144 = 12*12 = 3*4*3*4 = 3*2*2*3*2*2
Now let's reorder that in triplets:
144 = (2*2*2)(2)(3*3)
As noted above, for p^3 to be a perfect cube, it has to have triplets of primes. As we can see, 144 has a lone 2 and a pair of 3s. So, to guarantee that p^3 is a perfect cube we need to add two more 2s and 1 more 3, giving us:
(2*2*2)(2*2*2)(3*3*3)
rewriting with 1 prime from each bracket so we can see what our cube actually is:
(2*2*3)(2*2*3)(2*2*3) = 12^3
Accordingly, the smallest possible value for p^3 is 12^3 and the smallest possible value for p is 12. Therefore, the largest positive integer that MUST be a factor of p is 12: choose E!
Thanks again.
