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CONCEPT: Perfect squares have an even number of primes in their prime factorizations.sukhman wrote:Find the sum of divisors of 544 which are perfect squares
A. 32 B. 64 C.42 D. 21
Thanks Brent.. I was stuck with answer 20Brent@GMATPrepNow wrote:CONCEPT: Perfect squares have an even number of primes in their prime factorizations.sukhman wrote:Find the sum of divisors of 544 which are perfect squares
A. 32 B. 64 C.42 D. 21
For example: 36 = (2)(2)(2)(2)(3)(3). There are four 2's and two 3's
Likewise, 1600 = (2)(2)(2)(2)(2)(2)(5)(5). There are six 2's and two 5's
And 225 = (3)(3)(5)(5). There are two 3's and two 5's
Okay, now onto the question . . .
First find the prime factorization of 544.
544 = (2)(2)(2)(2)(2)(17)
So, each divisor of 544 will be the product of 1 or more of these prime factors.
The divisors that are perfect squares will have an EVEN number of prime factors. Since there's only one 17 in the prime factorization of 544, we can rule out using 17's
At this point, we can simply list the divisors that are perfect squares:
(2)(2) = 4, which is a perfect square
(2)(2)(2)(2) = 16, which is a perfect square
We can't forget 1, which is a divisor and a perfect square
That's it!
The sum = 4 + 16 + 1 = [spoiler]21 = D[/spoiler]
Cheers,
Brent
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