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Number Systems-Divisors and Multiples

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by Brent@GMATPrepNow » Fri Oct 11, 2013 7:42 am
sukhman wrote:Find the number of divisors of 544 which are greater than 3
A. 15 B. 10 C.12 D. None of these
If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = 5x4x2=40

544 = (2^5)(17^1)
So, the number of positive divisors of 544 = (5+1)(1+1) = 6 x 2 = 12

IMPORTANT: The correct answer is not 12, because we're asked to find the number of divisors that are greater than 3

We know that 1 and 2 are divisors of 544, and we know that 3 is NOT a divisor of 544. So, we must subtract 2 divisors from our total of 12 to get 10

Answer: B

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by [email protected] » Fri Oct 11, 2013 1:21 pm
Hi sukhman,

Brent provides an elegant solution to the problem, but if you don't have that knowledge, or forget how to do it, then you have to be ready for an alternative approach. Sometimes "brute force" is how you have to do things.

For this question, we can map out all of the possibilities. There are some specific Number Properties that can make this process go faster though.

So, let's make a list of all of the factors of 544:
1 and 544
2 and 272

3 is NOT divisible (you can use the "Rule of 3" to prove it; the digits 5 + 4 + 4 = 13 and 13 is NOT divisible by 3, so 544 is NOT divisible by 3). Since 3 is NOT divisible, then NO MULTIPLE OF 3 is divisible either, so we'll be able to ignore 3, 6, 9, 12, 15, 18, etc. as we work through the possibilities.

4 and 136 (notice how we "double the 2" and "halve the 272" to get this set of values? That's another Number Property shortcut for division).

5 is NOT divisible, so NO MULTIPLE OF 5 is divisible either. We can ignore 5, 10, 15, 20, etc. as we work through the possibilities.

7 requires a calculation, but you'll find it's not a factor
8 and 68 (double the 4; halve the 136)

16 and 34
32 and 17

You'd need to do a quick check for 11 and 13, but you'll find that neither is a factor.

With this list, you'll then have to "count up" the number of terms that are greater than 3:

4, 8, 16, 17, 32, 34, 68, 136, 272, 544 is a total of 10 terms.

While this is not "pretty", it can get the job done. On Test Day, staring at the screen is NOT an option. Sometimes you just have to put the pen on the pad and "brute force" the solution.

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by Jeff@TargetTestPrep » Mon Dec 11, 2017 4:33 pm
sukhman wrote:Find the number of divisors of 544 which are greater than 3
A. 15 B. 10 C.12 D. None of these
We can start by breaking 544 into prime factors.

544 = 4 x 136 = 4 x 4 x 34 = 2^2 x 2^2 x 2 x 17 = 2^5 x 17^1

The factors that are greater than 3 are as follows:

2^2, 2^3, 2^4, 2^5, 17, 2 x 17, 2^2 x 17, 2^3 x 17, 2^4 x 17, and 2^5 x 17.

Thus, there are 10 divisors of 544 greater than 3.

Alternate solution:

We can start by breaking 544 into prime factors.

544 = 4 x 136 = 4 x 4 x 34 = 2^2 x 2^2 x 2 x 17 = 2^5 x 17^1

The complete factorization (as opposed to prime factorization) requires that the number 1 (which is not prime) also be included in the complete factorization: 1 x 2^5 x 17^1

Recall that the number of divisors of a number can be found by using the following method:

1) Add 1 to each of the exponents (of the prime factors) in the number's complete factorization.
2) Multiply the resulting numbers.

Thus, the number of divisors of 544 is (5 + 1) x (1 + 1) = 6 x 2 = 12. Of the 12 divisors of 544, only 1 and 2 are not greater than 3, so 10 of them will be greater than 3.

Answer: B

Jeffrey Miller
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[email protected]

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