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Problem solving question

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by towerSpider » Sat Jan 29, 2011 12:28 am
shoot4greatness wrote:If n=4p where pis a prime number greater than 2, how many different positive even divisors does n have, including n?
a. two
b. three
c. four
d. six
e. eight

can anyone explain this question?
Let p = 3, then:

n = 12 (1, 2, 3, 4, 6, 13)

Let p = 5 then:

n = 20 (1, 2, 4, 5, 10, 20)

So answer is six.

Doing it ONCE is sufficient.
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by towerSpider » Sat Jan 29, 2011 12:50 am
I would like someone to explain it through proper method of number properties.
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by November Rain » Sat Jan 29, 2011 8:12 am
Hi guys,
I would solve it like this:

N is the result of the multiplication between 4 and P, where P is prime odd number (with no divisors except 1 and itself).

So, trough the number properties N = 2(^2) * P(^1). And here's a little trick: whenever you need to know how many different divisors a number have, you only need to add 1 to the exponents and then multiply them.

So the total number of divisors is (2+1)* (1+1) = 6. Now you need to take out two divisors: 1 and P (because they are both odd)


Therefore in total, we have 4 even divisors. Hence C
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by shoot4greatness » Sat Jan 29, 2011 9:49 am
Thank you November Rain,

A follow-up question is if n=4*n^2 than the number of divisor will be (2+1)*(2+1)= 9 and we would subtract 3 from 9 because there are 2 odd numbers and 1. So, the answer will be 6 even divisors(factors).

BTW, a huge fan of Slash afd-era. Was really tempted to put a deposit for the new Marshall AFD100.
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by towerSpider » Sat Jan 29, 2011 9:59 am
November Rain wrote:Hi guys,
I would solve it like this:

N is the result of the multiplication between 4 and P, where P is prime odd number (with no divisors except 1 and itself).

So, trough the number properties N = 2(^2) * P(^1). And here's a little trick: whenever you need to know how many different divisors a number have, you only need to add 1 to the exponents and then multiply them.

So the total number of divisors is (2+1)* (1+1) = 6. Now you need to take out two divisors: 1 and P (because they are both odd)


Therefore in total, we have 4 even divisors. Hence C
Correct me if I am wrong: you have to do this method only when a number is represented in prime factorized form, e.g. lets take 12:

12 = 2 . 6 = 2 . 2 . 3 = 2^2 . 3
No. of divisors = 3 . 2 = 6 (1, 2, 3, 4, 6, 12)

Now lets take 25:

5 . 5 = 5^2
No. of divisors: 2+1 = 3 (1, 5, 25)

By the way, should not answer be six in original question because it didn't ask to eliminate 1 and n from counting? In fact, question specifically asked to count n.
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by November Rain » Sat Jan 29, 2011 6:04 pm
Hi towerSpider
Correct me if I am wrong: you have to do this method only when a number is represented in prime factorized form, e.g. lets take 12:
Yes, you're right, altough you should see it the other way around - you need to represent a number in its prime factorized form in order to use this method.
When you factorize a number to its primes, the multiplication of each exponent (plus 1) represents the total number of divisors that the number has. By the way, you need to add 1 to each exponent because you need to count the divisibility of the number "1".
By the way, should not answer be six in original question because it didn't ask to eliminate 1 and n from counting? In fact, question specifically asked to count n.
I think it should be four, because the question asked to count even divisors only. Both "1" and "p" are odd, so they are not counted. "N" however, is an even number so I counted it.

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by November Rain » Sat Jan 29, 2011 6:07 pm
Hi
A follow-up question is if n=4*n^2 than the number of divisor will be (2+1)*(2+1)= 9 and we would subtract 3 from 9 because there are 2 odd numbers and 1. So, the answer will be 6 even divisors(factors).

I think you meant N=4*P^2. If that's the case then indeed you are right.

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by Adam@Knewton » Sun Jan 30, 2011 11:01 pm
Because the question asks for the number of even factors, applying the wonderful number-of-factors trick can be treacherous in this case. I did it like this:

Since an even factor must have a factor of 2, and p is an odd prime number, it should be easy to count up (the answer choices are small enough to do so) the even factors: 2, 4, 2p, and 4p. That's four.

By using reasoning, I applied my knowledge of number properties to get the question right quickly, not necessarily using a true Mathematics technique. While I agree that advanced knowledge of number properties and how numbers work is essential on the GMAT (and to my approach), this doesn't mean that the most math-y way is the best way on the actual test.

In fact, on the test, TowerSpider's method is the best and fastest. Since it's a Problem Solving question, you know that there's only one possible answer, even if through number theory you're not sure why. Plug in 3, count the even factors, pick the right answer and move on. Anyone who doesn't do it this way is asking for trouble on test day! The test does not reward your math skills -- it rewards your problem solving skills, and the quickest, most effective solution to the problem is always better than the more formal one, on the GMAT, in business school, and in life!
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