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ps - Divisor

Expert replies
by ccassel » Mon Apr 04, 2011 10:58 am
Hi,

How would you explain arriving at the answer to this question?

If the integer n has exactly three positive divisors, including 1 and n, how many positive divisors does n^2 have?

(a) 4
(b) 5
(c) 6
(d) 8
(e) 9

Cheers,
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Source: — Problem Solving |

by GMATGuruNY » Mon Apr 04, 2011 11:26 am
ccassel wrote:Hi,

How would you explain arriving at the answer to this question?

If the integer n has exactly three positive divisors, including 1 and n, how many positive divisors does n^2 have?

(a) 4
(b) 5
(c) 6
(d) 8
(e) 9

Cheers,
Plug in n=4. This works because 4 has three positive divisors: 1, 2, and 4.
n² = 4² = 16.
Divisors of 16: 1,2,4,8,16 = 5 divisors.

The correct answer is B.
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by ccassel » Mon Apr 04, 2011 11:32 am
Thanks.

How would you answer the questions if the obvious plug-in was not available?
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by GMATGuruNY » Mon Apr 04, 2011 12:08 pm
ccassel wrote:Thanks.

How would you answer the questions if the obvious plug-in was not available?
To determine the number of positive factors of an integer:

1) Prime-factorize the integer
2) Add 1 to each exponent
3) Multiply


For example:
72 = 2³ * 3².
Adding 1 to each exponent and multiplying, we get (3+1)*(2+1) = 12 factors.

Here's the reasoning. To determine how many factors can be created from 72 = 2³ * 3², we need to determine the number of choices we have of each prime factor:

For 2, we can use 2�, 2¹, 2², or 2³, giving us 4 choices.
For 3, we can use 3�, 3¹, or 3², giving us 3 choices.

Multiplying, we get 4*3 = 12 possible factors.

Thus, to count the total number of divisors of n, we would prime-factorize n, add 1 to each exponent, and multiply.
Thus, if n has exactly 3 divisors, we know that n = (prime number)². (Because if we add 1 to the exponent of 2, we'll get 2+1 = 3 factors.)
Thus, n² = (prime number²)² = (prime number)�.
Adding 1 to the exponent, we know that n² must have 4+1 = 5 factors.

I think that plugging in -- as I did in my original post -- is much easier.
Last edited by GMATGuruNY on Mon Apr 04, 2011 2:02 pm, edited 2 times in total.
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I unlock the best way for YOU to solve problems.

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by ccassel » Mon Apr 04, 2011 1:13 pm
Thanks againv Mitch. Pluging-In is defenetly easier but the reasoning will help confirm the answer.

One final note however, under your reasoning, should your example to determine the # of factors of 36^2 equal 25, not 12?

36^2 = 2^4*3^4 (Prime Factorize)
(4+1 = 5)(4+1 = 5) (Add 1 to each exponent)
(5)*(5)=25 (Multiply)

Thus, there are 25 factors of 36^2.

Cheers,
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by GMATGuruNY » Mon Apr 04, 2011 1:21 pm
ccassel wrote:Thanks againv Mitch. Pluging-In is defenetly easier but the reasoning will help confirm the answer.

One final note however, under your reasoning, should your example to determine the # of factors of 36^2 equal 25, not 12?

36^2 = 2^4*3^4 (Prime Factorize)
(4+1 = 5)(4+1 = 5) (Add 1 to each exponent)
(5)*(5)=25 (Multiply)

Thus, there are 25 factors of 36^2.

Cheers,
Yes, indeed. 36² has 25 factors. My second post had a typo -- instead of 36² I meant to type 72 -- which I've since corrected. Thanks for pointing out the typo.
Last edited by GMATGuruNY on Mon Apr 04, 2011 2:03 pm, edited 1 time in total.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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I unlock the best way for YOU to solve problems.

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by ccassel » Mon Apr 04, 2011 1:25 pm
You would do the same for me.

cheers,
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by manpsingh87 » Mon Apr 04, 2011 10:02 pm
ccassel wrote:Hi,

How would you explain arriving at the answer to this question?

If the integer n has exactly three positive divisors, including 1 and n, how many positive divisors does n^2 have?

(a) 4
(b) 5
(c) 6
(d) 8
(e) 9

Cheers,
since n has 3 divisors therefore it should be of the type n^2, therefore (n^2)^2 = n^4 will have (4+1) divisors...!!!
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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