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# How many two digit integers have exactly five divisors?

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How many two digit integers have exactly five divisors? [#permalink]

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28 Aug 2014, 10:58
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How many two digit integers have exactly five divisors?

A)Zero

B)One

C)Two

D)Three

E)Four
[Reveal] Spoiler: OA
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Joined: 02 Sep 2009
Posts: 43889
Re: How many two digit integers have exactly five divisors? [#permalink]

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28 Aug 2014, 11:09
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guerrero25 wrote:
How many two digit integers have exactly five divisors?

A)Zero

B)One

C)Two

D)Three

E)Four

Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

Back to the question:

Since 5 is a prime number, it cannot be the product of two integers greater than 1, which implies that a number having 5 factors must be of a form of (prime)^4 --> the number of factors = (4 + 1).

There are only 2 two-digit numbers which can be written this way: 2^4 = 16 and 3^4 = 81.

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Re: How many two digit integers have exactly five divisors? [#permalink]

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28 Aug 2014, 11:18
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Bunuel wrote:
guerrero25 wrote:
How many two digit integers have exactly five divisors?

A)Zero

B)One

C)Two

D)Three

E)Four

Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

Back to the question:

Since 5 is a prime number, it cannot be the product of two integers greater than 1, which implies that a number having 5 factors must be of a form of (prime)^4 --> the number of factors = (4 + 1).

There are only 2 two-digit numbers which can be written this way: 2^4 = 16 and 3^4 = 81.

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Hope this helps.
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How many two digit integers have exactly five divisors? [#permalink]

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19 Jan 2015, 19:40
Quote:

Yet another one:

Question: How many factors does the integer 9999 have?

99999 is the same as 10^4 - 1. This allows us to use the difference of squares to our advantage as follows:

(100+1) (100-1) = 101 * 99.

With this we know that 101 is prime, and 99 can be expressed as (3)(3)(11), which will result in the following factorization: 3*3*11*101 = (3^2)*11*101.

using the explanation in the first post of the thread we can get to the answer of 12 unique factors.
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Re: How many two digit integers have exactly five divisors? [#permalink]

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19 Jan 2015, 21:47
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guerrero25 wrote:
How many two digit integers have exactly five divisors?

A)Zero

B)One

C)Two

D)Three

E)Four

Note that if we generalize this question, it could become cumbersome - i.e. if we change it to "How many two digit integers have exactly four divisors?", it will involve quite a bit of work. There is something special about 5 divisors which makes it easy to solve if you understand the fundamentals.
If a number has 3 factors, it means it is a perfect square. If it has 5 factors, it means it is a power of 4 of a prime number. If it has 7 factors, it means it is a power of 6 of a prime number. If it has 9 factors, it means it is either a power of 8 of a prime number or has squares of 2 prime numbers.
You must understand why each of these is true. To do so, check out Bunuel's explanation above or these posts:
http://www.veritasprep.com/blog/2010/12 ... ly-number/
http://www.veritasprep.com/blog/2010/12 ... t-squares/

Here, since the number has 5 factors, it must be a power of 4 of a prime number. The fourth power of 2 is 16 (a two digit number). The fourth power of 3 is 81 (a two digit number). The fourth power of 5 is 625 (a three digit number so not acceptable). All other prime numbers will have fourth power higher than 625 so ignore them.

So the number could be either 16 or 81 i.e. two values.

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 20 Oct 2012 Posts: 15 Location: United States Concentration: Entrepreneurship, Strategy Re: How many two digit integers have exactly five divisors? [#permalink] ### Show Tags 21 Jul 2015, 23:35 Bunuel wrote: guerrero25 wrote: How many two digit integers have exactly five divisors? A)Zero B)One C)Two D)Three E)Four Finding the Number of Factors of an Integer First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers. The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$ Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors. Back to the question: Since 5 is a prime number, it cannot be the product of two integers greater than 1, which implies that a number having 5 factors must be of a form of (prime)^4 --> the number of factors = (4 + 1). There are only 2 two-digit numbers which can be written this way: 2^4 = 16 and 3^4 = 81. Answer: C. shouldn't the answer be 4?: 32,81,-32,-81 -32 is a 2 digit number with 5 divisors -1,-2,-4,-8,-32 the same with -81. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7951 Location: Pune, India Re: How many two digit integers have exactly five divisors? [#permalink] ### Show Tags 22 Jul 2015, 00:45 1 This post received KUDOS Expert's post nvad5955 wrote: Bunuel wrote: guerrero25 wrote: How many two digit integers have exactly five divisors? A)Zero B)One C)Two D)Three E)Four Finding the Number of Factors of an Integer First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers. The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$ Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors. Back to the question: Since 5 is a prime number, it cannot be the product of two integers greater than 1, which implies that a number having 5 factors must be of a form of (prime)^4 --> the number of factors = (4 + 1). There are only 2 two-digit numbers which can be written this way: 2^4 = 16 and 3^4 = 81. Answer: C. shouldn't the answer be 4?: 32,81,-32,-81 -32 is a 2 digit number with 5 divisors -1,-2,-4,-8,-32 the same with -81. When we talk about factors/divisors, it is assumed that we are talking about positive integers only. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: How many two digit integers have exactly five divisors? [#permalink]

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23 Jan 2016, 03:33
the original questions reads "positive divisors"
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Re: How many two digit integers have exactly five divisors? [#permalink]

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06 Apr 2016, 17:43
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only perfect squares have odd numbers of factors.
and only perfect squares of 4, 5, 6, 7, 8, and 9 are two digit.
so: 16, 25, 36, 49, 64, 81
now..16=2^4. 4+1 = 5 factors, so works.
25 - 5^2 -> 3 factors, out.
36 = 2^2 * 3^2 = 3x3=9 factors, out.
49 = 7^2 = 3 factors, out.
64 = 2^6 -> 7 factors, out.
81 = 3^4 => 5 factors.

we have 2 two digit numbers with 5 factors.
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Re: How many two digit integers have exactly five divisors? [#permalink]

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20 Dec 2016, 06:20
mvictor wrote:
only perfect squares have odd numbers of factors.
and only perfect squares of 4, 5, 6, 7, 8, and 9 are two digit.
so: 16, 25, 36, 49, 64, 81
now..16=2^4. 4+1 = 5 factors, so works.
25 - 5^2 -> 3 factors, out.
36 = 2^2 * 3^2 = 3x3=9 factors, out.
49 = 7^2 = 3 factors, out.
64 = 2^6 -> 7 factors, out.
81 = 3^4 => 5 factors.

we have 2 two digit numbers with 5 factors.

I did it in exactly the same way. I know it is application of brute force but it atleast helped me in getting the correct answer within the time frame.
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Re: How many two digit integers have exactly five divisors? [#permalink]

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19 Feb 2017, 06:37
VeritasPrepKarishma

Can you please share an instance from the official material that supports the same? It seems to me that +2 should be considered a divisor of -32.
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Re: How many two digit integers have exactly five divisors?   [#permalink] 19 Feb 2017, 06:37
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