Last visit was: 28 Mar 2025, 09:50 It is currently 28 Mar 2025, 09:50
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
avatar
YTT
Joined: 05 Aug 2009
Last visit: 31 Aug 2010
Posts: 5
Own Kudos:
430
 [388]
Posts: 5
Kudos: 430
 [388]
20
Kudos
Add Kudos
365
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 28 March 2025
Posts: 100,138
Own Kudos:
Given Kudos: 92,749
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 100,138
Kudos: 711,537
 [156]
52
Kudos
Add Kudos
104
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 28 March 2025
Posts: 100,138
Own Kudos:
Given Kudos: 92,749
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 100,138
Kudos: 711,537
 [146]
59
Kudos
Add Kudos
87
Bookmarks
Bookmark this Post
User avatar
mbaiseasy
Joined: 13 Aug 2012
Last visit: 29 Dec 2013
Posts: 324
Own Kudos:
1,938
 [36]
Given Kudos: 11
Concentration: Marketing, Finance
GPA: 3.23
Posts: 324
Kudos: 1,938
 [36]
22
Kudos
Add Kudos
14
Bookmarks
Bookmark this Post
My Solution:

Try increasing prime numbers with length 6:
Trial 1: \(2^6=64\) Valid
Trial 2: \(3^6 =729\) Invalid

This means our candidate 2-digit numbers have combinations of \(2\) and \(3\)

\(2^6=64\)
\(2^5x3^1=96\)
\(2^4x3^2=144\) Invalid

Answer: Two
General Discussion
avatar
YTT
Joined: 05 Aug 2009
Last visit: 31 Aug 2010
Posts: 5
Own Kudos:
430
 [1]
Posts: 5
Kudos: 430
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
That's brilliant!!! I especially love the part where I could take 5 away. This really save tons of time! Thanks!! BTW, Thanks so much for the prompt response!
User avatar
mmphf
Joined: 18 Mar 2010
Last visit: 28 Oct 2012
Posts: 46
Own Kudos:
101
 [5]
Given Kudos: 5
Location: United States
Posts: 46
Kudos: 101
 [5]
4
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
The best (quickest) way I can think of to get the answer, is start with 2^6, then move on from there.

2^6=64
2^5*3=96

Obviously 2^5*5 will be more than 2 digits, as will 2^4*3^2. So 64 and 96 are it. Answer is 2 (C).

You may have been looking for something even faster, but this is fast enough for me. Unless someone has a better way.
avatar
Spunkerspawn
Joined: 30 Apr 2010
Last visit: 02 Jun 2023
Posts: 14
Own Kudos:
87
 [5]
Given Kudos: 2
Posts: 14
Kudos: 87
 [5]
4
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Try the smallest possible value first: In this case it is 2^6 which equals 64.

If we replace the last 2 with 3, then we have 2^5*3 = 96

From here we can positively assume that any other number will have more than 2 digits. So the answer is (C) 2 numbers that have length 6 and are only 2 digits.
User avatar
Daddydekker
Joined: 27 Sep 2013
Last visit: 21 Nov 2014
Posts: 8
Own Kudos:
16
 [7]
Location: Netherlands
Posts: 8
Kudos: 16
 [7]
5
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Just writing it out took me .46 sec.:

length of 6, lets take the lowest prime factor 2

2x2x2x2x2x2 = 64

Now substitute the last 2 by a 3, and see the solution gets 96. We can think of what will happen when we substitute another 2 for a three.

Hence, C (2)
avatar
OptimusPrepJanielle
Joined: 06 Nov 2014
Last visit: 08 Sep 2017
Posts: 1,782
Own Kudos:
1,422
 [1]
Given Kudos: 23
Expert
Expert reply
Posts: 1,782
Kudos: 1,422
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
enigma123
For any positive integer n, the length of n is defined as number of prime factors whose product is n, For example, the length of 75 is 3, since 75=3*5*5. How many two-digit positive integers have length 6?

A. 0
B. 1
C. 2
D. 3
E. 4

I need to understand the concept behind solving this question please.
For the length to be 6, the number of prime factors should be maximum. Hence we need to use maximum 2's

The numbers can be 2^6 = 64 and 2^5*3 = 96
For any other number less than 100, the length will be less than 6

Correct option: C
User avatar
JeffTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 3,008
Own Kudos:
7,593
 [7]
Given Kudos: 1,646
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert
Expert reply
Posts: 3,008
Kudos: 7,593
 [7]
6
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
enigma123
For any positive integer n, the length of n is defined as number of prime factors whose product is n, For example, the length of 75 is 3, since 75=3*5*5. How many two-digit positive integers have length 6?

A. 0
B. 1
C. 2
D. 3
E. 4

We need to determine how many 2-digit integers have a length of 6, or in other words how many 2-digit integers are made up of 6 prime factors. Let’s start with the smallest possible numbers:

2^6 = 64 (has a length of 6)

2^5 x 3^1 = 96 (has a length of 6)

Since 2^4 x 3^2 = 144 and 2^5 x 5^1 = 160 are greater than 99, there are no more 2-digit numbers that have a length of 6.

Answer: C
User avatar
abhishekdadarwal2009
Joined: 04 Sep 2015
Last visit: 07 Dec 2022
Posts: 530
Own Kudos:
Given Kudos: 123
Location: India
WE:Information Technology (Computer Software)
Products:
Posts: 530
Kudos: 456
Kudos
Add Kudos
Bookmarks
Bookmark this Post
For any positive integer n, the length of n is defined as number of prime factors whose product is n, For example, the length of 75 is 3, since 75=3*5*5. How many two-digit positive integers have length 6?

A. 0
B. 1
C. 2
D. 3
E. 4

So it is evident that the length can only include 2 and 3 also when we try with 2 we find that length longer than 6 is not possible to be a 2 digit number.

and only one 2 can be replaced by 3 and we get 96,

this can be solved by hit and trial by starting from the lower prime and then moving up.



64 and 96 are the only two numbers possible to have length 6.
avatar
matthewsmith_89
Joined: 06 Dec 2016
Last visit: 06 Oct 2017
Posts: 194
Own Kudos:
Given Kudos: 10
Posts: 194
Kudos: 62
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
You made this question look easy. Thanks for your explanation.
User avatar
sashiim20
Joined: 04 Dec 2015
Last visit: 05 Jun 2024
Posts: 616
Own Kudos:
1,739
 [2]
Given Kudos: 276
Location: India
Concentration: Technology, Strategy
WE:Information Technology (Consulting)
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
YTT
For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two-digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four
Lets start with smallest prime number \(2\).

\(2^6 = 64\) ---------- (Length \(= 6\))

\(2^7\) is three digit number hence cannot be \(n\).

Therefore lets move to next prime number \(3\).

\(2^5*3^1 = 32*3 = 96\) ---------- (Length \(= 6\))

\(2^4*3^2\) will be three digit number, hence cannot be \(n\).

Therefore we have \("Two"\) two-digit positive integers which have length \(6\) \(= 64\) and \(96\)

Answer (C)...
User avatar
vivapopo
Joined: 19 Jul 2017
Last visit: 06 Jun 2020
Posts: 32
Own Kudos:
Given Kudos: 1,158
Posts: 32
Kudos: 6
Kudos
Add Kudos
Bookmarks
Bookmark this Post
YTT
For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two-digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four
2^6=64
2^5*3=96
2^4*3^2=144 out
So ans.C :thumbup:
User avatar
dave13
Joined: 09 Mar 2016
Last visit: 23 Nov 2024
Posts: 1,120
Own Kudos:
Given Kudos: 3,851
Posts: 1,120
Kudos: 1,071
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four

Basically the length of the integer is the sum of the powers of its prime factors.

Length of six means that the sum of the powers of primes of the integer (two digit) must be \(6\). First we can conclude that \(5\) can not be a factor of this integer as the smallest integer with the length of six that has \(5\) as prime factor is \(2^5*5=160\) (length=5+1=6), not a two digit integer.

The above means that the primes of the two digit integers we are looking for can be only \(2\) and/or \(3\). \(n=2^p*3^q\), \(p+q=6\) max value of \(p\) and \(q\) is \(6\).

Let's start with the highest value of \(p\):
\(n=2^6*3^0=64\) (length=6+0=6);
\(n=2^5*3^1=96\) (length=5+1=6);

\(n=2^4*3^2=144\) (length=4+2=6) not good as 144 is a three digit integer.

With this approach we see that actually \(5<=p<=6\).

Answer: C.

Hope it helps.


Bunuel but \(2^6\) is already 64 and if we multiply it by 3 we get 192

\(n=2^6*3^0=64\) how can it be equal 64 :? (length=6+0=6);
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 28 March 2025
Posts: 100,138
Own Kudos:
711,537
 [1]
Given Kudos: 92,749
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 100,138
Kudos: 711,537
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
dave13
Bunuel
For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four

Basically the length of the integer is the sum of the powers of its prime factors.

Length of six means that the sum of the powers of primes of the integer (two digit) must be \(6\). First we can conclude that \(5\) can not be a factor of this integer as the smallest integer with the length of six that has \(5\) as prime factor is \(2^5*5=160\) (length=5+1=6), not a two digit integer.

The above means that the primes of the two digit integers we are looking for can be only \(2\) and/or \(3\). \(n=2^p*3^q\), \(p+q=6\) max value of \(p\) and \(q\) is \(6\).

Let's start with the highest value of \(p\):
\(n=2^6*3^0=64\) (length=6+0=6);
\(n=2^5*3^1=96\) (length=5+1=6);

\(n=2^4*3^2=144\) (length=4+2=6) not good as 144 is a three digit integer.

With this approach we see that actually \(5<=p<=6\).

Answer: C.

Hope it helps.


Bunuel but \(2^6\) is already 64 and if we multiply it by 3 we get 192

\(n=2^6*3^0=64\) how can it be equal 64 :? (length=6+0=6);

We are not multiplying it by 3, we are multiplying by 3^0, which is 1.
User avatar
arjunnath
Joined: 16 Jul 2017
Last visit: 15 Nov 2021
Posts: 37
Own Kudos:
Given Kudos: 347
Location: India
Concentration: Marketing, General Management
GMAT 1: 590 Q44 V27
GPA: 2.94
WE:Marketing (Advertising and PR)
GMAT 1: 590 Q44 V27
Posts: 37
Kudos: 31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We start with taking the least possible two digit number with the length of 6.

In such a case it is,

2*2*2*2*2*2 = 64

We slowly move upwards by increasing the prime factor

2*2*2*2*2*3 = 96

Since taking any other greater prime factor, (5) our number will become 160 which is a three digit number, we stop at 3.

Hence, the answer is TWO

C
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,769
Own Kudos:
33,159
 [1]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,769
Kudos: 33,159
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
YTT
For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two-digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four

Let's first find the smallest value with length 6.
This is the case when each prime factor is 2.
We get 2 x 2 x 2 x 2 x 2 x 2 = 64. This is a 2-digit positive integer. PERFECT

To find the next largest number with length 6, we'll replace one 2 with a 3
We get 3 x 2 x 2 x 2 x 2 x 2 = 96. This is a 2-digit positive integer. PERFECT

To find the third largest number with length 6, we'll replace another 2 with a 3
We get 3 x 3 x 2 x 2 x 2 x 2 = 144. This is a 3-digit positive integer. NO GOOD

So there are only two two-digit positive integers with length 6.

Answer: C

Cheers,
Brent
User avatar
prateekchugh
Joined: 05 Aug 2017
Last visit: 27 Sep 2021
Posts: 360
Own Kudos:
512
 [1]
Given Kudos: 277
Location: India
Concentration: Strategy, Marketing
WE:Engineering (Energy)
Posts: 360
Kudos: 512
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two-digit positive integers have length 6?

According to the question,
n=\( a^p*b^q*c^r\)
Length = p+q+r

We need to find two-digit positive integers whose length=6; means, p+q=6

Case 1) \(2^6\); length=6; n=64
Case 2) \(2^5 * 3^1\); length=6; n=96
Case 3) \(2^4 * 3^2\); length=6; n=144 Three-Digit Incorrect
Case 5) \(2^3 * 3^3\); length=6; n=216 Three- Digit Incorrect
As we increase, the value of n will increase beyond 2-digit positive integer


A. None
B. One
C. Two Answer
D. Three
E. Four
avatar
tkorzhan1995
Joined: 16 Oct 2021
Last visit: 30 Aug 2022
Posts: 136
Own Kudos:
Given Kudos: 22
Location: Canada
Posts: 136
Kudos: 15
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATNinja, I determined that the following options are possible:
2^5, ---> (5+1)=6
2^2*3^1--> (2+1)(1+1)=6
2*3^2--> (1+1)*(2+1)=6

Hence, there are 3 possibilities in total. Could you please explain where I am getting wrong with my analysis?
 1   2   
Moderators:
Math Expert
100138 posts
PS Forum Moderator
522 posts