Last visit was: 20 Jun 2025, 16:11 It is currently 20 Jun 2025, 16:11
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
User avatar
sparky
Joined: 18 Apr 2005
Last visit: 30 Jul 2005
Posts: 321
Own Kudos:
100
 [82]
Location: Canuckland
Posts: 321
Kudos: 100
 [82]
6
Kudos
Add Kudos
76
Bookmarks
Bookmark this Post
Most Helpful Reply
avatar
PareshGmat
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,538
Own Kudos:
7,857
 [18]
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,538
Kudos: 7,857
 [18]
9
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
avatar
niel1989
Joined: 24 Sep 2015
Last visit: 21 Jun 2024
Posts: 3
Own Kudos:
23
 [18]
Given Kudos: 3
Posts: 3
Kudos: 23
 [18]
9
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
General Discussion
User avatar
Vithal
Joined: 01 Feb 2003
Last visit: 02 Jan 2020
Posts: 406
Own Kudos:
714
 [9]
Location: Hyderabad
 Q49  V35
Posts: 406
Kudos: 714
 [9]
5
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
sparky
How many digits 2^100 has?

A) 31
B) 35
C) 50
D) 99
E) 101


hmm....am sure there is a better way to do this, but -

2^10 = 1.024 * 10^3 => 2^100 = (1.024)^10 * 10^30

therefore 31 digits would be my best guess
User avatar
sparky
Joined: 18 Apr 2005
Last visit: 30 Jul 2005
Posts: 321
Own Kudos:
100
 [5]
Location: Canuckland
Posts: 321
Kudos: 100
 [5]
2
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
Vithal
sparky
How many digits 2^100 has?

A) 31
B) 35
C) 50
D) 99
E) 101

hmm....am sure there is a better way to do this, but -

2^10 = 1.024 * 10^3 => 2^100 = (1.024)^10 * 10^30

therefore 31 digits would be my best guess


cool way to do it, i didn't think about this one

I did it like this

at max 1 digit can be added after every third power, eg 2 4 8, 16 32 64, 128 etc
so max possible digit based on this observation can be (100 -1)/3 + 1 = 34
so the answer 31 comes naturally.
User avatar
AJB77
Joined: 30 May 2005
Last visit: 23 Sep 2008
Posts: 236
Own Kudos:
51
 [1]
Posts: 236
Kudos: 51
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
I solved it like sparky did - 2^100 = (2^3)^33 * 2

Number of digits is less than 34. So A is the only choice by elimination.
avatar
aishwarya276981
Joined: 29 Mar 2015
Last visit: 15 Dec 2015
Posts: 3
Own Kudos:
3
 [3]
Given Kudos: 191
Posts: 3
Kudos: 3
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
2^2=1 digit
2^3=1 digit
2^4=1 digit
2^5=2 digit
2^6=2 digit
2^7=3 digit
2^8=3 digit
2^9=3 digit
2^10=4 digits - so 2^20 will have 7 digits which is 3 digits difference -> 2^30 will have 10 digits and so on.. so 2^100 will have 31 digits :lol:
User avatar
narendran1990
Joined: 24 May 2014
Last visit: 09 Jun 2024
Posts: 78
Own Kudos:
24
 [1]
Given Kudos: 989
Location: India
GMAT 1: 640 Q42 V35 (Online)
GRE 1: Q159 V151
GRE 2: Q159 V153
GPA: 2.9
GMAT 1: 640 Q42 V35 (Online)
GRE 1: Q159 V151
GRE 2: Q159 V153
Posts: 78
Kudos: 24
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PareshGmat

Can you elaborate as to how (10^3+24)^10 equates to the last step.
User avatar
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,331
Own Kudos:
3,773
 [2]
Given Kudos: 816
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
GMAT 1: 750 Q49 V44
Posts: 2,331
Kudos: 3,773
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
narendran1990
PareshGmat

Can you elaborate as to how (10^3+24)^10 equates to the last step.

Let me try to explain.

\((10^3+24)^{10}\) , by binomial expansion of \((a+b)^n\) = \(a^n+a^{n-1}*b......+a*b^{n-1}+b^n\)

Similarly,

\((10^3+24)^{10} = 10^{30}+....+10*24^9+24^{10}\)

\(10^{29}*24+.....+24^{10}\) will have digits of the same order as that for \(10^{30}\) to reach a total of 31 digits.

Options C-E are way too much for this question and hence are easily eliminated.
User avatar
saroshgilani
Joined: 21 Mar 2014
Last visit: 20 Sep 2016
Posts: 15
Own Kudos:
16
 [1]
Given Kudos: 13
Posts: 15
Kudos: 16
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
2^1 = 1 digit
2^2 = 1 digit
2^3 = 1 digit
2^4, 2^5, 2^6 has 2 digits
2^7, 2^8, 2^9 has 3 digits
there is a repetition of 3.
is there a way to solve this question through this approach?

my answer comes out to be 34 though....
100/3 +1
User avatar
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,331
Own Kudos:
3,773
 [4]
Given Kudos: 816
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
GMAT 1: 750 Q49 V44
Posts: 2,331
Kudos: 3,773
 [4]
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
saroshgilani
Bunuel
2^1 = 1 digit
2^2 = 1 digit
2^3 = 1 digit
2^4, 2^5, 2^6 has 2 digits
2^7, 2^8, 2^9 has 3 digits
there is a repetition of 3.
is there a way to solve this question through this approach?

my answer comes out to be 34 though....
100/3 +1

The mistake you are doing is that you are assuming that the 'cyclicity' is actually 3. This is wrong

2^6 = 2 digits, 2^7 to 2^9= 3 digits BUT 2^10 to 2^14 = 4 digits (this breaks the cyclicity).

Thus, you can not use cyclicity for this question.

Easiest way is to use log function as mentioned above (but GMAT doesnt want you to know how log function works. If you do know it, becomes very straightforward). If not, use binomial theorem as mentioned in my post how-many-digits-2-100-has-17192.html#p1579232\

Alternately, you can solve it as :

2^100 = (2^10)^10 = 1024^10 = (1.024)^10* (1000)^10 = (a value just slightly greater than 1)*10^30, giving you a number with 31 digits.

Hope this helps.
User avatar
GMATDemiGod
Joined: 23 Sep 2015
Last visit: 05 Feb 2017
Posts: 64
Own Kudos:
52
 [6]
Given Kudos: 213
Concentration: General Management, Finance
GMAT 1: 680 Q46 V38
GMAT 2: 690 Q47 V38
GPA: 3.5
GMAT 2: 690 Q47 V38
Posts: 64
Kudos: 52
 [6]
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
yeah i used the log function for this as well.

Very quick and easy

quickly Log2^100 is 100Log2 = 100* Log(2)


I would remember a few common logs in case a question like this pops up

Log(2) = 0.305
Log(3)=0.477
Log(4)=Log(4^2)=2Log(2)
Log(5)=0.698
Log(6)=Log(2*3)=Log(2)=Log(3)
Log(7)=0.845

Now, as we can seem any simple common base we can use this trick for, unless they give us something a bit more complicated like 13 or 17, then I am out of tricks. But this is good for numerous ways this question can be asked.
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 20 Jun 2025
Posts: 16,059
Own Kudos:
73,819
 [2]
Given Kudos: 472
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,059
Kudos: 73,819
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
saroshgilani
Bunuel
2^1 = 1 digit
2^2 = 1 digit
2^3 = 1 digit
2^4, 2^5, 2^6 has 2 digits
2^7, 2^8, 2^9 has 3 digits
there is a repetition of 3.
is there a way to solve this question through this approach?

my answer comes out to be 34 though....
100/3 +1

First of all, your AP is 1, 4, 7, 10, 13 ... 100.
So number of terms will be (100 - 1)/3 + 1 = 34. It will not be 100/3 + 1

Next what you need to understand here is that the 34 gives you the maximum number of digits that \(2^{100}\) can have.
Look at it this way. 128 -> 256 -> 512 -> 4 digit number
But this is the quickest that the number of digits can change. For some digits, you could actually have 4 numbers.
1024 -> 2048 -> 4096 ->8192 -> 5 digit number

Fortunately, the only option that is less than 34 (but more than 25) is 31. Hence it must be the answer.
avatar
williambarberjr
Joined: 13 Jan 2016
Last visit: 30 Aug 2022
Posts: 2
Own Kudos:
Given Kudos: 38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Looking through the solutions, logarithms appear to be the fastest and most accurate way to solve this problem assuming you've memorized some basic logarithms. I haven't been a member long enough or posted enough to include a link but if you search youtube for a video titled "Logarithms Example - 4 / Find The Number Of Digits using Logarithms - Maths Arithmetic" you can see a good example of using logarithms do determine the number of digits in an exponent that would take far too long to compute manually.
User avatar
dabral
User avatar
Tutor
Joined: 19 Apr 2009
Last visit: 29 Nov 2024
Posts: 557
Own Kudos:
661
 [1]
Given Kudos: 19
Affiliations: GMATQuantum
Expert
Expert reply
Posts: 557
Kudos: 661
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I am no longer active on this forum.
avatar
deeksha6
Joined: 13 Sep 2015
Last visit: 30 Apr 2017
Posts: 11
Own Kudos:
Given Kudos: 256
Posts: 11
Kudos: 3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
First of all, your AP is 1, 4, 7, 10, 13 ... 100.
So number of terms will be (100 - 1)/3 + 1 = 34. It will not be 100/3 + 1

Karishma kindly explain this step.
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 20 Jun 2025
Posts: 16,059
Own Kudos:
73,819
 [1]
Given Kudos: 472
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,059
Kudos: 73,819
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
deeksha6
First of all, your AP is 1, 4, 7, 10, 13 ... 100.
So number of terms will be (100 - 1)/3 + 1 = 34. It will not be 100/3 + 1

Karishma kindly explain this step.


How do you find the number of terms in an AP?

Number of terms (n) = (Last term - First term)/Common difference + 1

It is derived from

Last term = First term + (n - 1) * Common difference

So it will be
n = (100 - 1)/3 + 1 = 34

Check out this post for more on AP formulas:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/03 ... gressions/
User avatar
rahul16singh28
Joined: 31 Jul 2017
Last visit: 09 Jun 2020
Posts: 431
Own Kudos:
492
 [1]
Given Kudos: 752
Location: Malaysia
GPA: 3.95
WE:Consulting (Energy)
Posts: 431
Kudos: 492
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sparky
How many digits 2^100 has?

A) 31
B) 35
C) 50
D) 99
E) 101

I solved it this way -

\(2^{100} = (1000 + 24)^{10}\).... Now, when you see the first term i.e. 1000 and its power 10.. the answer should lie close to 30.... 35 isn't possible because \(24^{10}\) will always be less than \(1000^{10}\).
User avatar
HWPO
Joined: 11 May 2020
Last visit: 12 May 2024
Posts: 121
Own Kudos:
Given Kudos: 146
Posts: 121
Kudos: 17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel where can I find more questions like this?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 June 2025
Posts: 102,214
Own Kudos:
734,016
 [2]
Given Kudos: 93,961
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,214
Kudos: 734,016
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
HWPO
Bunuel where can I find more questions like this?


The questions are listed below in increasing order of difficulty. Take your time to work through each question and try different methods until you find the solution. Don't be discouraged if you struggle with some of the questions - remember that practice is key to improving your problem-solving abilities.

Problem Solving:


Data Sufficiency:


Check more similar lists in Special Questions Directory.

Hope it helps.
 1   2   
Moderators:
Math Expert
102214 posts
PS Forum Moderator
653 posts