GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 21 Jun 2018, 03:36

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If M is a positive integer, then M^3 has how many digits?

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
SVP
SVP
avatar
Joined: 04 May 2006
Posts: 1787
Schools: CBS, Kellogg
Premium Member
If M is a positive integer, then M^3 has how many digits? [#permalink]

Show Tags

New post 01 Jun 2009, 18:49
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

If M is a positive integer, then M^3 has how many digits?
(1) M has 3 digits.
(2) M^2 has 5 digits

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________

GMAT Club Premium Membership - big benefits and savings

1 KUDOS received
Manager
Manager
User avatar
Joined: 08 Feb 2009
Posts: 140
Schools: Anderson
Re: Digits [#permalink]

Show Tags

New post 01 Jun 2009, 19:47
1
(1) Let M = 100, \(M^3\) = 1000000 (7 digits)
M = 500, \(M^3\) = 125000000 (9 digits)


(2) M = 100, \(M^2\) = 10000 (5 digits), \(M^3\) = 1000000 (7 digits)
M = 300, \(M^2\) = 90000 (5 digits), \(M^3\) = 27000000 (8 digits)

Combining,

Still NOT SUFFICIENT.

Thus, E.
Manager
Manager
User avatar
Joined: 14 May 2009
Posts: 188
Schools: Stanford, Harvard, Berkeley, INSEAD
Re: Digits [#permalink]

Show Tags

New post 01 Jun 2009, 22:52
sondenso wrote:
If M is a positive integer, then M^3 has how many digits?
(1) M has 3 digits.
(2) M^2 has 5 digits


(1)

X,Y,Z are the 3 digits of M respectively, X being the hundreds, Y the tens and Z the unit.

\(M = 100X + 10Y + Z\)

\(M^{3} = (100X+10Y+Z)^{3}\)

uh oh... here we go...I did the factoring on paper

\(M^{3} = 10^{6}(X^{3}) + 10^{5}(3X^{2}Y) + 10^{4}(3X^{2}Z+3XY^{2}) + 10^{3}(6XYZ+y^{3}) + 10^2(3XZ^{2}+3Y^{2}Z) + 10^{1}(3YZ^{2}) + 10^{0}(Z^{3})\)

10^6 represents the 7th digit-- but if \(X^{3}>=10\), then we really have the 8th digit, as we'd be multiplying 10^6 by 10, which would make it the 8th digit.

So if we look at strictly the digit X, and say it's 1, then X^3 would be 1, and we'd have a 1 in the 7th digit since we're multiplying by 10^6 and putting it into the 7th digit. Now the thing we have to watch out for is in the 6th digit spot, 10^5*(3X^{2}Y), and if X=1 then it's 10^5*(3Y), but \(0<=Y<=9\), so the most we could have inside there would be 27, which would make it 10^{5}(27) = 10^{6}(2.7)
adding it together we'd get 10^{6}(1) + 10^{6}(2.7 = 10^{6}(3.7), so for sure if X is 1 we'd have 7 or fewer digits. The other digits in the sum are irrelevant, they wouldn't be big enough to affect it.

To put this into perspective,

Say M = 100 (\(X=1,Y=0,Z=0\)), then...

\(M^{3} = 10^{6}(1) + 10^{5}(0) + 10^{4}(0) + 10^{3}(0) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)
\(M^{3} = 10^{6}(1) + 0 + 0 + 0 + 0 + 0 + 0\)
\(M^{3} = 10^{6}\)
\(M^{3} = 1000000\)

M has 7 digits.

Say M = 220 (\(X=2,Y=2,Z=0\)), then...

\(M^{3} = 10^{6}(8) + 10^{5}(24) + 10^{4}(24) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{6}(10.4) + 10^{5}(2.4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{6}(10 + 4/10) + 10^{5}(2 + 4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{6}(10) + 10^{6}(4/10) + 10^{5}(2) + 10^{5}(4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{7}(1) + 10^{5}(4) + 10^{5}(2) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{7}(1) + 10^{5}(6) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

So M = 10648000

8 digits.

INSUFFICIENT (lol).

(2) \(M^2\) has 5 digits.

This one we can quickly check.

M=90 M^2 = 8100.
M=99 M^2 = 9801
M=100 M^2 = 10000
M=300 M^2 = 90000
M=400 M^2 = 160000

So M is between 100 & 400.

But now looking at the 2 examples from (1), we see that 100^3 has 7 digits 220^3 has 8 digits

INSUFFICIENT

\(M^2 = 10^{4}(X^{2}) + 10^{3}(2XY) + 10^{2}(2XZ+Y^{2}) + 10^{1}(2YZ) + 10^{0}(Z^{2})\)

(1&2)

From 2 we have that \(100<=M<400\)

From (2) we have a range for X
\(1<=X<=3\)

We can just reuse the 2 examples from (1), since they're still valid when you assume (2) to be true. From the long ass equation from (1), we can see that if M=100, M^3 has 7 digits, but M=220 has 8 digits.

INSUFFICIENT

Final Answer, E.
_________________

Hades

Intern
Intern
avatar
Joined: 12 May 2009
Posts: 46
Location: Mumbai
Re: Digits [#permalink]

Show Tags

New post 02 Jun 2009, 01:05
IMO E, INsufficient even after combining
SVP
SVP
avatar
Joined: 04 May 2006
Posts: 1787
Schools: CBS, Kellogg
Premium Member
Re: Digits [#permalink]

Show Tags

New post 02 Jun 2009, 01:09
Thanks guys, OA is E
_________________

GMAT Club Premium Membership - big benefits and savings

SVP
SVP
User avatar
Joined: 29 Aug 2007
Posts: 2427
Re: Digits [#permalink]

Show Tags

New post 02 Jun 2009, 07:33
sondenso wrote:
If M is a positive integer, then M^3 has how many digits?
(1) M has 3 digits.
(2) M^2 has 5 digits


(1) If M has 3 digits, M could be 100 or 999.
If M = 100, M^3 = 1,000,000. So 7 digits.
If M = 999, M^3 is close to = 1000,000,000. So 9 digits.

NSF.
(2) If M^2 has 5 digits, M could be 100 or 999.

If M = 100, M^2 = 10,000. So 5 digits.
If M = 999, M^2 is close to = 1,000,000. So 5 digits.

Again M could be 100 or 999. NSF>
Together also same. E.

Hades wrote:
sondenso wrote:
If M is a positive integer, then M^3 has how many digits?
(1) M has 3 digits.
(2) M^2 has 5 digits


(1)

X,Y,Z are the 3 digits of M respectively, X being the hundreds, Y the tens and Z the unit.

\(M = 100X + 10Y + Z\)

\(M^{3} = (100X+10Y+Z)^{3}\)

uh oh... here we go...I did the factoring on paper

\(M^{3} = 10^{6}(X^{3}) + 10^{5}(3X^{2}Y) + 10^{4}(3X^{2}Z+3XY^{2}) + 10^{3}(6XYZ+y^{3}) + 10^2(3XZ^{2}+3Y^{2}Z) + 10^{1}(3YZ^{2}) + 10^{0}(Z^{3})\)

10^6 represents the 7th digit-- but if \(X^{3}>=10\), then we really have the 8th digit, as we'd be multiplying 10^6 by 10, which would make it the 8th digit.

So if we look at strictly the digit X, and say it's 1, then X^3 would be 1, and we'd have a 1 in the 7th digit since we're multiplying by 10^6 and putting it into the 7th digit. Now the thing we have to watch out for is in the 6th digit spot, 10^5*(3X^{2}Y), and if X=1 then it's 10^5*(3Y), but \(0<=Y<=9\), so the most we could have inside there would be 27, which would make it 10^{5}(27) = 10^{6}(2.7)
adding it together we'd get 10^{6}(1) + 10^{6}(2.7 = 10^{6}(3.7), so for sure if X is 1 we'd have 7 or fewer digits. The other digits in the sum are irrelevant, they wouldn't be big enough to affect it.

To put this into perspective,

Say M = 100 (\(X=1,Y=0,Z=0\)), then...

\(M^{3} = 10^{6}(1) + 10^{5}(0) + 10^{4}(0) + 10^{3}(0) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)
\(M^{3} = 10^{6}(1) + 0 + 0 + 0 + 0 + 0 + 0\)
\(M^{3} = 10^{6}\)
\(M^{3} = 1000000\)

M has 7 digits.

Say M = 220 (\(X=2,Y=2,Z=0\)), then...

\(M^{3} = 10^{6}(8) + 10^{5}(24) + 10^{4}(24) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{6}(10.4) + 10^{5}(2.4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{6}(10 + 4/10) + 10^{5}(2 + 4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{6}(10) + 10^{6}(4/10) + 10^{5}(2) + 10^{5}(4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{7}(1) + 10^{5}(4) + 10^{5}(2) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{7}(1) + 10^{5}(6) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

So M = 10648000

8 digits.

INSUFFICIENT (lol).

(2) \(M^2\) has 5 digits.

This one we can quickly check.

M=90 M^2 = 8100.
M=99 M^2 = 9801
M=100 M^2 = 10000
M=300 M^2 = 90000
M=400 M^2 = 160000

So M is between 100 & 400.

But now looking at the 2 examples from (1), we see that 100^3 has 7 digits 220^3 has 8 digits

INSUFFICIENT

\(M^2 = 10^{4}(X^{2}) + 10^{3}(2XY) + 10^{2}(2XZ+Y^{2}) + 10^{1}(2YZ) + 10^{0}(Z^{2})\)

(1&2)

From 2 we have that \(100<=M<400\)

From (2) we have a range for X
\(1<=X<=3\)

We can just reuse the 2 examples from (1), since they're still valid when you assume (2) to be true. From the long ass equation from (1), we can see that if M=100, M^3 has 7 digits, but M=220 has 8 digits.

INSUFFICIENT

Final Answer, E.



Thats definitely good effort and nice way to solve the question but is little lengthy and time consuming. GMAT questions such as this one should be solved in around 2 minuets.

I like it. Good job and keep it up. 8-)
_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

SVP
SVP
avatar
Joined: 04 May 2006
Posts: 1787
Schools: CBS, Kellogg
Premium Member
Re: Digits [#permalink]

Show Tags

New post 02 Jun 2009, 18:58
GMAT TIGER wrote:
sondenso wrote:
If M is a positive integer, then M^3 has how many digits?
(1) M has 3 digits.
(2) M^2 has 5 digits


(1) If M has 3 digits, M could be 100 or 999.
If M = 100, M^3 = 1,000,000. So 7 digits.
If M = 999, M^3 is close to = 1000,000,000. So 9 digits.

NSF.
(2) If M^2 has 5 digits, M could be 100 or 999.

If M = 100, M^2 = 10,000. So 5 digits.
If M = 999, M^2 is close to = 1,000,000. So 5 digits.

Again M could be 100 or 999. NSF>
Together also same. E.

Hades wrote:
sondenso wrote:
If M is a positive integer, then M^3 has how many digits?
(1) M has 3 digits.
(2) M^2 has 5 digits


(1)

X,Y,Z are the 3 digits of M respectively, X being the hundreds, Y the tens and Z the unit.

\(M = 100X + 10Y + Z\)

\(M^{3} = (100X+10Y+Z)^{3}\)

uh oh... here we go...I did the factoring on paper

\(M^{3} = 10^{6}(X^{3}) + 10^{5}(3X^{2}Y) + 10^{4}(3X^{2}Z+3XY^{2}) + 10^{3}(6XYZ+y^{3}) + 10^2(3XZ^{2}+3Y^{2}Z) + 10^{1}(3YZ^{2}) + 10^{0}(Z^{3})\)

10^6 represents the 7th digit-- but if \(X^{3}>=10\), then we really have the 8th digit, as we'd be multiplying 10^6 by 10, which would make it the 8th digit.

So if we look at strictly the digit X, and say it's 1, then X^3 would be 1, and we'd have a 1 in the 7th digit since we're multiplying by 10^6 and putting it into the 7th digit. Now the thing we have to watch out for is in the 6th digit spot, 10^5*(3X^{2}Y), and if X=1 then it's 10^5*(3Y), but \(0<=Y<=9\), so the most we could have inside there would be 27, which would make it 10^{5}(27) = 10^{6}(2.7)
adding it together we'd get 10^{6}(1) + 10^{6}(2.7 = 10^{6}(3.7), so for sure if X is 1 we'd have 7 or fewer digits. The other digits in the sum are irrelevant, they wouldn't be big enough to affect it.

To put this into perspective,

Say M = 100 (\(X=1,Y=0,Z=0\)), then...

\(M^{3} = 10^{6}(1) + 10^{5}(0) + 10^{4}(0) + 10^{3}(0) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)
\(M^{3} = 10^{6}(1) + 0 + 0 + 0 + 0 + 0 + 0\)
\(M^{3} = 10^{6}\)
\(M^{3} = 1000000\)

M has 7 digits.

Say M = 220 (\(X=2,Y=2,Z=0\)), then...

\(M^{3} = 10^{6}(8) + 10^{5}(24) + 10^{4}(24) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{6}(10.4) + 10^{5}(2.4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{6}(10 + 4/10) + 10^{5}(2 + 4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{6}(10) + 10^{6}(4/10) + 10^{5}(2) + 10^{5}(4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{7}(1) + 10^{5}(4) + 10^{5}(2) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

\(M^{3} = 10^{7}(1) + 10^{5}(6) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)\)

So M = 10648000

8 digits.

INSUFFICIENT (lol).

(2) \(M^2\) has 5 digits.

This one we can quickly check.

M=90 M^2 = 8100.
M=99 M^2 = 9801
M=100 M^2 = 10000
M=300 M^2 = 90000
M=400 M^2 = 160000

So M is between 100 & 400.

But now looking at the 2 examples from (1), we see that 100^3 has 7 digits 220^3 has 8 digits

INSUFFICIENT

\(M^2 = 10^{4}(X^{2}) + 10^{3}(2XY) + 10^{2}(2XZ+Y^{2}) + 10^{1}(2YZ) + 10^{0}(Z^{2})\)

(1&2)

From 2 we have that \(100<=M<400\)

From (2) we have a range for X
\(1<=X<=3\)

We can just reuse the 2 examples from (1), since they're still valid when you assume (2) to be true. From the long ass equation from (1), we can see that if M=100, M^3 has 7 digits, but M=220 has 8 digits.

INSUFFICIENT

Final Answer, E.



Thats definitely good effort and nice way to solve the question but is little lengthy and time consuming. GMAT questions such as this one should be solved in around 2 minuets.

I like it. Good job and keep it up. 8-)


For the basic guys like me, this approach is very preferable! Thanks Hades
_________________

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Senior Manager
User avatar
Joined: 15 Jan 2008
Posts: 263
Re: Digits [#permalink]

Show Tags

New post 03 Jun 2009, 23:45
My approach..

1. m is of 3 digits.

cube 100 has 7 digits.
cube 200 has 7 digits but cube 300 has 8 digits.. as 27 adds 2 didits...

2. square M has 5 digits..

square 100 has 5 sigits.. cube 100 has 7 digits..
sqauree 300 has 5 digits.. but cube 300 has 8 digits..

combined 1 and 2, ( actually the second condition already combines the first one..)

still E.
GMAT Instructor
avatar
B
Joined: 04 Jul 2006
Posts: 1253
Location: Madrid
CAT Tests
Re: Digits [#permalink]

Show Tags

New post 05 Jun 2009, 13:53
sondenso wrote:
If M is a positive integer, then M^3 has how many digits?
(1) M has 3 digits.
(2) M^2 has 5 digits


(1) 100 < = m < 1000
Thus 10^6 <= m < 10^9

m^3 could have 7,8 or 9 digits

Not suff

(2) 10000 < = M^2 < 100000
10^2 < M < = sqrt (10) x 10^2
10^6 < = m^3 < 10sqrt(10) x 10^6

m^3 could have 7 or 8 digits

Not suff

Together not suff
1 KUDOS received
Manager
Manager
avatar
Joined: 11 Aug 2008
Posts: 133
Re: Digits [#permalink]

Show Tags

New post 09 Jun 2009, 00:42
1
goldeneagle94 wrote:
(1) Let M = 100, \(M^3\) = 1000000 (7 digits)
M = 500, \(M^3\) = 125000000 (9 digits)


(2) M = 100, \(M^2\) = 10000 (5 digits), \(M^3\) = 1000000 (7 digits)
M = 300, \(M^2\) = 90000 (5 digits), \(M^3\) = 27000000 (8 digits)

Combining,

Still NOT SUFFICIENT.

Thus, E.

In this kind of question, just give example... + 1 Kudo for this post

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: Digits   [#permalink] 09 Jun 2009, 00:42
Display posts from previous: Sort by

If M is a positive integer, then M^3 has how many digits?

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.