Last visit was: 23 Jul 2024, 13:41 It is currently 23 Jul 2024, 13:41
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
SORT BY:
Date
Tags:
Show Tags
Hide Tags
User avatar
Director
Director
Joined: 14 May 2006
Posts: 706
Own Kudos [?]: 810 [35]
Given Kudos: 0
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 94589
Own Kudos [?]: 643354 [9]
Given Kudos: 86728
Send PM
General Discussion
User avatar
VP
VP
Joined: 20 Nov 2005
Posts: 1487
Own Kudos [?]: 1143 [3]
Given Kudos: 0
Concentration: Strategy, Entrepreneurship
Schools:Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
 Q50  V34
Send PM
User avatar
Senior Manager
Senior Manager
Joined: 06 May 2006
Posts: 387
Own Kudos [?]: 77 [1]
Given Kudos: 2
Send PM
[#permalink]
1
Kudos
Should this be E?

(Unit digit of p^3) - (Unit digit of p^2) = 0 if the units digit is 0, 1, 5, 6.

The answer in these cases would be 3, 4, 8, 9.

Even if 0 is neither posive or negative, we still have 3 options left.

However, only 9 is mentioned from among these choices, so could be D as well... any suggestions? :roll:
User avatar
Director
Director
Joined: 02 Jun 2006
Posts: 664
Own Kudos [?]: 213 [0]
Given Kudos: 0
Send PM
[#permalink]
E It cannot be determined from the given information.

As p is even,
p = 10x+y

y = {0,2,4,6,8}

Unit's digit of p^3 = {0, 4, 6}
Unit's digit of p^2 = {0,2,4, 6, 8}

As unit's digit p^3 -unit's digit p^2 =0

Common units digit between p^3 and p^2 = {0, 4,6}

We can get {0,4,6} in p^2 or p^3 from {0,2,4,6,8}

So cannot determine.
Answer: E
User avatar
VP
VP
Joined: 20 Nov 2005
Posts: 1487
Own Kudos [?]: 1143 [2]
Given Kudos: 0
Concentration: Strategy, Entrepreneurship
Schools:Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
 Q50  V34
Send PM
[#permalink]
2
Kudos
ps_dahiya wrote:
A

Let x is the unit digit of p. Then

Unit digit of p^3 = unit digit of x^3
Unit digit of p^2 = unit digit of x^2

Their difference is 0 (As given). This is possible only if unit digit is 1 or 0. But since p is an even integer. So unit digit of p =0
hence unit digit of p+3 = 3

What the heck I was thinking :wall

Answer should be D because last digit of p must be +ve and even.
User avatar
Director
Director
Joined: 29 Dec 2005
Posts: 562
Own Kudos [?]: 176 [0]
Given Kudos: 0
Send PM
Re: PS: Digits again [#permalink]
u2lover wrote:
Given that p is a positive even integer with a positive units digit, if the units digit of p^3 minus the units digit of p^2 is equal to 0, what is the units digit of p + 3?


A 3
B 6
C 7
D 9
E It cannot be determined from the given information.

please explain


unit digits could be 0, 1, 5 and 6 but only 6 satisfies the conditions given in the question. so D make sense..
User avatar
Manager
Manager
Joined: 26 Jun 2006
Posts: 69
Own Kudos [?]: 14 [1]
Given Kudos: 0
Send PM
[#permalink]
1
Kudos
From the condition,

p is even and not divisible by 10 (because units digits is positive)

so p=2k, and k can not be multiple of 5 or 10

p^3-p^2 = p^2(p-1) = (2k)^2*(2k-1)

Since we know that units digit of p^3-p^2 is 0 and k con not be 5 or 10,
the only way to get 0 at the end of the number is if 2k-1=5

=> 2k=6 => p=6 => p+3=9

(D)
User avatar
Director
Director
Joined: 02 Jun 2006
Posts: 664
Own Kudos [?]: 213 [0]
Given Kudos: 0
Send PM
[#permalink]
Hey U2...
Can you tell me where I missed this question?

My solution is below... goes without saying got it wrong...

haas_mba07 wrote:
E It cannot be determined from the given information.

As p is even,
p = 10x+y

y = {0,2,4,6,8}

Unit's digit of p^3 = {0, 4, 6}
Unit's digit of p^2 = {0,2,4, 6, 8}

As unit's digit p^3 -unit's digit p^2 =0

Common units digit between p^3 and p^2 = {0, 4,6}

We can get {0,4,6} in p^2 or p^3 from {0,2,4,6,8}

So cannot determine.
Answer: E
User avatar
Manager
Manager
Joined: 20 Mar 2006
Posts: 104
Own Kudos [?]: 10 [0]
Given Kudos: 0
Send PM
Re: PS: Digits again [#permalink]
u2lover wrote:
Given that p is a positive even integer with a positive units digit, if the units digit of p^3 minus the units digit of p^2 is equal to 0, what is the units digit of p + 3?


A 3
B 6
C 7
D 9
E It cannot be determined from the given information.

please explain


If units digit = 6 then units(216)-units(36) = 0
p+3 = 9

Hence D

Heman
User avatar
Manager
Manager
Joined: 20 Mar 2006
Posts: 104
Own Kudos [?]: 10 [2]
Given Kudos: 0
Send PM
[#permalink]
1
Kudos
1
Bookmarks
haas_mba07 wrote:
Hey U2...
Can you tell me where I missed this question?

My solution is below... goes without saying got it wrong...

haas_mba07 wrote:
E It cannot be determined from the given information.

As p is even,
p = 10x+y

y = {0,2,4,6,8}

Unit's digit of p^3 = {0, 4, 6}
Unit's digit of p^2 = {0,2,4, 6, 8}

As unit's digit p^3 -unit's digit p^2 =0

Common units digit between p^3 and p^2 = {0, 4,6}

We can get {0,4,6} in p^2 or p^3 from {0,2,4,6,8}

So cannot determine.
Answer: E


Since Q says unit digit has to be +ve 0 is out
if 4 is units digit and say # is 4--> 4^3 = 64 & 4^2=16 . Subrtacting units digit does not result in 0.

Try the rest. Only when 6 is in the units digit will 6^3= 216 , 6^2=36. Subtracting units digit will result in zero.

Hence reqd answer = 6+3 = 9 Choice D

Heman
User avatar
Director
Director
Joined: 14 May 2006
Posts: 706
Own Kudos [?]: 810 [0]
Given Kudos: 0
Send PM
[#permalink]
yes... zero is neither positive nor negative so you must leave it alone

I just picked p=2, 4, 6 and got p^3-p^2= 4, 8, 0

p=6 and p+3=9
User avatar
Director
Director
Joined: 30 Mar 2006
Posts: 894
Own Kudos [?]: 604 [0]
Given Kudos: 0
Send PM
[#permalink]
P is a even number with a positive number in the unit digit
Hence unit digit can be 2,4,6 or 8

2^2 = 4 2^3 = 8
4^2 = 6 4^3 = 4
6^2 = 6 6^3 = 6
8^2 = 4 8^3 = 2

Hence only 6 in the unit digit satisfies the condition given.

hence unit digit of p+3 = 9
User avatar
Intern
Intern
Joined: 24 Mar 2013
Posts: 49
Own Kudos [?]: 17 [0]
Given Kudos: 10
Send PM
Re: Given that p is a positive even integer with a positive [#permalink]
Argh, stung again by not reading the question correctly.

Thought it could have been 5 or 6, so picked e.

Until I reviewed the answer and re-read them q stem.
avatar
SVP
SVP
Joined: 27 Dec 2012
Status:The Best Or Nothing
Posts: 1558
Own Kudos [?]: 7303 [0]
Given Kudos: 193
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Send PM
Re: Given that p is a positive even integer with a positive [#permalink]
actionj wrote:
Argh, stung again by not reading the question correctly.

Thought it could have been 5 or 6, so picked e.

Until I reviewed the answer and re-read them q stem.



As stated above; numbers ending with 0, 1 , 5 , 6 have squares, cubes or any high power returning the same units digit at the end

Example:

5 ^ 10000 would also have 5 at the end
User avatar
Tutor
Joined: 20 Aug 2015
Posts: 349
Own Kudos [?]: 1410 [0]
Given Kudos: 10
Location: India
GMAT 1: 760 Q50 V44
Send PM
Given that p is a positive even integer with a positive [#permalink]
Expert Reply
Given: P is an even positive number (i)
and units digit of p^3 - units digit of p^2= 0 (ii)

Required: Units digit of p+3

For (i), The units digit can be 2,4,6 or 8

For (ii), the numbers for which the units digit of cube and square is same are {0, 1, 5, 6}
But we know that p is positive and even.
Hence p = 6

p+3 = 9 Option D
Intern
Intern
Joined: 26 Jul 2016
Posts: 35
Own Kudos [?]: 22 [0]
Given Kudos: 21
Location: India
Send PM
Re: If positive even integer p has a positive units digit and the units di [#permalink]
Units digit of cube and square should be equal.
3 units digit satisfy the above condition- 1, 5, 6
p+3 for above units digit are 4,8,9.
Among the options 9 is there.
Hence answer D) 9
Manager
Manager
Joined: 13 Apr 2010
Posts: 67
Own Kudos [?]: 48 [0]
Given Kudos: 16
Send PM
Re: If positive even integer p has a positive units digit and the units di [#permalink]
obliraj wrote:
Units digit of cube and square should be equal.
3 units digit satisfy the above condition- 1, 5, 6
p+3 for above units digit are 4,8,9.
Among the options 9 is there.
Hence answer D) 9



I understand that units digits must end in 1 , 5 , and 6 .
Therefore, P + 3 will be 4 , 8 , and 9
But as there are three options , we will not be able to determine a unique answer .
So the answer must be E even though one of the options (that is, 9) is mentioned .
If E would have be any other digit than 4 or 8 then answer would have been D .

Your comments .
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11485
Own Kudos [?]: 34565 [1]
Given Kudos: 325
Send PM
Re: If positive even integer p has a positive units digit and the units di [#permalink]
1
Kudos
Expert Reply
ziyuenlau wrote:
If positive even integer p has a positive units digit and the units digit of (\(p^{3} – p^{2}\)\(\)) is equal to 0, what is the units digit of the quantity \(p + 3\)?

A. 3
B. 6
C. 7
D. 9
E. It cannot be determined from the given information.



Hi sb0541 and obliraj

There is only one possible answer out of 1,5, and 6..
Do not forget the info given.. p is a positive even integer so 6 is the only possibility
And thus answer is 6+3=9..

\(p^3-p^2\) has units digit as 0... So p^3 and p^2 should have same digit..
Possible values 1, 5, 0, and 6...
we are looking for POSITIVE even integers
0 and 6 are even and only 6 is positive and even..
So p=6, then p+3=6+3=9

D
Current Student
Joined: 14 Nov 2016
Posts: 1169
Own Kudos [?]: 20984 [0]
Given Kudos: 926
Location: Malaysia
Concentration: General Management, Strategy
GMAT 1: 750 Q51 V40 (Online)
GPA: 3.53
Send PM
If positive even integer p has a positive units digit and the units di [#permalink]
ziyuenlau wrote:
If positive even integer p has a positive units digit and the units digit of (\(p^{3} – p^{2}\)\(\)) is equal to 0, what is the units digit of the quantity \(p + 3\)?

A. 3
B. 6
C. 7
D. 9
E. It cannot be determined from the given information.


Official answer from Manhattan Prep.

The variable p is a positive even integer and the units digit of this number is also positive. Therefore, the units digit has four possible values: 2, 4, 6, or 8. (O is not a possibility, since 0 is not positive.)

This is useful information because the units digit of a number is the only value that determines the new units digit when that number is squared or cubed. For instance \(1,394^{2} =\) some very large number whose units digit is 6, because \(4^{2} = 16\).

In other words, only the units digit of \(p\) will determine the units digit of \(p^{3}\) and \(p^{2}\) . Further, since the units digit of \(p^{3} – p^{2} = 0\), you’re looking for a circumstance in which the square of the units digit is equal to the cube of the units digit.

When does this occur? Test the possible units digits 2, 4, 6, and 8 to see.

Only when \(p = 6\) are the units digits of \(p^{3}\) and \(p^{2}\) equal—and therefore the units digit of  \(p^{3} – p^{2} = 0\). As a result, the units digit of p must equal 6, and the units digit of \(p + 3\) must equal 9.

The correct answer is (D).
Attachments

Untitled.jpg
Untitled.jpg [ 20.11 KiB | Viewed 12309 times ]

GMAT Club Bot
If positive even integer p has a positive units digit and the units di [#permalink]
 1   2   
Moderator:
Math Expert
94589 posts