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

 It is currently 16 Oct 2019, 08:28 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

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.  If p is a positive integer, what is the greatest integer that must be

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Manager  S
Joined: 09 Jul 2018
Posts: 63
If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

8 00:00

Difficulty:   85% (hard)

Question Stats: 41% (02:10) correct 59% (01:58) wrong based on 159 sessions

HideShow timer Statistics

If p is a positive integer, what is the greatest integer that must be a factor of p^6 - p^2?

A. 5
B. 12
C. 19
D. 25
E. 16

Originally posted by ritu1009 on 12 Dec 2018, 04:17.
Last edited by ritu1009 on 12 Dec 2018, 04:25, edited 2 times in total.
Director  V
Status: Manager
Joined: 27 Oct 2018
Posts: 673
Location: Egypt
Concentration: Strategy, International Business
GPA: 3.67
WE: Pharmaceuticals (Health Care)
If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

I started plugging numbers:

if p = 1, the result would be zero, and any number can be claimed as a factor of zero.

if p = 2, 2^6 - 2^2 = 60, which eliminates options C,D,E instantly.
5 and 12 are factors of 60, but 12 is bigger.

so B.
_________________
Thanks for Kudos

Originally posted by MahmoudFawzy on 12 Dec 2018, 13:29.
Last edited by MahmoudFawzy on 15 Dec 2018, 11:26, edited 1 time in total.
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

1
1
Hi there!

Let me contribute with some observations in this matter:

01. It is easy to prove that the expression given may be factorized as p^2 (p^2+1) (p+1) (p-1) and from that, to conclude that this expression is always divisible by 12 (for p positive integer).

02. The fact is that p^6 - p^2 = p (p^5-p) and it can be proved that p^5 - p is always divisible by 5 (for p positive integer) (*), therefore the correct answer is 12*5 = 60, not among the alternative choices.

Mahmoudfawzy83 : 3^6 -3^2 is 720

Regards,
Fabio.

P.S.: (*) Just consider the factorization of (p^5 - p) and take into account that p must be equal to 5M, 5M+1, 5M+2, 5M+3 or 5M+4 , where M is integer. All cases are trivial.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Director  V
Status: Manager
Joined: 27 Oct 2018
Posts: 673
Location: Egypt
Concentration: Strategy, International Business
GPA: 3.67
WE: Pharmaceuticals (Health Care)
Re: If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

fskilnik wrote:
Hi there!

Let me contribute with some observations in this matter:

01. It is easy to prove that the expression given may be factorized as p^2 (p^2+1) (p+1) (p-1) and from that, to conclude that this expression is always divisible by 12 (for p positive integer).

02. The fact is that p^6 - p^2 = p (p^5-p) and it can be proved that p^5 - p is always divisible by 5 (for p positive integer) (*), therefore the correct answer is 12*5 = 60, not among the alternative choices.

Mahmoudfawzy83 : 3^6 -3^2 is 720

Regards,
Fabio.

P.S.: (*) Just consider the factorization of (p^5 - p) and take into account that p must be equal to 5M, 5M+1, 5M+2, 5M+3 or 5M+4 , where M is integer. All cases are trivial.

I answered in right by mistake _________________
Thanks for Kudos
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

Mahmoudfawzy83 wrote:

I answered in right by mistake Hello again , Mahmoudfawzy83 !

Thanks for the kudos.

I would like to congratulate you for your approach, in my opinion the BEST way to manage GMAT Problem Solving problems like this one.

I explain:

You may (and SHOULD) suppose the problem is well-posed, in the sense that there is ONLY one CORRECT alternative choice among the five given.

In other words, your reasoning for refuting some alternative choices based on a particular case exploration is GREAT.

Among the "survivors", it is also commendable to FOCUS on the property asked (maximum value), but in this case just one small detail is important:

To be sure 12 "would be" the right answer, 12 must be a divisor of EVERY expression p^6-p^2, I mean, of p^6-p^2 for every value of positive integer p. That´s the case!

(The fact that 60 is not among the possibilities, is an examiner´s fault. It is his/her burden to guarantee that you, the test taker, is offered the right answer among the ones given!)

I hope you (and other readers) benefit from those details!

Regards and success in your studies,
Fabio.

P.S.: "any number can be claimed as a factor of zero." must be understood as "any integer different from zero can be claimed as a factor of zero".
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Intern  B
Joined: 11 Mar 2017
Posts: 11
If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

Hi,

Can you please explain solution "01"?

How were you able to determine that p^2 (p^2+1) (p+1) (p-1) will be divisble by 12?

i solved by expressing (p^2-1) (p^2) (p^2+1) where 25 can also fit in considering p may be 5.

Where am i wrong here.

Thanks

fskilnik wrote:
Hi there!

Let me contribute with some observations in this matter:

01. It is easy to prove that the expression given may be factorized as p^2 (p^2+1) (p+1) (p-1) and from that, to conclude that this expression is always divisible by 12 (for p positive integer).

02. The fact is that p^6 - p^2 = p (p^5-p) and it can be proved that p^5 - p is always divisible by 5 (for p positive integer) (*), therefore the correct answer is 12*5 = 60, not among the alternative choices.

Mahmoudfawzy83 : 3^6 -3^2 is 720

Regards,
Fabio.

P.S.: (*) Just consider the factorization of (p^5 - p) and take into account that p must be equal to 5M, 5M+1, 5M+2, 5M+3 or 5M+4 , where M is integer. All cases are trivial.
Director  V
Status: Manager
Joined: 27 Oct 2018
Posts: 673
Location: Egypt
Concentration: Strategy, International Business
GPA: 3.67
WE: Pharmaceuticals (Health Care)
If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

Hi,

Can you please explain solution "01"?

How were you able to determine that p^2 (p^2+1) (p+1) (p-1) will be divisble by 12?

i solved by expressing (p^2-1) (p^2) (p^2+1) where 25 can also fit in considering p may be 5.

Where am i wrong here.

Thanks

fskilnik wrote:
Hi there!

Let me contribute with some observations in this matter:

01. It is easy to prove that the expression given may be factorized as p^2 (p^2+1) (p+1) (p-1) and from that, to conclude that this expression is always divisible by 12 (for p positive integer).

02. The fact is that p^6 - p^2 = p (p^5-p) and it can be proved that p^5 - p is always divisible by 5 (for p positive integer) (*), therefore the correct answer is 12*5 = 60, not among the alternative choices.

Mahmoudfawzy83 : 3^6 -3^2 is 720

Regards,
Fabio.

P.S.: (*) Just consider the factorization of (p^5 - p) and take into account that p must be equal to 5M, 5M+1, 5M+2, 5M+3 or 5M+4 , where M is integer. All cases are trivial.

25 can fit if p = 5. that's correct , but can't always fit.
in the question: MUST is a keyword

I am tagging fskilnik as well so we can learn from him
_________________
Thanks for Kudos

Originally posted by MahmoudFawzy on 20 Dec 2018, 10:30.
Last edited by MahmoudFawzy on 20 Dec 2018, 13:30, edited 1 time in total.
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

1
2
Hi Sajjad1093 and Mahmoudfawzy83 ,

It is a pleasure to help you both understand my reasoning in my statement (1):

01. It is easy to prove that the expression given may be factorized as p^2 (p^2+1) (p+1) (p-1) and from that, to conclude that this expression is always divisible by 12 (for p positive integer).

Consider ANY positive integer p fixed. For this value of p, we have:

$${p^6} - {p^2} = {p^2}\left( {{p^4} - 1} \right) = {p^2}\left( {{p^2} + 1} \right)\left( {{p^2} - 1} \right) = {p^2}\left( {{p^2} + 1} \right)\left( {p + 1} \right)\left( {p - 1} \right) = A\left( p \right) \cdot B\left( p \right)$$

$$A\left( p \right) = \left( {p - 1} \right)p\left( {p + 1} \right)\,\,{\rm{is}}\,\,{\rm{even}}\,\,\left( {{\rm{consecutive}}\,\,{\rm{integers}}} \right)\,\,{\rm{and}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\left( {3\,\,{\rm{consecutive}}\,\,{\rm{integers}}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,A\left( p \right) = 6M,\,\,M\,\,{\mathop{\rm int}}$$

$$B\left( p \right) = p\left( {{p^2} + 1} \right)\,\,{\rm{is}}\,\,{\rm{even}}\,\,\,\,\left( {{\rm{if}}\,\,p\,\,{\rm{odd}}\,,\,\,{p^2}\,\,{\rm{is}}\,\,{\rm{odd}}\,\,{\rm{hence}}\,\,{p^2} + 1\,\,{\rm{is}}\,\,{\rm{even}}} \right)\,\,\,\, \Rightarrow \,\,\,\,B\left( p \right) = 2N,\,\,N\,\,{\mathop{\rm int}}$$

$${p^6} - {p^2} = 12MN\,\,,\,\,MN\,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,{p^6} - {p^2}\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,12\,\,\,\,$$

Please tag me again, if needed.

Regards and success in your studies,
Fabio.

P.S.: if you want my explanation for my statement (2), tag me about it, too.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Director  V
Status: Manager
Joined: 27 Oct 2018
Posts: 673
Location: Egypt
Concentration: Strategy, International Business
GPA: 3.67
WE: Pharmaceuticals (Health Care)
Re: If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

Thank fskilnik

please, can you explain how p^6 - P^4 is divisible by 5?
_________________
Thanks for Kudos
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

1
Mahmoudfawzy83 wrote:
Thank fskilnik

please, can you explain how p^6 - p^2 is divisible by 5?

Sure, let´s prove my statement 02., repeated below for convenience:

02. The fact is that p^6 - p^2 = p (p^5-p) and it can be proved that p^5 - p is always divisible by 5 (for p positive integer).

Consider ANY positive integer p fixed.

We know (from the Division Algorithm, or simply "remainders understanding") that p must be of one (only) of the following possibilities:
p = 5M , 5M+1 , 5M+2 , 5M+3 or 5M+4 , where M is a (nonnegative) integer.

$${p^5} - p = p\left( {{p^4} - 1} \right) = p\left( {{p^2} + 1} \right)\left( {p + 1} \right)\left( {p - 1} \right)$$

$$p = 5M\,\,\,\, \Rightarrow \,\,\,\,\,{p^5} - p = 5M\left( {{p^2} + 1} \right)\left( {p + 1} \right)\left( {p - 1} \right) = 5M \cdot {\mathop{\rm int}} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{done}}$$

$$p = 5M + 1\,\,\,\, \Rightarrow \,\,\,\,\,{p^5} - p = p\left( {{p^2} + 1} \right)\left( {p + 1} \right)\left( {5M} \right) = 5M \cdot {\mathop{\rm int}} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{done}}$$

$$p = 5M + 2\,\,\,\, \Rightarrow \,\,\,\,\,{p^5} - p = p \cdot \underbrace {\left[ {25{M^2} + 20M + 4 + 1} \right]}_{5\left( {5{M^2} + 4M + 1} \right)\,\, = \,5N\,,\,\,N\,\,{\mathop{\rm int}} }\left( {p + 1} \right)\left( {p - 1} \right) = 5N \cdot {\mathop{\rm int}} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{done}}$$

$$p = 5M + 3\,\,\,\, \Rightarrow \,\,\,\,\,{p^5} - p = p \cdot \underbrace {\left[ {25{M^2} + 30M + 9 + 1} \right]}_{5\left( {5{M^2} + 6M + 2} \right)\,\, = \,5N\,,\,\,N\,\,{\mathop{\rm int}} }\left( {p + 1} \right)\left( {p - 1} \right) = 5N \cdot {\mathop{\rm int}} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{done}}$$

$$p = 5M + 4\,\,\,\, \Rightarrow \,\,\,\,\,{p^5} - p = p\left( {{p^2} + 1} \right)\underbrace {\left( {5M + 5} \right)}_{5\left( {M + 1} \right)\,\, = 5N\,\,,\,\,N\,\,{\mathop{\rm int}} }\left( {p - 1} \right) = 5N \cdot {\mathop{\rm int}} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{done}}$$

Regards,
Fabio.

P.S.: if you consider learning this (and all other GMAT-focused-content) deeply to gain mathematical maturity and to be able to perform the quant section of the GMAT in a MUCH higher-level than the average, please check our free trial from the link given below!
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Manager  B
Joined: 13 Nov 2018
Posts: 62
Location: United States (PA)
GMAT 1: 650 Q39 V40 GMAT 2: 660 Q44 V37 GPA: 2.91
Re: If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

Could a kind soul help me out here?

How I went about thinking about p^2(P^2)(P-1)(P+1) is that no matter what the largest factor would be a positive root +1.

I've read each post and can't get it to click. What do I not see?
Intern  B
Joined: 18 Jun 2018
Posts: 2
Re: If p is a positive integer, what is the greatest integer that must be  [#permalink]

Show Tags

2
The expression p^6 - p^2 can be expressed as follow:
p^6 - p^2=(p^2 +1)*p^2*(p+1)*p*(p-1)

As we can see, (p+1)*p*(p-1) is a set of 3 consecutive number. Therefore, the product of these numbers is divisible for 1,2,3 (1)

Moreover, (p^2 +1)*p^2 is also a set of 2 consecutive number. Therefore, the product of these 2 numbers is divisible for 1 & 2 (2)

From (1) and (2), we can conclude that the product of these 5 numbers is divisible for 1,2,3,4,6 & 12. Re: If p is a positive integer, what is the greatest integer that must be   [#permalink] 21 Dec 2018, 10:45
Display posts from previous: Sort by

If p is a positive integer, what is the greatest integer that must be

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

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