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

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.

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!

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.

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".

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

Greetings Sajjad1093

for your second question about 25:
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

Thank fskilnik

please, can you explain how p^6 - P^4 is divisible by 5?

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?

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