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

Question

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

fskilnik · Accepted Answer

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.

MahmoudFawzy · Answer

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.

fskilnik · Answer

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.

MahmoudFawzy · Answer

I answered in right by mistake

fskilnik · Answer

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".

Sajjad1093 · Answer

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

MahmoudFawzy · Answer

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

MahmoudFawzy · Answer

Thank  fskilnik

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

fskilnik · Answer

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 }_{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 }_{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!

MWithrock · Answer

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?

HanhPham309 · Answer

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.

Szartaj · Answer

Here is another possible explanation to prove p^6-p^2 (or p^5-p) is divisible by 5:

Lets consider p^5 - p => p (p^4 - 1)  => p (p^2 - 1) (p^2 +1)

Now for any number to be divisible by 5, its unit digit has to be either 0 or 5.

Thus, for all p that'll end on 0 or 5, (p^5 - p) is always divisible by 5
Now considering p as any other number whose unit digit is not 0 or 5, p^2 will come into play. Since the square of such a number will always end with one of these unit digits: 1,4,9,6. Therefore, either (p^2 -1) or (p^2 +1) which are the other two factors of (p^5 - p) will result in a number with a unit digit 0 or 5 and thus divisible by 5.

Schloop · Answer

This question and all subsequent answers fail to account for the fact that 1 is a positive integer. 1^6-1^2 = 0, to which 12 is not a factor.

Bunuel · Answer

You are wrong. 0 is divisible by all integers, except 0 itself: 0/12 = 0 = integer.

ZERO:

1. Zero is an  INTEGER .

2. Zero is an  EVEN  integer.

3. Zero is  neither positive nor negative  (the only one of this kind)

4. Zero is  divisible by EVERY integer except 0 itself  ( \frac{0}{x} = 0 , so 0 is a divisible by every number, x).

5. Zero is a  multiple of EVERY integer  ( x*0 = 0 , so 0 is a multiple of any number, x)

6. Zero is  NOT a prime number  (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed:  anything/0 is undefined .

8. Any non-zero number  to the power of 0 equals 1  ( x^0 = 1 )

9.  0^0  case is  NOT tested on the GMAT .

10. If the exponent n is positive (n > 0),  0^n = 0 .

11. If the exponent n is negative (n < 0),  0^n  is undefined, because  0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0} , which is undefined.  You CANNOT take 0 to the negative power .

12.  0! = 1! = 1 .

Hope it helps.

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

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GMAT Problem Solving (PS) Questions

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

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