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If p is a positive integer, what is the greatest integer that must be
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Updated on: 12 Dec 2018, 03:25
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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
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Originally posted by ritu1009 on 12 Dec 2018, 03:17.
Last edited by ritu1009 on 12 Dec 2018, 03:25, edited 2 times in total.



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If p is a positive integer, what is the greatest integer that must be
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Updated on: 15 Dec 2018, 10:26
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.
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Re: If p is a positive integer, what is the greatest integer that must be
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14 Dec 2018, 12:27
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) (p1) 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^5p) 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.
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Re: If p is a positive integer, what is the greatest integer that must be
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15 Dec 2018, 10:27
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) (p1) 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^5p) 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
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Re: If p is a positive integer, what is the greatest integer that must be
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16 Dec 2018, 04:15
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 wellposed, 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^6p^2, I mean, of p^6p^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".
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If p is a positive integer, what is the greatest integer that must be
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20 Dec 2018, 08:49
Hi, Can you please explain solution "01"? How were you able to determine that p^2 (p^2+1) (p+1) (p1) will be divisble by 12? i solved by expressing (p^21) (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) (p1) 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^5p) 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.



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If p is a positive integer, what is the greatest integer that must be
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Updated on: 20 Dec 2018, 12:30
Sajjad1093 wrote: Hi, Can you please explain solution "01"? How were you able to determine that p^2 (p^2+1) (p+1) (p1) will be divisble by 12? i solved by expressing (p^21) (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) (p1) 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^5p) 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. Greetings Sajjad1093for 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
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If p is a positive integer, what is the greatest integer that must be
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20 Dec 2018, 11:52
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) (p1) 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.
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Re: If p is a positive integer, what is the greatest integer that must be
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20 Dec 2018, 12:27
Thank fskilnikplease, can you explain how p^6  P^4 is divisible by 5?
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Re: If p is a positive integer, what is the greatest integer that must be
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20 Dec 2018, 12:51
Mahmoudfawzy83 wrote: Thank fskilnikplease, 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^5p) 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 GMATfocusedcontent) deeply to gain mathematical maturity and to be able to perform the quant section of the GMAT in a MUCH higherlevel than the average, please check our free trial from the link given below!
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Re: If p is a positive integer, what is the greatest integer that must be
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20 Dec 2018, 14:20
Could a kind soul help me out here?
How I went about thinking about p^2(P^2)(P1)(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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Re: If p is a positive integer, what is the greatest integer that must be
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21 Dec 2018, 09:45
The expression p^6  p^2 can be expressed as follow: p^6  p^2=(p^2 +1)*p^2*(p+1)*p*(p1)
As we can see, (p+1)*p*(p1) 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 &nbs
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