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If P = m^3  m, where m and P are positive integers, and k is given by
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Updated on: 27 Apr 2018, 23:40
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If \(P = m^3  m\), where m and P are positive integers, and m is given by the expression \(m=(10x)^n+5^4\), where x and n are positive integers, then what is the remainder when P is divided by 4 ? A) 3 B) 0 C) 1 D) undetermined E) 2
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Originally posted by BOmoakin on 27 Apr 2018, 06:54.
Last edited by Bunuel on 27 Apr 2018, 23:40, edited 1 time in total.
Edited the question and moved to PS forum.



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Re: If P = m^3  m, where m and P are positive integers, and k is given by
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27 Apr 2018, 07:05
BOmoakin wrote: If P = m^3  m, where m and P are positive integers, and k is given by the expression m=(10x)^n+5^4 , where x and n are positive integers, then what is the remainder when P is divided by 4 ?
A) 3 B) 0 C) 1 D) undetermined E) 2 OA is B I believe it's a typo " and k is given by the". Answer would be B. P being multiplication of three consecutive integers (m1), m and (m+1). m is given as sum of one even number (10x^n) and one odd number (5^4), thus m is always odd. So P is multiplication of two even numbers and one odd number. So always divisible by 4. So remainder 0. BR Sent from my Redmi 4 using GMAT Club Forum mobile app



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Re: If P = m^3  m, where m and P are positive integers, and k is given by
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27 Apr 2018, 20:52
maneeshawl wrote: BOmoakin wrote: If P = m^3  m, where m and P are positive integers, and k is given by the expression m=(10x)^n+5^4 , where x and n are positive integers, then what is the remainder when P is divided by 4 ?
A) 3 B) 0 C) 1 D) undetermined E) 2 OA is B I believe it's a typo " and k is given by the". Answer would be B. P being multiplication of three consecutive integers (m1), m and (m+1). m is given as sum of one even number (10x^n) and one odd number (5^4), thus m is always odd. So P is multiplication of two even numbers and one odd number. So always divisible by 4. So remainder 0. BR Sent from my Redmi 4 using GMAT Club Forum mobile appWell explained thank you!



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Re: If P = m^3  m, where m and P are positive integers, and k is given by
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28 Apr 2018, 07:48
BOmoakin wrote: If \(P = m^3  m\), where m and P are positive integers, and m is given by the expression \(m=(10x)^n+5^4\), where x and n are positive integers, then what is the remainder when P is divided by 4 ?
A) 3 B) 0 C) 1 D) undetermined E) 2 \(P = m^3  m\) = \(m(m^2  1)\) = \(m (m  1)(m + 1)\) \(m=(10x)^n+5^4\), \(let\) \(x\) \(and\) \(p=1\), \(then\) \(P/4 =(634*635*636)/4\) \(reminder =0.\) \(Answer(B).\)



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Re: If P = m^3  m, where m and P are positive integers, and k is given by
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09 May 2018, 05:43
Hero8888 wrote: BOmoakin wrote: If \(P = m^3  m\), where m and P are positive integers, and m is given by the expression \(m=(10x)^n+5^4\), where x and n are positive integers, then what is the remainder when P is divided by 4 ?
A) 3 B) 0 C) 1 D) undetermined E) 2 \(P = m^3  m\) = \(m(m^2  1)\) = \(m (m  1)(m + 1)\) \(m=(10x)^n+5^4\), \(let\) \(x\) \(and\) \(p=1\), \(then\) \(P/4 =(634*635*636)/4\) \(reminder =0.\) \(Answer(B).\) sorry but I dont understand what let x and p be 1?



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Re: If P = m^3  m, where m and P are positive integers, and k is given by
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09 May 2018, 09:09
superbenson wrote: Hero8888 wrote: BOmoakin wrote: If \(P = m^3  m\), where m and P are positive integers, and m is given by the expression \(m=(10x)^n+5^4\), where x and n are positive integers, then what is the remainder when P is divided by 4 ?
A) 3 B) 0 C) 1 D) undetermined E) 2 \(P = m^3  m\) = \(m(m^2  1)\) = \(m (m  1)(m + 1)\) \(m=(10x)^n+5^4\), \(let\) \(x\) \(and\) \(p=1\), \(then\) \(P/4 =(634*635*636)/4\) \(reminder =0.\) \(Answer(B).\) sorry but I dont understand what let x and p be 1? When you are given variables, them the equation must work with any value of x and n. So just substitute any convenient real numbers for variables, in my case x=1, n=1 are a good fit.



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Re: If P = m^3  m, where m and P are positive integers, and k is given by
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29 May 2018, 07:36
P=m(m+1)(m1)
M= (10x)^n +5^4
M is always odd as 10x would always be even + +5^4 (odd)
So m+1 and m1 will always be even So P would always be divisible by 4 as product of two even number is always divisible by 4
So zero is remainder
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Re: If P = m^3  m, where m and P are positive integers, and k is given by
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25 Jun 2019, 02:00
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Re: If P = m^3  m, where m and P are positive integers, and k is given by
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