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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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If P = m^3 - m, where m and P are positive integers, and k is given by  [#permalink]

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New post 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

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  [#permalink]

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New post 27 Apr 2018, 07:05
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1
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 (m-1), 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

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Re: If P = m^3 - m, where m and P are positive integers, and k is given by  [#permalink]

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New post 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 (m-1), 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

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Well 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  [#permalink]

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New post 28 Apr 2018, 07:48
1
1
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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 29 May 2018, 07:36
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P=m(m+1)(m-1)

M= (10x)^n +5^4

M is always odd as 10x would always be even + +5^4 (odd)

So m+1 and m-1 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

Give kudos if it helps

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Re: If P = m^3 - m, where m and P are positive integers, and k is given by  [#permalink]

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Re: If P = m^3 - m, where m and P are positive integers, and k is given by   [#permalink] 25 Jun 2019, 02:00
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