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Intern  Joined: 27 Apr 2018
Posts: 1
If P = m^3 - m, where m and P are positive integers, and k is given by  [#permalink]

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11 00:00

Difficulty:   55% (hard)

Question Stats: 56% (01:56) correct 44% (02:09) wrong based on 94 sessions

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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.
Intern  B
Joined: 04 Oct 2017
Posts: 5
Re: If P = m^3 - m, where m and P are positive integers, and k is given by  [#permalink]

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3
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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Manager  S
Joined: 03 Apr 2016
Posts: 88
Location: India
Schools: ISB '19, IIMB EPGP"20
GMAT 1: 580 Q43 V27
GMAT 2: 650 Q32 V48
GRE 1: Q160 V151
GPA: 3.99
WE: Design (Consulting)
Re: If P = m^3 - m, where m and P are positive integers, and k is given by  [#permalink]

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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

Sent from my Redmi 4 using GMAT Club Forum mobile app

Well explained thank you!
Senior Manager  G
Joined: 29 Dec 2017
Posts: 365
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33 GMAT 2: 690 Q47 V37 GMAT 3: 710 Q50 V37 GPA: 3.25
WE: Marketing (Telecommunications)
Re: If P = m^3 - m, where m and P are positive integers, and k is given by  [#permalink]

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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

$$P = m^3 - m$$ = $$m(m^2 - 1)$$ = $$m (m - 1)(m + 1)$$
$$m=(10x)^n+5^4$$, $$let$$ $$x$$ $$and$$ $$p=1$$, $$then$$ $$P/4 =(634*635*636)/4$$ $$-reminder =0.$$ $$Answer(B).$$
Intern  B
Joined: 07 Jan 2018
Posts: 13
Re: If P = m^3 - m, where m and P are positive integers, and k is given by  [#permalink]

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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?
Senior Manager  G
Joined: 29 Dec 2017
Posts: 365
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33 GMAT 2: 690 Q47 V37 GMAT 3: 710 Q50 V37 GPA: 3.25
WE: Marketing (Telecommunications)
Re: If P = m^3 - m, where m and P are positive integers, and k is given by  [#permalink]

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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.
Director  P
Joined: 02 Oct 2017
Posts: 684
Re: If P = m^3 - m, where m and P are positive integers, and k is given by  [#permalink]

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2
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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