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Math Expert V
Joined: 02 Sep 2009
Posts: 60647
If P = k^3 – k, where k and P are positive integers, and k is given by  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 54% (02:07) correct 46% (02:18) wrong based on 56 sessions

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If $$P = k^3 – k$$, where k and P are positive integers, and k is given by the expression $$k = (10x)^{n} + 5^{4}$$, where x and n are positive integers, then what is the remainder when P is divided by 4 ?

A. 0
B. 1
C. 2
D. 3
E. Cannot be determined

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Senior Manager  P
Joined: 25 Jul 2018
Posts: 482
Re: If P = k^3 – k, where k and P are positive integers, and k is given by  [#permalink]

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P = k^{3} —k = k(k —1)(k+1)
P,k — positive integers

k = (10x)^{n} + 5^{4}
—> k = (10x)^{n} + 625

(x,n — positive integers )
—> (10x)^{n} — always even number
Even+ odd = Odd number
k — odd number

If k=1, then k(k—1)(k+1)= 0 ( P cannot be 0)

If k=3, then P= 3*2*4= 24.
If k=11, then P= 11*10*12

Any odd value of k (k >1) gives the positive value of P which is divisible by 4 ( remainder is 0(zero))

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Manager  G
Joined: 10 Mar 2018
Posts: 74
Location: India
Concentration: Entrepreneurship, Marketing
Schools: IIML IPMX "21
GMAT 1: 680 Q44 V38 WE: Design (Retail)
Re: If P = k^3 – k, where k and P are positive integers, and k is given by  [#permalink]

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1
3
Given that

k=$$(10x)^n+5^4$$

So basically
k = one even no. + one odd no. (Since 10 raised to any power will be even and 5 raised to power 4 is odd)

k is a odd no.

P = $$k^3$$-k

= k($$k^2$$-1)

= k (k+1)(k-1)

So P is the product of three consecutive nos.

SINCE K IS ODD
P = even no. * odd no. * even no.

Definitely P is divisible by two 2's or 4.

hence remainder will be 0

ANS (A)
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~ETERNAL~ Re: If P = k^3 – k, where k and P are positive integers, and k is given by   [#permalink] 07 Nov 2019, 03:16
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# If P = k^3 – k, where k and P are positive integers, and k is given by  