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If P = k^3 – k, where k and P are positive integers, and k is given by

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

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New post 07 Nov 2019, 02:29
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Question Stats:

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

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New post 07 Nov 2019, 02:52
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))

The answer is A

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

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New post 07 Nov 2019, 03:16
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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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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

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