GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 24 Jan 2020, 18:34

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

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

### Show Tags

07 Nov 2019, 02:29
00:00

Difficulty:

75% (hard)

Question Stats:

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

### HideShow timer Statistics

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

Are You Up For the Challenge: 700 Level Questions

_________________
Senior Manager
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]

### Show Tags

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))

Posted from my mobile device
Manager
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]

### Show Tags

07 Nov 2019, 03:16
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)
_________________
~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
Display posts from previous: Sort by