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

Question

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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Eternal · Accepted Answer

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)

lacktutor · Answer

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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metalhead2593 · Answer

k^3-k = k(k^2-1) = (k-1)k(k+1)

k = 10^n*x^n + 625. K is odd.
So the remainder of k when divided to 4 is either 1 or 3

* remainder = 1. K-1 = 4a
* remainder = 3. K+1 = 4a

IMO A.

carouselambra · Answer

So basically, k^3-k can be written as : k(k^2-1) = (k-1)(k)(k+1) - which is nothing but product of 3 consecutive numbers.
Now, an odd number will always leave a remainder of either 1 or 3 when divided by 4.
Eg : 15, 25,27..

Here, the product will always be divisible by 4 hence remainder is zero.
IMO A

Fdambro294 · Answer

If you can find the loophole you can be in and out in 1 minute.

P = (k)^3 - k

Since k = positive integer, P is the Product of 3 consecutive Integers, regardless of the value of X and N (positive integers) in:

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

P = (k)^3 - k

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

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

If K = odd integer, then the 3 Factors will be the product of Two Consecutive Even Integers, which will be divisible by 4

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

Since the variables x and n must be positive integers ———> the base of 10x raised to any positive integer power will be EVEN

and (5)^4 = ODD

Therefore:

k = Even + Odd = Odd Integer

Therefore, P will be the Product of 3 consecutive integers, the middle of which is ODD

P = (even integer) (even integer + 1) ( even integer + 2

P will be a multiple of 4 and the remainder will be 0

A

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Ekland · Answer

Quick one.
 P = k^3 – k  says that P is product of three consecutive integers. and  k = (10x)^{n} + 5^{4}  that the unit digit of k is 5. Thus P Must be a multiple of 8 since it's last two digits is a MULTIPLE OF 8.

4 is a factor of 8 thus A is the answer.

LET'S EXPLAIN IT ANOTHER WAY
The product of three consecitive positive integers must be a multiple of at least 6 and in one situation of 8. This is because any three consecutive postive integer must contain a multiple of 3(because you're counting up to three consecutiveness) and an even number which means a multiple of 2;  (3*2 = 6) 
The only situation where consecutive integers (k-1), k, and (k+1) must be a multiple of 8 is when k has a unit digit of 5, as in 5 or 15 or 25 etc.. thus multiplying whatever that ends in 5 with the preceding consecutive number and next consecutive must contain up to three different 2s which is 8 when multiplied.
 k = (10x)^{n} + 5^{4}  plain tells you the unit digit of k is 5. So P MUST be divisible by 8.

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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

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