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

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

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New post 19 Nov 2014, 01:00
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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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If P = k^3 – k, where k and P are positive integers, and k is given b  [#permalink]

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New post 01 Oct 2016, 04:37
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Answer is 0, but it took me about 5 minutes


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Question is very easy and can be solved in hardly 30 secs.

What you need to notice is that p is a product of 3 consecutive integers.

or p = \(k^3 - k\)

or p = (k-1)k(k+1)

Now, notice that we have been given k = \((10x)^n + 5^4\)

or I can say K is Even + ODD = ODD.

Hence, if the middle number if ODD, the other two will be even. Or when divided by four will give remainder 0. Hence, A.
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Re: If P = k^3 – k, where k and P are positive integers, and k is given b  [#permalink]

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New post 19 Nov 2014, 01:57
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Answer = A = 0

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

\(k = (10x)^n + 625\)

\(p = ((10x)^n + 625)*((10x)^n + 626)*((10x)^n + 624)\)

For any value of x,n > 1, any of the above 3 terms get divisible by 4

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

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New post 19 Nov 2014, 02:22
p=k^3-k
=k(k^2-1)
=(k-1)k(k+1)
as can be seen, p is a product of three consecutive integers. now k can be both even and odd.
now look into the given expression of k
\(k=(10x)^{n} + 5^{4}\)
\(5^{4}\) will always be odd. and \((10x)^{n}\) will always be even
thus k= even + odd =odd
Also, k =1 is not possible ( because p is a positive integer)
so lets try k=3
p=(3-1)(3)(3+1)
=2*3*4

which is clearly divisible by 4.
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If P = k^3 – k, where k and P are positive integers, and k is given b  [#permalink]

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New post 19 Nov 2014, 13:41
PareshGmat wrote:
Answer = A = 0

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

\(k = (10x)^n + 625\)

\(p = ((10x)^n + 625)*((10x)^n + 626)*((10x)^n + 624)\)

For any value of x,n > 1, any of the above 3 terms get divisible by 4

Answer = A


Is it divisible by 4 because 626 and 624 are each divisible by 2 and no matter what (10x)^n is, its units digit will still be 0 so 2 factors will always be divisible by 2 regardless of what x & n may be?
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= k^3 – k, where k and P are positive integers, and k is given b  [#permalink]

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New post 19 Nov 2014, 17:44
xserenity114 wrote:
PareshGmat wrote:
Answer = A = 0

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

\(k = (10x)^n + 625\)

\(p = ((10x)^n + 625)*((10x)^n + 626)*((10x)^n + 624)\)

For any value of x,n > 1, any of the above 3 terms get divisible by 4

Answer = A


Is it divisible by 4 because 626 and 624 are each divisible by 2 and no matter what (10x)^n is, its units digit will still be 0 so 2 factors will always be divisible by 2 regardless of what x & n may be?


No. First 626 or 624 would be added to (10x)^n expression & then divisibility would be checked. Then the denominator may get split in 2 * 2 to get the numerator divided completely by 4
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Re: If P = k^3 – k, where k and P are positive integers, and k is given b  [#permalink]

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New post 01 Oct 2016, 04:08
Answer is 0, but it took me about 5 minutes


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

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New post 15 Mar 2018, 18:17
P = (k+1)*k*(k-1)

Consecutive numbers, can either be Odd-Even-Odd or Even-Odd-Even

If, k is odd, then the other two numbers will be even, hence divisible by 2, twice. Remainder is 0.

If, k is even, then the other two numbers will be odd, and we cannot say anything.

Thankfully, it is given that k is odd. (5^4+10x*10x*10x.....)

So, other two numbers are even, each divisible by 2.

Hence, P is divisible by 4. Remainder is 0.

30 seconds.

manpreetsingh86 wrote:
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 b  [#permalink]

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New post 09 Apr 2018, 03:00
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What you need to notice is that p is a product of 3 consecutive integers.

or p = k3−kk3−k

or p = (k-1)k(k+1)

Now, notice that we have been given k = (10x)n+54(10x)n+54

or I can say K is Even + ODD = ODD.

Hence, if the middle number if ODD, the other two will be even. Or when divided by four will give remainder 0. Hence, A.
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Re: If P = k^3 – k, where k and P are positive integers, and k is given b &nbs [#permalink] 09 Apr 2018, 03:00
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