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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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19 Nov 2014, 02:00
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65% (hard)

Question Stats:

61% (02:12) correct 39% (02:31) wrong based on 251 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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If P = k^3 – k, where k and P are positive integers, and k is given b  [#permalink]

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01 Oct 2016, 05:37
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ProfX wrote:

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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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19 Nov 2014, 02:57
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$$p = k^3 - k = k (k^2 - 1) = k(k+1)(k-1)$$

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

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

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

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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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19 Nov 2014, 03:22
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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19 Nov 2014, 14:41
PareshGmat wrote:

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

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

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

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

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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19 Nov 2014, 18:44
xserenity114 wrote:
PareshGmat wrote:

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

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

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

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

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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01 Oct 2016, 05:08

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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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15 Mar 2018, 19:17
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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Posts: 24
Re: If P = k^3 – k, where k and P are positive integers, and k is given b  [#permalink]

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09 Apr 2018, 04:00
1
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.
Re: If P = k^3 – k, where k and P are positive integers, and k is given b   [#permalink] 09 Apr 2018, 04:00
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