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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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19 Nov 2014, 02:00
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65% (01:24) correct 35% (01:45) wrong based on 234 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
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01 Oct 2016, 05:37
ProfX wrote: Answer is 0, but it took me about 5 minutes Sent from my iPhone using GMAT Club Forum mobile appQuestion 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 = (k1)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
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19 Nov 2014, 02:57
Answer = A = 0 \(p = k^3  k = k (k^2  1) = k(k+1)(k1)\) \(k = (10x)^n + 625\) \(p = ((10x)^n + 625)*((10x)^n + 626)*((10x)^n + 624)\) 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
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19 Nov 2014, 03:22
p=k^3k =k(k^21) =(k1)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=(31)(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
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19 Nov 2014, 14:41
PareshGmat wrote: Answer = A = 0
\(p = k^3  k = k (k^2  1) = k(k+1)(k1)\)
\(k = (10x)^n + 625\)
\(p = ((10x)^n + 625)*((10x)^n + 626)*((10x)^n + 624)\)
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
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19 Nov 2014, 18:44
xserenity114 wrote: PareshGmat wrote: Answer = A = 0
\(p = k^3  k = k (k^2  1) = k(k+1)(k1)\)
\(k = (10x)^n + 625\)
\(p = ((10x)^n + 625)*((10x)^n + 626)*((10x)^n + 624)\)
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
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01 Oct 2016, 05:08
Answer is 0, but it took me about 5 minutes Sent from my iPhone using GMAT Club Forum mobile app



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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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15 Mar 2018, 19:17
P = (k+1)*k*(k1) Consecutive numbers, can either be OddEvenOdd or EvenOddEven 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
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09 Apr 2018, 04:00
What you need to notice is that p is a product of 3 consecutive integers.
or p = k3−kk3−k
or p = (k1)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 &nbs
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09 Apr 2018, 04:00






