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Bunuel
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If P = k^3 - k^2, where k and P are positive integers, is P divisible 28 Nov 2022, 10:15
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If \(P = k^3 - k^2\), where k and P are positive integers, is P divisible by 4?

(1) \(k = (10x)n + 5^4\) where x and n are positive integers and n > 1.
(2) \((2n + 1)k\) leaves a remainder when divided by 2; n is a positive integer
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gmatophobia
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If P = k^3 - k^2, where k and P are positive integers, is P divisible 28 Nov 2022, 13:09
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If \(P = k^3 - k^2\), where k and P are positive integers, is P divisible by 4?

(1) \(k = (10x)n + 5^4\) where x and n are positive integers and n > 1.
(2) \((2n + 1)k\) leaves a remainder when divided by 2; n is a positive integer
Show more

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

Inference : For P to be divisible by 4
  • If k = even, k^2 will be divisible by 4. Hence P is divisible by 4.
  • If k = odd, (k-1) should be divisible by 4.


Statement 1

\(k = (10x)n + 5^4\) where x and n are positive integers and n > 1.

k = even + odd = odd

We have established that k is odd, now we need to find if k - 1 is divisible by 4.

It's given that x & n > 1, so k-1 will always be an even integer which is divisible by 4.

Statement 1 is sufficient and we can eliminate B, C and E.

Statement 2

\((2n + 1)k\) leaves a remainder when divided by 2; n is a positive integer

If a number leaves a remainder when divided by 2, then the number is an odd.

Therefore, (2n+1)k is odd. (2n+1) is odd, therefore we can conclude that k is odd.

Now we need to determine if k-1 is divisible by 4.

Say k = 3; k - 1 = 2. k-1 is not divisible by 4.
k = 5; k - 1 = 4. k-1 is divisible by 4.

As we are getting two different answers, we can eliminate B.

Option A
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If P = k^3 - k^2, where k and P are positive integers, is P divisible 01 Dec 2022, 04:42
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Bunuel
If \(P = k^3 - k^2\), where k and P are positive integers, is P divisible by 4?

(1) \(k = (10x)n + 5^4\) where x and n are positive integers and n > 1.
(2) \((2n + 1)k\) leaves a remainder when divided by 2; n is a positive integer

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

Inference :
  • If k = even, k^2 will be divisible by 4. Hence P is divisible by 4.
  • If k = odd, (k-1) should be divisible by 4.


Statement 1

\(k = (10x)n + 5^4\) where x and n are positive integers and n > 1.

k = even + odd = odd

We have established that k is odd, now we need to find if k - 1 is divisible by 4.

It's given that x & n > 1, so k-1 will always be an even integer which is divisible by 4.

Statement 1 is sufficient and we can eliminate B, C and E.

Statement 2

\((2n + 1)k\) leaves a remainder when divided by 2; n is a positive integer

If a number leaves a remainder when divided by 2, then the number is an odd.

Therefore, (2n+1)k is odd. (2n+1) is odd, therefore we can conclude that k is odd.

Now we need to determine if k-1 is divisible by 4.

Say k = 3; k - 1 = 2. k-1 is not divisible by 4.
k = 5; k - 1 = 4. k-1 is divisible by 4.

As we are getting two different answers, we can eliminate B.

Option A
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how it is possible if k is odd

k-1 is divisible by 4

k=3
not possible

statement 1 and 2 both are not sufficient


in statement 1

k=(10x)n+ 5^4

if we put x=1 n=2 as per given statement

k=645
putting value

p=645^2(644)

divisible by 4

however for x=3 n=5
k=775

p=775^2(774)

not divisible by 4

hence insufficient
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If P = k^3 - k^2, where k and P are positive integers, is P divisible 01 Dec 2022, 04:57
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Rabab36

how it is possible if k is odd

k-1 is divisible by 4

Show more

Rabab36 - Not sure if this was a question for me, or if you intended to present your version of the solution.

However, as you quoted my response , responding to the above quoted question

From the question stem we know

\(P = k^3 - k^2\)

If we take \(k^2\) common

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

In a scenario when k is odd, for P to be divisible by 4, (k - 1) should be divisible by 4. This is because k is odd, so \(k^2\) will not be divisible by 4.

Ex. k = 5

P = \(5^2(5-1)\)

We can tell P is divisible by 4, as (5-1) is divisible by 4.

The inference does not intend to conclude that whenever k is odd, k - 1 is divisible by 4. I guess you inferred this from the statements in blue.

The inference is in conjunction with the question, "Is P divisible by 4". So if P were to be divisible by 4, and k was odd, then (k-1) would be divisible by 4.

Hope this clarifies.
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If P = k^3 - k^2, where k and P are positive integers, is P divisible 01 Dec 2022, 05:02
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gmatophobia
Rabab36

how it is possible if k is odd

k-1 is divisible by 4


Rabab36 - Not sure if this was a question for me, or you intended to present your version of the solution.

However, as you quoted the response , responding to the above quoted question

From the question stem we know

\(P = k^3 - k^2\)

If we take \(k^2\) common

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

In a scenario when k is odd, for P to be divisible by 4, (k - 1) should be divisible by 4. This is because k is odd, so \(k^2\) will not be divisible by 4.

Ex. k = 5

P = \(5^2(5-1)\)

We can tell P is divisible by 4, as (5-1) is divisible by 4.

The inference does not intend to conclude that whenever k is odd, k - 1 is divisible by 4. I guess you inferred this from the statements in blue.

The inference is in conjunction with the question, "Is P divisible by 4". So if P were to be divisible by 4, and k was odd, then (k-1) would be divisible by 4.

Hope this clarifies.
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then sir for x=5 n=3

statement is insufficient
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If P = k^3 - k^2, where k and P are positive integers, is P divisible 17 Feb 2025, 08:20
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K-1 is not necessarily divisible by 4 for odd value of k, which is why statement 1 is insufficient. Could someone pls explain why A is correct ?
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If P = k^3 - k^2, where k and P are positive integers, is P divisible 17 Feb 2025, 10:21
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Pablopikachu
K-1 is not necessarily divisible by 4 for odd value of k, which is why statement 1 is insufficient. Could someone pls explain why A is correct ?
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Hey buddy just calculate the value of K^2(K-1) and divide it by four you’ll get the answer

For statement 2 we just that K^2(K-1) is even which tell us nothing whether that even number is a multiple of 4 or not

so 1 alone is sufficient!
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If P = k^3 - k^2, where k and P are positive integers, is P divisible 01 Jun 2025, 11:52
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Hello gmatophobia In S1 "It's given that x & n > 1, so k-1 will always be an even integer which is divisible by 4.", how did you arrive at x>1? further, even if both x & n are >1, lets say x=3 and n=5, k = 775, then k-1=774 which although is even but not divisible by 4. Bunuel , chetan2u , any other expert please clarify
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Bunuel
If \(P = k^3 - k^2\), where k and P are positive integers, is P divisible by 4?

(1) \(k = (10x)n + 5^4\) where x and n are positive integers and n > 1.
(2) \((2n + 1)k\) leaves a remainder when divided by 2; n is a positive integer

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

Inference : For P to be divisible by 4

  • If k = even, k^2 will be divisible by 4. Hence P is divisible by 4.
  • If k = odd, (k-1) should be divisible by 4.


Statement 1

\(k = (10x)n + 5^4\) where x and n are positive integers and n > 1.

k = even + odd = odd

We have established that k is odd, now we need to find if k - 1 is divisible by 4.

It's given that x & n > 1, so k-1 will always be an even integer which is divisible by 4.

Statement 1 is sufficient and we can eliminate B, C and E.

Statement 2

\((2n + 1)k\) leaves a remainder when divided by 2; n is a positive integer

If a number leaves a remainder when divided by 2, then the number is an odd.

Therefore, (2n+1)k is odd. (2n+1) is odd, therefore we can conclude that k is odd.

Now we need to determine if k-1 is divisible by 4.

Say k = 3; k - 1 = 2. k-1 is not divisible by 4.
k = 5; k - 1 = 4. k-1 is divisible by 4.

As we are getting two different answers, we can eliminate B.

Option A
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If P = k^3 - k^2, where k and P are positive integers, is P divisible 01 Jun 2025, 23:02
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NoeticImbecile
Hello gmatophobia In S1 "It's given that x & n > 1, so k-1 will always be an even integer which is divisible by 4.", how did you arrive at x>1? further, even if both x & n are >1, lets say x=3 and n=5, k = 775, then k-1=774 which although is even but not divisible by 4. Bunuel , chetan2u , any other expert please clarify
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Hi

There is no reason for statement 1 to be sufficient.
Is P divisible by 4 means is k^2(k-1) divisible by 4.

Statement 1 tells us that k is odd, so the question now becomes- Is k-1 divisible by 4?
There is nothing in statement that can give a YES or NO to the question.
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