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For prime number y, y>3. Which of the following could be the remainder

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For prime number y, y>3. Which of the following could be the remainder  [#permalink]

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New post 24 Oct 2016, 23:22
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

79% (01:18) correct 21% (01:22) wrong based on 285 sessions

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Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

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New post 25 Oct 2016, 00:10
Bunuel wrote:
For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2
B. 3
C. 4
D. 5
E. 6



Checking with prime numbers greater than 3

@y=5, Rem (y^3 / 12) = Rem (5^3 / 12) = 5 (BINGO) :-D

Answer: Option D
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Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

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New post 25 Oct 2016, 08:15
Bunuel wrote:
For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2
B. 3
C. 4
D. 5
E. 6


y = { 5 , 7 , 11............ }

Plug in a number and check -

5^3/12 = Remainder 5

Thats in the given option , hit it...

Answer will be (D)
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For prime number y, y>3. Which of the following could be the remainder  [#permalink]

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New post 25 Oct 2016, 16:56
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Bunuel wrote:
For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2
B. 3
C. 4
D. 5
E. 6


Let’s test the first prime number greater than 3, which is 5.

5^3 = 125

125/12 = 10 remainder 5

Answer: D
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Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

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New post 27 Nov 2016, 09:26
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1
Bunuel wrote:
For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2
B. 3
C. 4
D. 5
E. 6


Prime number greater than 3 can be expressed as \(6k + 1\) or \(6k -1\).

\(\frac{(6k ± 1)^3}{12} = \frac{6^3k^3 ± 3*6^2k^2 + 3*6k ± 1}{12} = \frac{0 + 0 + 18k ± 1}{12} = \frac{6k ± 1}{12}\)

Let’s check first case \(6k + 1\)

k=1 (7/12) rem=7
k=2 (13/12) rem =1
k=3 (19/12) rem = 7
k =4 (25/12) rem = 1

we have cycle of remainders \(7\) and \(1\).

Next \(6k – 1\)

k=1 (5/12) rem=5
k=2 (11/12) rem =11
k=3 (17/12) rem = 5
k =4 (23/12) rem = 11

again cycle of only two remainders \(5\) and \(11\).

The only remainders after division by 12 cube of prime number \(y > 3\) can have are \(1\), \(5\), \(7\) and \(11\).

Checking our options.

Answer D.
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Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

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New post 19 Apr 2017, 11:41
vitaliyGMAT wrote:
Bunuel wrote:
For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2
B. 3
C. 4
D. 5
E. 6


Prime number greater than 3 can be expressed as \(6k + 1\) or \(6k -1\).

\(\frac{(6k ± 1)^3}{12} = \frac{6^3k^3 ± 3*6^2k^2 + 3*6k ± 1}{12} = \frac{0 + 0 + 18k ± 1}{12} = \frac{6k ± 1}{12}\)

Let’s check first case \(6k + 1\)

k=1 (7/12) rem=7
k=2 (13/12) rem =1
k=3 (19/12) rem = 7
k =4 (25/12) rem = 1

we have cycle of remainders \(7\) and \(1\).

Next \(6k – 1\)

k=1 (5/12) rem=5
k=2 (11/12) rem =11
k=3 (17/12) rem = 5
k =4 (23/12) rem = 11

again cycle of only two remainders \(5\) and \(11\).

The only remainders after division by 12 cube of prime number \(y > 3\) can have are \(1\), \(5\), \(7\) and \(11\).

Checking our options.

Answer D.


I have a doubt here please....prime numbers graeter than 3 can be expressed as 6k+1..putting k=4 we get 25, which is not a prime num,ber....am I missing something here?
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Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

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New post 19 Apr 2017, 11:51
saurabhsavant wrote:
vitaliyGMAT wrote:
Bunuel wrote:
For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2
B. 3
C. 4
D. 5
E. 6


Prime number greater than 3 can be expressed as \(6k + 1\) or \(6k -1\).

\(\frac{(6k ± 1)^3}{12} = \frac{6^3k^3 ± 3*6^2k^2 + 3*6k ± 1}{12} = \frac{0 + 0 + 18k ± 1}{12} = \frac{6k ± 1}{12}\)

Let’s check first case \(6k + 1\)

k=1 (7/12) rem=7
k=2 (13/12) rem =1
k=3 (19/12) rem = 7
k =4 (25/12) rem = 1

we have cycle of remainders \(7\) and \(1\).

Next \(6k – 1\)

k=1 (5/12) rem=5
k=2 (11/12) rem =11
k=3 (17/12) rem = 5
k =4 (23/12) rem = 11

again cycle of only two remainders \(5\) and \(11\).

The only remainders after division by 12 cube of prime number \(y > 3\) can have are \(1\), \(5\), \(7\) and \(11\).

Checking our options.

Answer D.


I have a doubt here please....prime numbers graeter than 3 can be expressed as 6k+1..putting k=4 we get 25, which is not a prime num,ber....am I missing something here?


Any prime number \(p>3\) when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).

So any prime number \(p>3\) could be expressed as \(p=6n+1\) or\(p=6n+5\) or \(p=6n-1\), where n is an integer >1.

But:
Not all number which yield a remainder of 1 or 5 upon division by 6 are prime, so vise-versa of above property is not correct. For example 25 yields a remainder of 1 upon division be 6 and it's not a prime number.

Hope it's clear.
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Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

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New post 23 May 2017, 13:55
prime number 5 , 5^3 /12 = 25*5 /12 => 1*5/12 ( since 25/12 will give rem as 1) => 5 is rem

Option D is answer
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Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

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Re: For prime number y, y>3. Which of the following could be the remainder   [#permalink] 10 Oct 2018, 07:42
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