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If k = 2n  1, where n is an integer, what is the remainder [#permalink]
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19 Jun 2010, 20:05
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If k = 2n  1, where n is an integer, what is the remainder of k^2/8? A. 1 B. 3 C. 5 D. 7 E. Cannot be determined from the information given.
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Re: Remainder when k^2/8? [#permalink]
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19 Jun 2010, 20:12
My initial thoughts were: 1) \(k = 2n 1\), so \(k\) must be odd 2) For \(k^2\) to be divisible by 8, \(k^2\) must contain at least 3 2's. Therefore, each k must contain 2 2's.
I'm not sure how to continue using my initial thoughts.
The official answer shows: i) Express \(k^2 = (2n1)^2 = 4n^2  4n + 1\)
ii) Factor to \(k = 4n(n1)+1\)
Step ii doesn't make sense to me. If you factor out \(4n\) on the right side, why would the left not still be \(k^2\)? If this is just a misprint, then the the next step makes sense to me.
iii) If \(n\) is even, then \(n1\) is odd, while if \(n\) is odd, then \(n1\) is even. Therefore no matter what integer \(n\) is, \(k\) will equal 4 * even * odd, plus 1. In other words, \(k\) will equal a multiple of 8, plus 1. Therefore, the remainder of \(k^2/8\) is 1.



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Re: Remainder when k^2/8? [#permalink]
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20 Jun 2010, 08:18
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jpr200012 wrote: If \(k = 2n  1\), where \(n\) is an integer, what is the remainder of \(k^2/8\)?
A. 1 B. 3 C. 5 D. 7 E. Cannot be determined from the information given. This one can be done very easily with number picking. As you correctly noted \(k = 2n  1\) means that k is an odd number (basically \(k = 2n  1\) is a formula of an odd number). Now let's try several odd numbers: k=1 > k^2=1 > remainder upon division of 1 by 8 is 1; k=3 > k^2=3 > remainder upon division of 9 by 8 is 1; k=5 > k^2=25 > remainder upon division of 25 by 8 is 1; At this point we can safely assume that this will continue for all odd numbers. But if you want algebraic approach, here you go: \(k = 2n  1\) > \(k^2=(2n1)^2=4n^24n+1=4n(n1)+1\) > what is a remainder when \(4n(n1)+1\) is divided by 8: Now either \(n\) or \(n1\) will be even so in any case \(4n(n1)=4*odd*even=multiple \ of \ 8\), so \(4n(n1)\) is divisible by 8, so \(4n(n1)+1\) divided by 8 gives remainder of 1. Answer: A. Hope it's clear.
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Re: Remainder when k^2/8? [#permalink]
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20 Jun 2010, 08:21
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I think picking a number is a lot easier on this one.



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Re: Remainder when k^2/8? [#permalink]
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20 Jun 2010, 22:10
Hi Bunuel, How come 1 div by 8 gives remainder as 1??? utin. Bunuel wrote: jpr200012 wrote: If \(k = 2n  1\), where \(n\) is an integer, what is the remainder of \(k^2/8\)?
A. 1 B. 3 C. 5 D. 7 E. Cannot be determined from the information given. This one can be done very easily with number picking. As you correctly noted \(k = 2n  1\) means that k is an odd number (basically \(k = 2n  1\) is a formula of an odd number). Now let's try several odd numbers: k=1 > k^2=1 > remainder upon division of 1 by 8 is 1; k=3 > k^2=3 > remainder upon division of 9 by 8 is 1; k=5 > k^2=25 > remainder upon division of 25 by 8 is 1; At this point we can safely assume that this will continue for all odd numbers. But if you want algebraic approach, here you go: \(k = 2n  1\) > \(k^2=(2n1)^2=4n^24n+1=4n(n1)+1\) > what is a remainder when \(4n(n1)+1\) is divided by 8: Now either \(n\) or \(n1\) will be even so in any case \(4n(n1)=4*odd*even=multiple \ of \ 8\), so \(4n(n1)\) is divisible by 8, so \(4n(n1)+1\) divided by 8 gives remainder of 1. Answer: A. Hope it's clear.



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Re: Remainder when k^2/8? [#permalink]
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20 Jun 2010, 22:27
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I don't think he meant that 1 is divisible by 8. I think he was referring to the term before: 4(n)(n1)
Either n or n1 must be even so we have odd*even*4 which gives you 8*number, so we have this term divisible by 8.
Let us assume that k = 4(n)(n1)
This is divisible by 8.
So k+1 when divided by 8 will give reminder 1.
For example, consider n = 2
We have 4*2*1 + 1 = 9
When we divide this by 8 we get reminder 1. And so on. Hope this explains.



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Re: Remainder when k^2/8? [#permalink]
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21 Jun 2010, 02:52
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Re: Remainder when k^2/8? [#permalink]
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01 Sep 2010, 00:17
jpr200012 wrote: I think picking a number is a lot easier on this one. Agree, but good to know that a square of odd number gives 1 when divided by 8.



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Re: Remainder when k^2/8? [#permalink]
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09 Aug 2013, 06:36
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K=2n1 K^2 = 4n^2 +1  4n Dividing both sides by '8'. => K^2/8 = 4(n^2n)/8 +1/8 => n(n1)/2 + 1/8 In both cases whether 'n' is EVEN or ODD n(n1) is divisible by '2'. Hence, Remainder is = 1/8 = 1
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Re: If k = 2n  1, where n is an integer, what is the remainder [#permalink]
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10 Aug 2013, 14:03
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jpr200012 wrote: If k = 2n  1, where n is an integer, what is the remainder of k^2/8?
A. 1 B. 3 C. 5 D. 7 E. Cannot be determined from the information given. ...... Finally after 3lines calculation, n(n1)/2 + 1/8 so remainder = 1
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Re: If k = 2n  1, where n is an integer, what is the remainder [#permalink]
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16 Oct 2016, 14:56
Here K^2= (2n1)^2 = 4n^24n+1=> 4n(n1)+1 Now using the property > " The product of n consecutive integers is always divisible by n! => K^2= 4*2p +1 => 8p+1 for some integer p hence remainder with 8 must be 1 Hence A
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Re: If k = 2n  1, where n is an integer, what is the remainder [#permalink]
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