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Re: Remainder when k^2/8? [#permalink]
19 Jun 2010, 19:12
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 \(k^2 = (2n-1)^2 = 4n^2 - 4n + 1\)
ii) Factor to \(k = 4n(n-1)+1\)
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 \(n-1\) is odd, while if \(n\) is odd, then \(n-1\) 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.
Re: Remainder when k^2/8? [#permalink]
20 Jun 2010, 07:18
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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: \(k = 2n - 1\) --> \(k^2=(2n-1)^2=4n^2-4n+1=4n(n-1)+1\) --> what is a remainder when \(4n(n-1)+1\) is divided by 8:
Now either \(n\) or \(n-1\) will be even so in any case \(4n(n-1)=4*odd*even=multiple \ of \ 8\), so \(4n(n-1)\) is divisible by 8, so \(4n(n-1)+1\) divided by 8 gives remainder of 1.
Re: Remainder when k^2/8? [#permalink]
20 Jun 2010, 21:10
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: \(k = 2n - 1\) --> \(k^2=(2n-1)^2=4n^2-4n+1=4n(n-1)+1\) --> what is a remainder when \(4n(n-1)+1\) is divided by 8:
Now either \(n\) or \(n-1\) will be even so in any case \(4n(n-1)=4*odd*even=multiple \ of \ 8\), so \(4n(n-1)\) is divisible by 8, so \(4n(n-1)+1\) divided by 8 gives remainder of 1.
Re: Remainder when k^2/8? [#permalink]
21 Jun 2010, 01:52
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utin wrote:
Hi Bunuel,
How come 1 div by 8 gives remainder as 1???
utin.
THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).
So when divisor (8 in our case) is more than dividend (1 in our case) then the reminder equals to the dividend:
1 divided by 8 yields a reminder of 1 --> \(1=0*8+1\); or:
5 divided by 6 yields a reminder of 5 --> \(5=0*6+5\). _________________
Re: Remainder when k^2/8? [#permalink]
09 Aug 2013, 05:36
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K=2n-1
K^2 = 4n^2 +1 - 4n
Dividing both sides by '8'.
=> K^2/8 = 4(n^2-n)/8 +1/8
=> n(n-1)/2 + 1/8
In both cases whether 'n' is EVEN or ODD n(n-1) is divisible by '2'.
Hence,
Remainder is = 1/8 = 1 _________________
Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________
Re: If k = 2n - 1, where n is an integer, what is the remainder [#permalink]
03 Sep 2014, 05:18
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