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If k = 2n - 1, where n is an integer, what is the remainder

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If k = 2n - 1, where n is an integer, what is the remainder [#permalink]

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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.
[Reveal] Spoiler: OA

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Re: Remainder when k^2/8? [#permalink]

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New post 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 \(k^2 = (2n-1)^2 = 4n^2 - 4n + 1\)

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.

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Re: Remainder when k^2/8? [#permalink]

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New post 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:
\(k = 2n - 1\) --> \(k^2=(2n-1)^2=4n^2-4n+1=4n(n-1)+1\) --> 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.

Answer: A.

Hope it's clear.
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Re: Remainder when k^2/8? [#permalink]

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New post 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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New post 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:
\(k = 2n - 1\) --> \(k^2=(2n-1)^2=4n^2-4n+1=4n(n-1)+1\) --> 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.

Answer: A.

Hope it's clear.

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Re: Remainder when k^2/8? [#permalink]

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New post 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)(n-1)

Either n or n-1 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)(n-1)

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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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 --> \(5=0*6+5\).
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Re: Remainder when k^2/8? [#permalink]

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New post 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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New post 09 Aug 2013, 06:36
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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
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Re: If k = 2n - 1, where n is an integer, what is the remainder [#permalink]

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New post 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(n-1)/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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New post 16 Oct 2016, 14:56
Here K^2= (2n-1)^2 = 4n^2-4n+1=> 4n(n-1)+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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Re: If k = 2n - 1, where n is an integer, what is the remainder   [#permalink] 11 Nov 2017, 05:48
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