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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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19 Jun 2010, 19:05
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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.
[Reveal] Spoiler: OA
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Re: Remainder when k^2/8? [#permalink]

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

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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.

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

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20 Jun 2010, 07:21
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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, 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.

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

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20 Jun 2010, 21:27
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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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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 --> $$5=0*6+5$$.
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Re: Remainder when k^2/8? [#permalink]

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31 Aug 2010, 23:17
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, 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
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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, 13:03
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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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03 Sep 2014, 05:18
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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]

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16 Oct 2016, 13:56
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] 16 Oct 2016, 13:56
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