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# Remainder when k^2/8?

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

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]  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.

Hope it's clear.
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Re: Remainder when k^2/8? [#permalink]  20 Jun 2010, 08:21
I think picking a number is a lot easier on this one.
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Re: Remainder when k^2/8? [#permalink]  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.

Hope it's clear.
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Re: Remainder when k^2/8? [#permalink]  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]  21 Jun 2010, 02:52
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]  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.
Re: Remainder when k^2/8?   [#permalink] 01 Sep 2010, 00:17
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