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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

1

This post received KUDOS

Expert's post

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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This post received KUDOS

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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