GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 19 Oct 2019, 10:20

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
Joined: 30 May 2010
Posts: 165
If k = 2n - 1, where n is an integer, what is the remainder  [#permalink]

Show Tags

New post 19 Jun 2010, 20:05
30
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

70% (01:27) correct 30% (01:31) wrong based on 870 sessions

HideShow timer Statistics

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.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58445
Re: Remainder when k^2/8?  [#permalink]

Show Tags

New post 20 Jun 2010, 08:18
10
9
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.
_________________
General Discussion
Manager
Manager
avatar
Joined: 30 May 2010
Posts: 165
Re: Remainder when k^2/8?  [#permalink]

Show Tags

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.
Manager
Manager
avatar
Joined: 30 May 2010
Posts: 165
Re: Remainder when k^2/8?  [#permalink]

Show Tags

New post 20 Jun 2010, 08:21
1
I think picking a number is a lot easier on this one.
Manager
Manager
avatar
Joined: 27 Mar 2010
Posts: 77
Re: Remainder when k^2/8?  [#permalink]

Show Tags

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.
SVP
SVP
User avatar
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1824
Concentration: General Management, Nonprofit
GMAT ToolKit User
Re: Remainder when k^2/8?  [#permalink]

Show Tags

New post 20 Jun 2010, 22:27
2
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.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58445
Re: Remainder when k^2/8?  [#permalink]

Show Tags

New post 21 Jun 2010, 02:52
2
2
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 --> \(5=0*6+5\).
_________________
Manager
Manager
avatar
Joined: 06 Apr 2010
Posts: 64
GMAT ToolKit User
Re: Remainder when k^2/8?  [#permalink]

Show Tags

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.
Director
Director
avatar
Joined: 03 Aug 2012
Posts: 660
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Re: Remainder when k^2/8?  [#permalink]

Show Tags

New post 09 Aug 2013, 06:36
1
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
Senior Manager
Senior Manager
avatar
Joined: 10 Jul 2013
Posts: 289
Re: If k = 2n - 1, where n is an integer, what is the remainder  [#permalink]

Show Tags

New post 10 Aug 2013, 14:03
1
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
_________________
Asif vai.....
Current Student
User avatar
D
Joined: 12 Aug 2015
Posts: 2567
Schools: Boston U '20 (M)
GRE 1: Q169 V154
GMAT ToolKit User
Re: If k = 2n - 1, where n is an integer, what is the remainder  [#permalink]

Show Tags

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
_________________
Manager
Manager
avatar
G
Joined: 13 Nov 2018
Posts: 118
Location: India
GMAT 1: 700 Q51 V32
CAT Tests
Re: If k = 2n - 1, where n is an integer, what is the remainder  [#permalink]

Show Tags

New post 26 Nov 2018, 08:00
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.



Pick any number for n=1, 2, 3, or 4

Lets say n=2
so k-4-1=3

so remainder for 9/8=1

correct OA:A
VP
VP
User avatar
D
Joined: 09 Mar 2016
Posts: 1230
Re: If k = 2n - 1, where n is an integer, what is the remainder  [#permalink]

Show Tags

New post 26 Nov 2018, 08:14
From the given expression it can be clearly seen that K is an ODD number.

let K be 7 then \(7= 2*4 - 1\)

\(\frac{7^2}{8}\) yields remainder 1

let K be 9 then \(10= 2*5 - 1\)

\(\frac{9^2}{8}\) yields remainder 1

A

:)
Manager
Manager
avatar
B
Joined: 29 May 2017
Posts: 125
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability
Re: If k = 2n - 1, where n is an integer, what is the remainder  [#permalink]

Show Tags

New post 01 Dec 2018, 01:41
Can we also do the following:

k^2/8 = (2n-1)^2/8 = (4n^2 +1 - 4n)/8

the individual remainders of each term are: 4, 1, -4 and we add these up, we 1.
GMAT Club Bot
Re: If k = 2n - 1, where n is an integer, what is the remainder   [#permalink] 01 Dec 2018, 01:41
Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne