Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 23 May 2017, 03:18

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# remainder

Author Message
Senior Manager
Joined: 02 Dec 2007
Posts: 450
Followers: 3

Kudos [?]: 222 [0], given: 6

### Show Tags

23 Jul 2008, 05:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Guys can anyone explain how to approach such questions
If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

3^3 +2 / 5 = 4 . Is this approach correct?
Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10
Followers: 8

Kudos [?]: 56 [0], given: 0

### Show Tags

23 Jul 2008, 05:58
Nihit wrote:
Guys can anyone explain how to approach such questions
If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

3^3 +2 / 5 = 4 . Is this approach correct?

4 is indeed the correct answer.

The remainder of 3^1 when divided by 5 is 3
The remainder of 3^2 when divided by 5 is 4
The remainder of 3^3 when divided by 5 is 2
The remainder of 3^4 when divided by 5 is 1

When you reach "1" it means that the period you described is going to repeat over and over.

Therefore:

The remainder of 3^(4k+1) when divided by 5 is 3, with k any non-negative integer
The remainder of 3^(4k+2) when divided by 5 is 4, with k any non-negative integer
The remainder of 3^(4k+3) when divided by 5 is 2, with k any non-negative integer
The remainder of 3^(4k+4) when divided by 5 is 1, with k any non-negative integer

Here: 3^(8n+3) = 3^(4*(2n)+3) so the remainder when divided by 5 is 2

Add 2 to this number and the remainder when divided by 5 becomes 4

Senior Manager
Joined: 06 Apr 2008
Posts: 437
Followers: 1

Kudos [?]: 155 [0], given: 1

### Show Tags

23 Jul 2008, 07:33
Nihit wrote:
Guys can anyone explain how to approach such questions
If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

3^3 +2 / 5 = 4 . Is this approach correct?

The way I would tackle this Q is following

If you see 3^n then unit digit gets repeated after every four numbers

3, 9 , 27, 81, 243

Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9. And remainder of any number with unit digit 9 is 4
Retired Moderator
Joined: 18 Jul 2008
Posts: 975
Followers: 10

Kudos [?]: 219 [0], given: 5

### Show Tags

23 Jul 2008, 17:12
Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9

I do not understand why this is done (the purpose) - can you assist? thx

nmohindru wrote:
Nihit wrote:
Guys can anyone explain how to approach such questions
If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

3^3 +2 / 5 = 4 . Is this approach correct?

The way I would tackle this Q is following

If you see 3^n then unit digit gets repeated after every four numbers

3, 9 , 27, 81, 243

Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9. And remainder of any number with unit digit 9 is 4
Intern
Joined: 20 Jul 2008
Posts: 24
Followers: 0

Kudos [?]: 43 [0], given: 0

### Show Tags

23 Jul 2008, 20:43
nmohindru wrote:
Nihit wrote:
Guys can anyone explain how to approach such questions
If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

3^3 +2 / 5 = 4 . Is this approach correct?

The way I would tackle this Q is following

If you see 3^n then unit digit gets repeated after every four numbers

3, 9 , 27, 81, 243

Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9. And remainder of any number with unit digit 9 is 4

Can you please explain the second part?
Manager
Joined: 27 May 2008
Posts: 141
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

23 Jul 2008, 20:47
Its a typical Remainder theorem problem.

Here's my approach
Remainder
3^1/5 = 3
3^2/5 = 4
3^3/5 = 2
3^4/5 = 1

Now every power of the order of 3^4n will give a remainder of 1.

3^(8n+3) = 3^8n. 3^3

Now as per the remainder theorem:

the remainders could of 3^8n and 3^3 can be multiplied.

3^8n will give a remainder of 1, 3^3 will give a remainder of 2.

1*2+2/5 will give a remainder of 4.
Senior Manager
Joined: 19 Mar 2008
Posts: 352
Followers: 1

Kudos [?]: 62 [0], given: 0

### Show Tags

24 Jul 2008, 00:20
Consider the unit digit of the index of 3,
3^1 is 3
3^2 is 9
3^3 is 7
3^4 is 1

and then repeating 3, 9, 7, 1

So the unit digit of 3^(8n+3) is 7

So the unit digit of 3^(8n+3) +2 is 9

Which gives the reminder of 4 when divided by 5

(E)
Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10
Followers: 8

Kudos [?]: 56 [0], given: 0

### Show Tags

24 Jul 2008, 00:24
The "unit digit repeats" method only works because the exercise asks to divide by 5. If it were by 4, 6, 7, etc... it wouldn't work.

That's why I think the "reminder repeats" method is better since it is a general method.
Manager
Joined: 14 Jun 2008
Posts: 161
Followers: 1

Kudos [?]: 34 [0], given: 0

### Show Tags

24 Jul 2008, 03:47
3^(8n+3) + 2

= 3 ^(4N) * 3 * (3) + 2
= (3 ^ 4)^N * 9 + 2
= (81)^N * 3 * 9 + 2

81 ^ N will always have a 1 in the units place.
so when multiplied by 3, will have 3 in the units place
which
so when multiplied by 9, will have 7 in the units place

adding two will give a 9 in the units place.

any integer with a 9 in the units place, when divided by 5 will give a remainder of 4

Hence D

OA?
Retired Moderator
Joined: 18 Jul 2008
Posts: 975
Followers: 10

Kudos [?]: 219 [0], given: 5

### Show Tags

24 Jul 2008, 06:11
rahulgoyal1986 -

1*2+2/5 will give a remainder of 4.

But:

2/5 gives remainder of 3.

3+2 = 5

Did I miss something?

rahulgoyal1986 wrote:
Its a typical Remainder theorem problem.

Here's my approach
Remainder
3^1/5 = 3
3^2/5 = 4
3^3/5 = 2
3^4/5 = 1

Now every power of the order of 3^4n will give a remainder of 1.

3^(8n+3) = 3^8n. 3^3

Now as per the remainder theorem:

the remainders could of 3^8n and 3^3 can be multiplied.

3^8n will give a remainder of 1, 3^3 will give a remainder of 2.

1*2+2/5 will give a remainder of 4.
Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10
Followers: 8

Kudos [?]: 56 [0], given: 0

### Show Tags

24 Jul 2008, 06:58
rahulgoyal1986 -

1*2+2/5 will give a remainder of 4.

But:

2/5 gives remainder of 3.

3+2 = 5

Did I miss something?

This is just that 2/5 gives a remainder of 2 and not 3

2 = 0*5 + 2
Manager
Joined: 27 May 2008
Posts: 141
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

24 Jul 2008, 07:34
judokan wrote:
Consider the unit digit of the index of 3,
3^1 is 3
3^2 is 9
3^3 is 7
3^4 is 1

and then repeating 3, 9, 7, 1

So the unit digit of 3^(8n+3) is 7

So the unit digit of 3^(8n+3) +2 is 9

Which gives the reminder of 4 when divided by 5

(E)

Wrong approach:

Here's the way to proove it

Say 4^1/7 = 4
4^2 = 16
4^3= 64

Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.
Manager
Joined: 27 May 2008
Posts: 141
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

24 Jul 2008, 07:36
Oski wrote:
rahulgoyal1986 -

1*2+2/5 will give a remainder of 4.

But:

2/5 gives remainder of 3.

3+2 = 5

Did I miss something?

This is just that 2/5 gives a remainder of 2 and not 3

2 = 0*5 + 2

Oski

1*2+2/5 means 4 on Nr and 5 on Dr. thus a remainder of 4.

If still not clear feel free to quote again
Manager
Joined: 27 May 2008
Posts: 141
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

24 Jul 2008, 07:37
sset009 wrote:
3^(8n+3) + 2

= 3 ^(4N) * 3 * (3) + 2
= (3 ^ 4)^N * 9 + 2
= (81)^N * 3 * 9 + 2

81 ^ N will always have a 1 in the units place.
so when multiplied by 3, will have 3 in the units place
which
so when multiplied by 9, will have 7 in the units place

adding two will give a 9 in the units place.

any integer with a 9 in the units place, when divided by 5 will give a remainder of 4

Hence D

OA?

Wrong approach:

Here's the way to proove it

Say 4^1/7 = 4
4^2 = 16
4^3= 64

Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.
Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10
Followers: 8

Kudos [?]: 56 [0], given: 0

### Show Tags

24 Jul 2008, 08:21
rahulgoyal1986 wrote:
Oski

1*2+2/5 means 4 on Nr and 5 on Dr. thus a remainder of 4.

If still not clear feel free to quote again

I was just quoting the "2/5 gives remainder of 3"

2/5 gives a remainder of 2, and not 3 (that's what I was saying above )

Last edited by Oski on 24 Jul 2008, 08:52, edited 1 time in total.
Retired Moderator
Joined: 18 Jul 2008
Posts: 975
Followers: 10

Kudos [?]: 219 [0], given: 5

### Show Tags

24 Jul 2008, 08:35
Thanks gang I see my mistake.
Manager
Joined: 14 Jun 2008
Posts: 161
Followers: 1

Kudos [?]: 34 [0], given: 0

### Show Tags

24 Jul 2008, 12:13
rahulgoyal1986 wrote:

Wrong approach:
Here's the way to proove it

Say 4^1/7 = 4
4^2 = 16
4^3= 64

Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.

thanx, i think i see the mistake

could you please state the remainder theorem?
cheers
Senior Manager
Joined: 06 Apr 2008
Posts: 437
Followers: 1

Kudos [?]: 155 [0], given: 1

### Show Tags

25 Jul 2008, 00:06
rahulgoyal1986 wrote:
nmohindru wrote:
Nihit wrote:
Guys can anyone explain how to approach such questions
If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

3^3 +2 / 5 = 4 . Is this approach correct?

The way I would tackle this Q is following

If you see 3^n then unit digit gets repeated after every four numbers

3, 9 , 27, 81, 243

Wrong approach:

Here's the way to proove it

Say 4^1/7 = 4
4^2 = 16
4^3= 64

Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.

Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9. And remainder of any number with unit digit 9 is 4

Well hold on ... where did I say that 4^n gets repeated ?

Each specific number has its own properties e.g. if number has to be divisible by 3 then total of digits should be divisible by 3. This does not mean that if total of digits is divisible by 4 then it is divisible by 4 as well. 4 has its own properties.

I hope you get what I mean
Manager
Joined: 27 May 2008
Posts: 141
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

25 Jul 2008, 00:19
@ nmohindru

Frd

I cud not understand your approach. Kindly solve 4^24/7 for the remainder from your approach and post. Probably you have an easier and shorter way.
Senior Manager
Joined: 06 Apr 2008
Posts: 437
Followers: 1

Kudos [?]: 155 [0], given: 1

### Show Tags

25 Jul 2008, 00:34
rahulgoyal1986 wrote:
@ nmohindru

Frd

I cud not understand your approach. Kindly solve 4^24/7 for the remainder from your approach and post. Probably you have an easier and shorter way.

Rahul you are trying to compare apples with oranges

When divisor is 5 no matter what is tens digit the units digit always determine the remainder. In case of 7 this does not hold true so approach will be different.

Each number's specific properties regarding their squares, remainders etc are applied after seeing a particular Q and is only used to speed up that particular Q and should not be used to solve every Q of that nature.

Take a look at example below

If Q is whether 249678 is divisible by 11 then I would find sum of each even digit and compare with sum of each odd digit. If sum is equal that means number is divisible by 11. Now if Q is whether same number is divisible by 9 then I will sum each digit and see whether sum is divisible by 9 or not. So different approaches allow me to get the solution faster.

A consistent approach will be to divide the number and see which will give you correct answers for all the similar Qs.
Re: remainder   [#permalink] 25 Jul 2008, 00:34

Go to page    1   2    Next  [ 23 posts ]

Display posts from previous: Sort by