It is currently 21 Sep 2017, 20:02

### 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

# If n is a positive integer, what is the remainder when

Author Message
Manager
Joined: 31 Oct 2007
Posts: 112

Kudos [?]: 72 [0], given: 0

Location: Frankfurt, Germany
If n is a positive integer, what is the remainder when [#permalink]

### Show Tags

17 Jul 2008, 02:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Show the quickest way to solve this question.

Thks
_________________

Jimmy Low, Frankfurt, Germany
Blog: http://mytrainmaster.wordpress.com
GMAT Malaysia: http://gmatmalaysia.blogspot.com

Kudos [?]: 72 [0], given: 0

Current Student
Joined: 11 May 2008
Posts: 555

Kudos [?]: 216 [0], given: 0

Re: Math Set 1: Q13 [#permalink]

### Show Tags

17 Jul 2008, 03:41
3^(8n+3)+2 = 3^8n*3^3+2.
3 ^( anything to the form 4n where n=1,2,3,4) will end with 1.
(8n=4*2n)
3^3 ends with 7
so ....1*....7=no. ending with 7.
so (no. ending with 7) +2=no. ending with 9.
when divided by 5,remaider is 4

Kudos [?]: 216 [0], given: 0

Director
Joined: 14 Aug 2007
Posts: 726

Kudos [?]: 208 [0], given: 0

Re: Math Set 1: Q13 [#permalink]

### Show Tags

17 Jul 2008, 04:39
Another E.

arjtryarjtry has already given a great explanation!

arjtryarjtry wrote:
3^(8n+3)+2 = 3^8n*3^3+2.
3 ^( anything to the form 4n where n=1,2,3,4) will end with 1.
(8n=4*2n)
3^3 ends with 7
so ....1*....7=no. ending with 7.
so (no. ending with 7) +2=no. ending with 9.
when divided by 5,remaider is 4

Kudos [?]: 208 [0], given: 0

Manager
Joined: 15 Jul 2008
Posts: 206

Kudos [?]: 66 [0], given: 0

Re: Math Set 1: Q13 [#permalink]

### Show Tags

17 Jul 2008, 05:50
In case you are not familiar with rules governing such powers, just write down last digits for powers of 3.

3^1 -- last digit =3
3^2 -- last digit = 9
3^3 -- last digit = 7
3^4 -- last digit = 1
3^5 -- last digit = 3 (starts repeating from here...)

so 8n+3 will always lead you to the 3 step in the iteration. 3,11,19..etc. so last digit 7+2 =9.
when divided by 5, remainder always is 4.

Kudos [?]: 66 [0], given: 0

Senior Manager
Joined: 07 Jan 2008
Posts: 287

Kudos [?]: 48 [0], given: 0

Re: Math Set 1: Q13 [#permalink]

### Show Tags

17 Jul 2008, 10:45
jimmylow wrote:
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

Show the quickest way to solve this question.

Thks

Powers of 3: 1 2 3 4
3 9 7 1 [unit digit]

for n=1,2,3,... we are looking for 3^11, 3^19 and 3^27. They all present unit digit of 7
7+2=9
9/5 => remainder = 4

Kudos [?]: 48 [0], given: 0

Manager
Joined: 11 Apr 2008
Posts: 128

Kudos [?]: 58 [0], given: 0

Location: Chicago
Re: Math Set 1: Q13 [#permalink]

### Show Tags

06 Aug 2008, 16:48
arjtryarjtry wrote:
3^(8n+3)+2 = 3^8n*3^3+2.
3 ^( anything to the form 4n where n=1,2,3,4) will end with 1.
(8n=4*2n)
3^3 ends with 7
so ....1*....7=no. ending with 7.
so (no. ending with 7) +2=no. ending with 9.
when divided by 5,remaider is 4

This method works with a denominator of 5 with any number that ends with a 7 (as noted above):
(27+2)/5 = 5 remainder of 4
(37+2)/5 = 7 remainder of 4
(47+2)/5 = 9 remainder of 4 ---> all remainders of 4

but what if the denominator was 6:
(27+2)/6 = 4 remainder of 5
(37+2)/6 = 6 remainder of 3 ---> you get 2 different remainders. Am I missing something in the method above??
_________________

Factorials were someone's attempt to make math look exciting!!!

Kudos [?]: 58 [0], given: 0

VP
Joined: 17 Jun 2008
Posts: 1378

Kudos [?]: 392 [0], given: 0

Re: Math Set 1: Q13 [#permalink]

### Show Tags

06 Aug 2008, 18:01
jimmylow wrote:
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

Show the quickest way to solve this question.

Thks

i took n=1, $$3^{8n+3} + 2$$ = (27(3^8)+2)/5 = 5+((2(3^8)+2)/5)
3^8=6561 => 5+(((6561*2)+2 )/2)=> remainder = 4

Hence this is the method i used.Here there is no question of what can be the remainder its just what is the remnainder which means take any value of n and solve
_________________

cheers
Its Now Or Never

Kudos [?]: 392 [0], given: 0

SVP
Joined: 29 Aug 2007
Posts: 2473

Kudos [?]: 832 [0], given: 19

Re: Math Set 1: Q13 [#permalink]

### Show Tags

07 Aug 2008, 11:03
jimmylow wrote:
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

Show the quickest way to solve this question.

Thks

the quickest, hm....

$$3^{1}$$ = 3
$$3^{2}$$ = 9
$$3^{3}$$ = 27
$$3^{4}$$ = 81
$$3^{5}$$ = 243

see the pattern, so:

= $$3^{8n+3} + 2$$
= $$3^{8n} . 3^{3} + 2$$

since 8n is a multiple of 4 and therefore no matter the value of n, we can drop $$3^{8n}$$
8n:

= $$3^{3} + 2$$
= 29

so 4 is the reminder
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Kudos [?]: 832 [0], given: 19

Re: Math Set 1: Q13   [#permalink] 07 Aug 2008, 11:03
Display posts from previous: Sort by