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

 It is currently 14 Aug 2018, 08:32

### 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

# If you divide 7^131 by 5, which remainder do you get?

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

### Hide Tags

Intern
Joined: 01 Sep 2009
Posts: 6
If you divide 7^131 by 5, which remainder do you get?  [#permalink]

### Show Tags

Updated on: 19 Oct 2012, 03:39
7
20
00:00

Difficulty:

35% (medium)

Question Stats:

66% (00:52) correct 34% (00:52) wrong based on 1355 sessions

### HideShow timer Statistics

If you divide 7^131 by 5, which remainder do you get?

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

Originally posted by blackcrow on 01 Sep 2009, 04:06.
Last edited by Bunuel on 19 Oct 2012, 03:39, edited 1 time in total.
Renamed the topic and edited the question.
##### Most Helpful Expert Reply
Math Expert
Joined: 02 Sep 2009
Posts: 47898
Re: question about remainders  [#permalink]

### Show Tags

19 Feb 2011, 03:26
4
5
144144 wrote:
jeeteshsingh wrote:
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?

7^131 Mod 5
= (7 Mod 5) ^ 131
= 2 ^ 131
= (2 Mod 5)^ 130 * (2 Mod 5)
= (4 Mod 5)^65 * 2
= (-1)^65 * 2
= -2

Therefore remainder = 5-2 = 3 (since -2 obtained is negative)

Therefore D

can someone plz explain me this other way? thanks.

Check this first: 218-if-x-and-y-are-positive-integers-what-is-the-remainder-109636.html#p875157

If you divide 7^131 by 5, which remainder do you get?
A. 0
B. 1
C. 2
D. 3
E. 4

Last digit of 7^(positive integer) repeats in blocks of 4: {7, 9, 3, 1} - {7, 9, 3, 1} - ... (cyclicity of 7 in power is 4).

As the remainder upon division 131 by 4 (cyclicity) is 3 then the last digit of 7^131 is the same as that of 7^3 so 3 (the third digit from the pattern {7, 9, 3, 1}). Now, any positive integer ending with 3 upon division by 5 yields the remainder of 3.

Answer: D.

Hope it's clear.
_________________
##### Most Helpful Community Reply
Manager
Joined: 25 Aug 2009
Posts: 168
Re: question about remainders  [#permalink]

### Show Tags

01 Sep 2009, 05:31
2
4
whenever you come across a question like this :-- a number raised to higher power of integer divided by 2,3,4,5,6,7,8,9..
and asking for remainder...then, the question is actually asking for the unit digit of the numerator..

now,
3^4 has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 3.

6 to the power anything, always has a unit digit of 6

7^4 also has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 7

now, 151 = 148 + 3

7^148 * 7^3 = unit digit of 1 * unit digit of 3 = unit digit of 3..

any number divided by 5 gives remainder = unit digit of the number, when unit digit is less than 5

and (unit digit - 5) , when unit digit is greator than 5
##### General Discussion
Director
Joined: 04 Jan 2008
Posts: 818
Re: question about remainders  [#permalink]

### Show Tags

01 Sep 2009, 04:18
1
2
Same as asking

If you divide 7^3 by 5, which remainder do you get?
hence OA must be d) 3

because 131=4(K) + 3

now where from I am getting these
follow the pattern

when you place 1,you get 7^1 and results in 7 in unit digit

1---7
2---49--hence 9
3---343---hence 3
4--1

now its repeated

5-7
6-9
7-3
8-1
so on..

131=128+3=4(32)+3

hope i am making sense:)

blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?

_________________

http://gmatclub.com/forum/math-polygons-87336.html
http://gmatclub.com/forum/competition-for-the-best-gmat-error-log-template-86232.html

Intern
Joined: 24 Aug 2009
Posts: 34
Re: question about remainders  [#permalink]

### Show Tags

01 Sep 2009, 08:50
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?

If you look at the power of 7, it shows a repeated trend at the Unit's digit. For e.g.:
7^1 = 7 (7 at unit place)
7^2 = 49 (9 at unit place)
7^3 = 343 (3 at unit place)
7^4 = 2401 (1 at unit place)
7^5 = 16807 (7 at unit place)

So if you see, this trend of 7,9,3,1......7,9,3,1.......repeats itself.
7^131 will give 3 as unit's digit and when it is divided by 5, it will give the remainder as 3.

Hope it helps.
Senior Manager
Joined: 23 May 2008
Posts: 385
Re: question about remainders  [#permalink]

### Show Tags

01 Sep 2009, 09:27
2
when you divide 7 by 5 we get 2 as remainder

so now we can write it as 7^131 \ 5 = 2 ^131 \ 5

= 2^ 120 * 2^ 11 \ 5
= (2^4)^30 * 2048 ( when we multiply 2^11 we will get the value as 2048)
= 16^30 * 2048
= 1^ 30 * 2048 ( when we divide 16 by 5 the remainder is 1 so 1^30 = 1

finally divide 2048 by 5 and u get the remainder = 3

hence 1 * 3 = 3

So answer is D
Senior Manager
Joined: 23 May 2008
Posts: 385
Re: question about remainders  [#permalink]

### Show Tags

01 Sep 2009, 09:36
5
there is another easy method

cyclicity of 7 = 4

so when we divide 131 by 4 we get 3 as remainder

therefore 7^3 = 343

now divide 343 by 5 and u get the remainder as 3

so answer is D........

Note : cyclicity is used to find the last digit of the number i.e in case of powers it is specially usefull
Manager
Joined: 12 Aug 2009
Posts: 87
Re: question about remainders  [#permalink]

### Show Tags

01 Sep 2009, 20:19
7*7*7*7= xxx1 -> only last digit is important.

you can use this for 32 time and get to 7^128 and last digit will be 1.

1 *7*7*7 = 343 and divided by 5 leaves a remainder of 3.
Senior Manager
Joined: 22 Dec 2009
Posts: 320
Re: question about remainders  [#permalink]

### Show Tags

31 Jan 2010, 07:24
1
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?

7^131 Mod 5
= (7 Mod 5) ^ 131
= 2 ^ 131
= (2 Mod 5)^ 130 * (2 Mod 5)
= (4 Mod 5)^65 * 2
= (-1)^65 * 2
= -2

Therefore remainder = 5-2 = 3 (since -2 obtained is negative)

Therefore D
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Intern
Joined: 06 Dec 2010
Posts: 3
Re: question about remainders  [#permalink]

### Show Tags

07 Dec 2010, 11:18
7^1 = 7 => 7/5 has R2
7^2 = 49 => 49/5 has R4
7^3 = 343 => 343/5 has R3
7^4 = 2401 =>2401/5 has R1
7^5 = 16807 => 16807/5 has R2

So it repeats every 4. 131/4 = 128 R3
So 7^128 / 5 has the same remainder as 7^4 /5
7^129 / 5 has the same remainder as 7^1 / 5
7^130 / 5 has the same remainder as 7^2 / 5
7^131 / 5 has the same remainder as 7^3 / 5, which is 3.

The answer is 3.
Math Expert
Joined: 02 Sep 2009
Posts: 47898
If you divide 7^131 by 5, which remainder do you get?  [#permalink]

### Show Tags

09 Mar 2014, 13:08
For more on this kind of questions check Units digits, exponents, remainders problems collection.
_________________
SVP
Joined: 06 Sep 2013
Posts: 1852
Concentration: Finance
Re: If you divide 7^131 by 5, which remainder do you get?  [#permalink]

### Show Tags

01 Apr 2014, 05:37
Actually in my sleep yesterday it occurred to me that something such as the following could be done.
Please advice is this method is not indeed flawed

Whenever we have 5 in the denominator we can multiply both numerator and denominator by 2 so we get 10 in the denominator and therefore we can just find the units digit in the numerator

It would be something like this

7^131/5 = 7^131*2/10

7 has cyclisity of 7,9,3,1 therefore units digit is 3*2=6 but since we multiplied by 2, then 3

Answer is thus 3

* Remember that the remainder of a number when divided by 2 is the units digit. Likewise, the remainder of a number when divided by 100 are the last two digits

Hope it makes sense
Cheers
J
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8184
Location: Pune, India
Re: If you divide 7^131 by 5, which remainder do you get?  [#permalink]

### Show Tags

01 Apr 2014, 22:31
jlgdr wrote:
Actually in my sleep yesterday it occurred to me that something such as the following could be done.
Please advice is this method is not indeed flawed

Whenever we have 5 in the denominator we can multiply both numerator and denominator by 2 so we get 10 in the denominator and therefore we can just find the units digit in the numerator

It would be something like this

7^131/5 = 7^131*2/10

7 has cyclisity of 7,9,3,1 therefore units digit is 3*2=6 but since we multiplied by 2, then 3

Answer is thus 3

* Remember that the remainder of a number when divided by 2 is the units digit. Likewise, the remainder of a number when divided by 100 are the last two digits

Hope it makes sense
Cheers
J

Your method is correct but for this question. If you generalize it, it could be flawed. The reason is this: if there is a number with units digit as 6 (e.g. .......6), when you divide it by 2, the last digit could be 3 but it could also be 8. Here we know that we multiplied a power of 7 so the last digit CANNOT be 8 so your method is fine but be careful when you try to generalize it.

Also, you don't need to multiply by 2 to make the denominator 10. Even when the denominator is 5, the last digit is enough to give you the remainder. If the last digit is from 0 to 4, the remainder is the same as the last digit. If the last digit is from 5 to 9, remainder is (last digit - 5).
7^131 ends with 3 so remainder when divided by 5 must be 3.
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 12 May 2013 Posts: 73 Re: question about remainders [#permalink] ### Show Tags 29 Apr 2014, 11:36 Check this first: 218-if-x-and-y-are-positive-integers-what-is-the-remainder-109636.html#p875157 If you divide 7^131 by 5, which remainder do you get? A. 0 B. 1 C. 2 D. 3 E. 4 Last digit of 7^(positive integer) repeats in blocks of 4: {7, 9, 3, 1} - {7, 9, 3, 1} - ... (cyclicity of 7 in power is 4). As the remainder upon division 131 by 4 (cyclicity) is 3 then the last digit of 7^131 is the same as that of 7^3 so 3 (the third digit from the pattern {7, 9, 3, 1}). Now, any positive integer ending with 3 upon division by 5 yields the remainder of 3. Answer: D. Hope it's clear.[/quote] bunuel, as per the theory in gmat club math book "When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller integer." i tried solving the above sum and got the correct answer just want to know whether the method is correct or not , as 3 is smaller than 5 so the remainder would be 3 thanks Math Expert Joined: 02 Sep 2009 Posts: 47898 Re: question about remainders [#permalink] ### Show Tags 30 Apr 2014, 07:41 adymehta29 wrote: bunuel, as per the theory in gmat club math book "When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller integer." i tried solving the above sum and got the correct answer just want to know whether the method is correct or not , as 3 is smaller than 5 so the remainder would be 3 thanks Yes, 3 divided by 5 gives the remainder o f 3. I don't understand the rest of your post though... _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8184 Location: Pune, India Re: question about remainders [#permalink] ### Show Tags 30 Apr 2014, 23:27 1 adymehta29 wrote: Check this first: 218-if-x-and-y-are-positive-integers-what-is-the-remainder-109636.html#p875157 If you divide 7^131 by 5, which remainder do you get? A. 0 B. 1 C. 2 D. 3 E. 4 Last digit of 7^(positive integer) repeats in blocks of 4: {7, 9, 3, 1} - {7, 9, 3, 1} - ... (cyclicity of 7 in power is 4). As the remainder upon division 131 by 4 (cyclicity) is 3 then the last digit of 7^131 is the same as that of 7^3 so 3 (the third digit from the pattern {7, 9, 3, 1}). Now, any positive integer ending with 3 upon division by 5 yields the remainder of 3. Answer: D. Hope it's clear. bunuel, as per the theory in gmat club math book "When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller integer." i tried solving the above sum and got the correct answer just want to know whether the method is correct or not , as 3 is smaller than 5 so the remainder would be 3 thanks Hey Bunuel, If you don't mind, I think I see the problem Ady is facing so I will take it up. Note that the gmat club math book says "When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller integer." So when 3 is divided by 5, remainder is 3. When 10 is divided by 39, remainder is 10 and so on... But in this question, you have $$7^{131}$$ divided by 5. You find that $$7^{131}$$ ends in 3. This means it is a number which looks something like this: $$7^{131} = 510320.....75683$$ (a huge number that ends in 3. Other than 3, I have used some random digits.) So the remainder is 3 not because 3 is smaller than 5. The actual number is much much bigger than 5. For example, if you have 276543 divided by 5, will you say that the number is smaller than 5 and hence the remainder is 3? No. The remainder is 3 because the number ends in 3 and when you divide such a number by 5, you know that the number which is 3 steps before it i.e. 276540 in this case, is a multiple of 5 (every number ending in 0 or 5 is a multiple of 5). That is the reason that you will be left with 3 when you divide this number by 5. Similarly, 4367 divided by 5 leaves remainder 2 because 4365 is divisible by 5 and so on... The first part of this post discusses remainders in case of division by 5: http://www.veritasprep.com/blog/2014/03 ... emainders/ _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Intern
Joined: 03 Nov 2014
Posts: 17
If you divide 7^131 by 5, which remainder do you get?  [#permalink]

### Show Tags

28 Mar 2018, 00:51
hi guys, the cycle is 7,9,3,1 and unite digit is 3 here so when we divide 3/5 the remainder is 3, how come it comes 3? please explain
Math Expert
Joined: 02 Sep 2009
Posts: 47898
Re: If you divide 7^131 by 5, which remainder do you get?  [#permalink]

### Show Tags

28 Mar 2018, 02:35
1
imtisalsiddiqui wrote:
hi guys, the cycle is 7,9,3,1 and unite digit is 3 here so when we divide 3/5 the remainder is 3, how come it comes 3? please explain

Your question is not clear. Any positive integer ending with 3 upon division by 5 yields the remainder of 3:

3 divided by 5 gives the remainder of 3;
13 divided by 5 gives the remainder of 3;
23 divided by 5 gives the remainder of 3;
...
_________________
Intern
Joined: 03 Nov 2014
Posts: 17
Re: If you divide 7^131 by 5, which remainder do you get?  [#permalink]

### Show Tags

28 Mar 2018, 03:09
Thanks for clearing up my confusion, understood!
VP
Status: It's near - I can see.
Joined: 13 Apr 2013
Posts: 1211
Location: India
Concentration: International Business, Operations
GMAT 1: 480 Q38 V22
GPA: 3.01
WE: Engineering (Consulting)
Re: If you divide 7^131 by 5, which remainder do you get?  [#permalink]

### Show Tags

29 Mar 2018, 23:55
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

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

7 has cyclicity of 4 = 7, 9, 3, 1

Dividing 131 by 4, we get remainder 3

In cyclicity, third term we have is "3"

Hence (D)
_________________

"Do not watch clock; Do what it does. KEEP GOING."

Re: If you divide 7^131 by 5, which remainder do you get? &nbs [#permalink] 29 Mar 2018, 23:55

Go to page    1   2    Next  [ 22 posts ]

Display posts from previous: Sort by

# If you divide 7^131 by 5, which remainder do you get?

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

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.