If you divide 7^131 by 5, which remainder do you get? : GMAT Problem Solving (PS) - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 04:29

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

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 62

Kudos [?]: 592 [0], given: 355

Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]

### Show Tags

01 Apr 2014, 04:37
Actually in my sleep yesterday it occurred to me that something such as the following could be done.

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

* 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: 7119
Location: Pune, India
Followers: 2130

Kudos [?]: 13629 [0], given: 222

Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]

### Show Tags

01 Apr 2014, 21:31
jlgdr wrote:
Actually in my sleep yesterday it occurred to me that something such as the following could be done.

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

* 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
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 12 May 2013 Posts: 84 Followers: 2 Kudos [?]: 36 [0], given: 12 Re: question about remainders [#permalink] ### Show Tags 29 Apr 2014, 10:36 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: 36531 Followers: 7070 Kudos [?]: 92962 [0], given: 10541 Re: question about remainders [#permalink] ### Show Tags 30 Apr 2014, 06:41 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: 7119 Location: Pune, India Followers: 2130 Kudos [?]: 13629 [1] , given: 222 Re: question about remainders [#permalink] ### Show Tags 30 Apr 2014, 22:27 1 This post received KUDOS Expert's post 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Intern
Joined: 14 May 2014
Posts: 45
Followers: 0

Kudos [?]: 40 [0], given: 1

Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]

### Show Tags

21 May 2014, 06:19
If we raise the power of 7 by 2 , last digit will be 9 (7X7 = 49)
If we raise the power of 7 by 3, last digit will be 3 (7X7X7 = 343)
If we raise the power of 7 by 4 , last digit will be 1 (A9 X A9 = XYZ1; Here A is 49)
If we raise the power of 7 by multiple of 4 , last digit will be always 1 .

Now 131 = 4X32 + 3

Raising the power of 7 by 131 is equal to raising the power by 4X32 and then multiplying by 7^3

last digit of first term will be 1 and second term will be 3

Hence last digit of 7^131 will be 1X3 = 3

Since last digit is less than 5, if we divide the number by 5 we will get 3 as remainder ( If last digit is 5 or greater than 5 we have to subtract 5 from last digit to get the remainder)

_________________

Help me with Kudos if it helped you "

Mathematics is a thought process.

Current Student
Joined: 02 Jul 2012
Posts: 215
Location: India
Schools: IIMC (A)
GMAT 1: 720 Q50 V38
GPA: 2.6
WE: Information Technology (Consulting)
Followers: 15

Kudos [?]: 224 [0], given: 84

Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]

### Show Tags

20 Oct 2014, 01:25
If you divide 7^131 by 5, which remainder do you get?

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

We can solve this problem using the two ways:

1. This method is limited to this question or any question which has 5 or 10 or their power in the denominator.

If I can know the last 2 digits of this, I can know what will be the remainder of 7^131 when divided by 5, by dividing the last 2 digits by 5.

To calculate the last two digits, we need to see the pattern of the digits -
7
49
43
01
07
49

The cyclicity is 4. Now dividing 131 by cliclicity to know the 2 digit number. The remainder is 3 thus the last 3 digits will be 03.

Now a number with last 2 digits as 03, will give remainder 3 when divided by 5.

Thus 3.

The second method is generic and will work with all the possible scenarios.

When 7 is divided by 5, the remainder is 2.
2 has a cyclicity of 4 with 131, we'll be left with 3.
2^3 = 8
when we divide this by 5, we get the remainder 3.

Thus Ans = 3 (D)
_________________

Give KUDOS if the post helps you...

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13423
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]

### Show Tags

26 Jan 2016, 20:49
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 29 May 2015
Posts: 6
Followers: 0

Kudos [?]: 0 [0], given: 161

Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]

### Show Tags

26 Jan 2016, 20:56
VeritasPrepKarishma
Bunuel
I started doing this question a couple of wks after going through remainders and binomial theorem.
My first instinct was to use the binomial theorem, and I failed to remember it/do it correctly.

Can you help me understand when I would use binomial theorem, vs. the simple solution that is used here of just finding the pattern?

My suspicion is that binomial theorem will be used if we have a denominator >10. Please let me know if my understanding is correct. Thank you!!
BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 1886
Followers: 46

Kudos [?]: 353 [0], given: 447

Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]

### Show Tags

21 Mar 2016, 08:34
Cyclicity of 7 is 4 => hence the answer => 3 i.e Option => D
_________________

Mock Test -1 (Integer Properties Basic Quiz) ---> http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1676182

Mock Test -2 (Integer Properties Advanced Quiz) --->http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1765951

Give me a hell yeah ...!!!!!

Re: If you divide 7^131 by 5, which remainder do you get?   [#permalink] 21 Mar 2016, 08:34

Go to page   Previous    1   2   [ 30 posts ]

Similar topics Replies Last post
Similar
Topics:
3 When the integer n is divided by 8, the remainder is 5. Which of the 8 23 Jul 2015, 02:11
4 When integer n is divided by 15, the remainder is 5. Which of the foll 4 25 Jun 2015, 00:11
3 When the integer k is divided by 7, the remainder is 5. Which of the f 4 04 Dec 2014, 06:40
19 What is the remainder when you divide 2^200 by 7? 16 16 Oct 2012, 19:42
10 What is the remainder when you divide 2^200 by 7? 8 02 Oct 2011, 10:50
Display posts from previous: Sort by