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

 It is currently 18 Aug 2018, 21:17

### 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

# When positive integer n is divided by 5, the remainder is 1.

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 47983
When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

30 Jan 2014, 01:51
8
49
00:00

Difficulty:

15% (low)

Question Stats:

81% (01:38) correct 19% (01:54) wrong based on 1238 sessions

### HideShow timer Statistics

The Official Guide For GMAT® Quantitative Review, 2ND Edition

When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35

Problem Solving
Question: 68
Category: Arithmetic Properties of numbers
Page: 70
Difficulty: 650

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 47983
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

30 Jan 2014, 01:51
17
26
SOLUTION

When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35

Positive integer n is divided by 5, the remainder is 1 --> $$n=5q+1$$, where $$q$$ is the quotient --> 1, 6, 11, 16, 21, 26, 31, ...
Positive integer n is divided by 7, the remainder is 3 --> $$n=7p+3$$, where $$p$$ is the quotient --> 3, 10, 17, 24, 31, ....

There is a way to derive general formula for $$n$$ (of a type $$n=mx+r$$, where $$x$$ is divisor and $$r$$ is a remainder) based on above two statements:

Divisor $$x$$ would be the least common multiple of above two divisors 5 and 7, hence $$x=35$$.

Remainder $$r$$ would be the first common integer in above two patterns, hence $$r=31$$.

Therefore general formula based on both statements is $$n=35m+31$$. Thus the smallest positive integer k such that k+n is a multiple of 35 is 4 --> $$n+4=35k+31+4=35(k+1)$$.

More about deriving general formula for such problems at: manhattan-remainder-problem-93752.html#p721341
_________________
Manager
Joined: 04 Oct 2013
Posts: 155
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

Updated on: 29 Jan 2015, 22:26
5
4
When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35

Method 1

n is divided by 5, the remainder is 1 ---> $$n= 5x + 1$$
or, n + k = 5x + (1 + k)
So, n + k is divisible by 5, when (1+ k) is a multiple of 5.
Or, Possible values of k are 4, 9, 14,19, 24, 29, 33,.....

n is divided by 7, the remainder is 3 ----> $$n=7y + 3$$
Or, n + k = 7y +(3 + k)
So, n + k is divisible by 7, when (3+ k) is a multiple of 7.
Or, Possible values of k are 4, 11, 18, 25, 32, 39,.....

As the lowest common value is 4, the answer is (B).

Originally posted by arunspanda on 31 Jan 2014, 10:06.
Last edited by arunspanda on 29 Jan 2015, 22:26, edited 1 time in total.
##### General Discussion
Director
Joined: 25 Apr 2012
Posts: 701
Location: India
GPA: 3.21
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

30 Jan 2014, 03:07
3
1
When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35

Sol: Given n=5a+1 where a is any non-negative integer and also n=7b+3 where b is any non-negative integer.....so n is of the form

Possible values of n in case 1 : 1,6,11,16,21,26,31....
Possible value of n in case 2 : 3,10,17, 24,31...

So, n=35C+ 31....Now for K+ n to be multiple of 35 K needs to be 4 so that k+n = 35C+31+4 or 35(c+1)

Ans B.

650 level is okay
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Manager
Joined: 20 Dec 2013
Posts: 245
Location: India
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

30 Jan 2014, 03:26
1
1
Ans.B
n=5p+1=>n=6,11,16,21,26,31,36...
n=7q+3=>n=10,17,24,31,...
First no. common to both the series=31.We have to add 4.

I know there is a better way of doing this kind of question.
Math Expert
Joined: 02 Sep 2009
Posts: 47983
When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

01 Feb 2014, 08:49
Manager
Joined: 18 Oct 2013
Posts: 77
Location: India
Concentration: Technology, Finance
GMAT 1: 580 Q48 V21
GMAT 2: 530 Q49 V13
GMAT 3: 590 Q49 V21
WE: Information Technology (Computer Software)
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

14 Feb 2014, 09:59
1
According to me simplest way to solve this problem is given below:

Let N=5P+1 or N=7Q+3

So 5P+1=7Q+3 => P=Q+((2Q+2)/5) .. Check for value of 2Q+2 divisible by 5 i.e. 4 and hence P=6
So smallest value of N=31 for P=6 and Q=4.

So, N is 31 and smallest number K , which add to N to make it multiple of 35 is 4. so K=4

Intern
Joined: 27 Mar 2014
Posts: 24
When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

24 Jun 2014, 02:27
1
n+k is a multiple of 35 then n+k = 35x => n = 35x - k.
When positive integer n is divided by 5, the remainder is 1 => when integer k is divided by 5, the remainder is 4 => B

[or
When n is divided by 7, the remainder is 3 => when integer is divided by 7, the remainder is 4 => B or C, but we do not need to check this case]
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1835
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

02 Oct 2014, 03:10
1
2
Picked up 7 & added 3 = 10

10 gives a remainder 3 when divided by 7

Wrote down similar numbers. We are looking for numbers which would either end by 1 or 6 (so they would provide a remainder 1 when divided by 5)

10
17
24
31 ........ stop

31 is the number when divided by 5, provides remainder 1 (Its already tested that it provides remainder 3 when divided by 7)

First available multiple of 35 is 35, which is just 4 away from 31

_________________

Kindly press "+1 Kudos" to appreciate

Intern
Joined: 26 Jan 2010
Posts: 17
Location: chile
WE 1:
WE 2:
WE 3:
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

26 Nov 2014, 12:23
REMAINDER

It is suggested to use the division algorithm.

By dividing a positive integer N for 5, residue 1 is obtained.

The algorithm is associated 5X + 1 = N, where N is the dividend, the divisor, and X is 1 is the remainder.

By dividing the same N for 7, remainder 3 is obtained.

The associated algorithm is N = 7Y + 3, where N is the dividend and the divisor and 3 is the remainder.

You have:

N = 5X + 1 and N = 7Y +3

Since N = N, equaling have:

5X +3 +1 = 7Y

Grouping variables, with positive results, it:

5X - 7Y = 2

It is suggested to start giving positive integers X, then watch multiple of 7 is closer to 5X, and meets equality:

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 35, 50, .....

Multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, ....

The condition is:

Multiple of 5 - multiple of 7 = 2

30-28 = 2

Then X = 6 and Y = 4

Thus we obtain N = 31

Smaller value for K, if (K + N) is a multiple of 35?

This value is 4, because 4 + 31 = 35, and 35 is a multiple of 35.

claudio hurtado GMAT GRE SAT math classes part
_________________

Private lessons GMAT QUANT GRE QUANT SAT QUANT
Classes group of 6 students GMAT QUANT GRE QUANT SAT QUANT
Distance learning courses GMAT QUANT GRE QUANT SAT QUANT

Website http://www.gmatchile.cl
Whatsapp +56999410328
Email clasesgmatchile@gmail.com
Skype: clasesgmatchile@gmail.com
Address Avenida Hernando de Aguirre 128 Of 904, Tobalaba Metro Station, Santiago Chile.

Intern
Joined: 13 Mar 2015
Posts: 4
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

24 Mar 2015, 13:24
The fastest way here is to realize that 35 is a factor of both 5 and 7. So you need to add a number to the remainder of both situations so that each number a factor of 5 and 7. By adding 4 to 1 it becomes a factor of 5, and by adding 4 to 3 it becomes a factor of 7. So B must be true
Director
Joined: 10 Mar 2013
Posts: 561
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

26 Aug 2015, 14:41
2
Hi all, actually we don't need pluging all those values for x,y...
n=5x+1 and n=7y+3 --> n+k=> (5x+1+k)/35 so 1+k must be a multiple of 5 if we want this expression to yield an integer so k=4
Use same logic here (7y+3+k)/35 -> 3+k must be a multiple of 7, so k=4
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Retired Moderator
Joined: 29 Oct 2013
Posts: 272
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

01 Dec 2015, 14:44
1
N+4 is multiple of both 5 and 7 so N+4 must be an LCM of 5 and 7 i.e. 35. So the smallest value needed to make it multiple of 35 is 4
_________________

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2759
When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

03 Aug 2016, 09:45
2
Quote:
When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35

We can find the value of n first by just strategically find values that when divided by 5 have a reminder of 1. For example, since the remainder is 1 when n is divided by 5, n will be a [(multiple of 5) + 1] and thus must be one of the following numbers:

1, 6, 11, 16, 21, 26, 31, …

Now we have to find out which of these numbers when divided by 7, have a remainder of 3.

1/7 = 0 remainder 1

6/7 = 0 remainder 6

11/7 = 0 remainder 6

6/7 = 1 remainder 4

16/7 = 2 remainder 2

21/7 = 3 remainder 0

26/7 = 3 remainder 5

31/7 = 4 remainder 3

We can see that 31 is the smallest value of n that satisfies the requirement. So we must determine the value of k such that k + n is a multiple of 35. Obviously, since 4 + 31 = 35 and 35 is a multiple of 35, then the smallest positive integer value of k is 4.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

VP
Joined: 07 Dec 2014
Posts: 1069
When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

Updated on: 12 Jul 2018, 18:02
When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35

n=5q+1
n=7p+3
5q-2=7p
the least multiple of 5 that is two greater than a multiple of 7 is 30
q=6
p=4
n=31
35-31=4
B

Originally posted by gracie on 31 Aug 2016, 16:42.
Last edited by gracie on 12 Jul 2018, 18:02, edited 3 times in total.
Intern
Joined: 13 Nov 2016
Posts: 2
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

19 Dec 2016, 03:01
A number which leaves a remainder x-a when divided by x and leaves a remainder y-a when divided by y can be written as : LCM(x,y)-a
Here the number leaves a remainder 1 when divided by 5 , thus x=5 x-a=1 , therfore a=4
the same number leaves a remainder 3 when divided by 7 , thus y-7 and x-a=3 thus a=4

So the number is LCM(5,7)- 4 which is 35-4 =31 , the smallest number required to get a multiple of 35 is 31+4 =35

Therefore the answer is 4 which is B
CEO
Joined: 12 Sep 2015
Posts: 2706
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

30 Nov 2017, 11:30
Top Contributor
Bunuel wrote:
When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35

There's a nice rule that says, If, when N is divided by D, the remainder is R, then the possible values of N include: R, R+D, R+2D, R+3D,. . .
For example, if k divided by 6 leaves a remainder of 2, then the possible values of k are: 2, 2+6, 2+(2)(6), 2+(3)(6), 2+(4)(6), . . . etc.

When n is divided by 5, the remainder is 1.
So, possible values of n are 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, etc.

When n is divided by 7, the remainder is 3.
So, possible values of n are 3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, etc.

So, we can see that n could equal 31, or 66, or an infinite number of other values.

Important: Since the Least Common Multiple of 7 and 5 is 35, we can conclude that if we list the possible values of n, each value will be 35 greater than the last value.
So, n could equal 31, 66, 101, 136, and so on.

Answer choice A: If we add 3 to any of these possible n-values, the sum is NOT a multiple of 35.
ELIMINATE A

Answer choice B: if we take ANY of these possible n-values, and add 4, the sum will be a multiple of 35.

So, the smallest value of k is 4 such that k+n is a multiple of 35.

RELATED VIDEO FROM OUR COURSE

_________________

Brent Hanneson – Founder of gmatprepnow.com

Intern
Joined: 07 Jun 2015
Posts: 2
When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

13 Jan 2018, 23:57
Bunuel wrote:
SOLUTION

When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35

Positive integer n is divided by 5, the remainder is 1 --> $$n=5q+1$$, where $$q$$ is the quotient --> 1, 6, 11, 16, 21, 26, 31, ...
Positive integer n is divided by 7, the remainder is 3 --> $$n=7p+3$$, where $$p$$ is the quotient --> 3, 10, 17, 24, 31, ....

There is a way to derive general formula for $$n$$ (of a type $$n=mx+r$$, where $$x$$ is divisor and $$r$$ is a remainder) based on above two statements:

Divisor $$x$$ would be the least common multiple of above two divisors 5 and 7, hence $$x=35$$.

Remainder $$r$$ would be the first common integer in above two patterns, hence $$r=31$$.

Therefore general formula based on both statements is $$n=35m+31$$. Thus the smallest positive integer k such that k+n is a multiple of 35 is 4 --> $$n+4=35k+31+4=35(k+1)$$.

Hi Bunuel,

Would you please have a look at my solution? I've got the same answer yet a different path. Can you point it out what's wrong with my thought? Many thanks.

Question:
What is the smallest value of K, when N+K is the multiple of 35?

My solution process:
step 1. n+k=x(35), I was assuming that if x=1(because the hint is 'the smallest integer value'), then n+k=35, so k=35-n
step 2. how to get n? (the same process as you've stated above) n=31
step 3. k=35-n, and n=31, therefore k=4
Math Expert
Joined: 02 Sep 2009
Posts: 47983
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

14 Jan 2018, 00:16
1
bingzhang wrote:
Bunuel wrote:
SOLUTION

When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35

Positive integer n is divided by 5, the remainder is 1 --> $$n=5q+1$$, where $$q$$ is the quotient --> 1, 6, 11, 16, 21, 26, 31, ...
Positive integer n is divided by 7, the remainder is 3 --> $$n=7p+3$$, where $$p$$ is the quotient --> 3, 10, 17, 24, 31, ....

There is a way to derive general formula for $$n$$ (of a type $$n=mx+r$$, where $$x$$ is divisor and $$r$$ is a remainder) based on above two statements:

Divisor $$x$$ would be the least common multiple of above two divisors 5 and 7, hence $$x=35$$.

Remainder $$r$$ would be the first common integer in above two patterns, hence $$r=31$$.

Therefore general formula based on both statements is $$n=35m+31$$. Thus the smallest positive integer k such that k+n is a multiple of 35 is 4 --> $$n+4=35k+31+4=35(k+1)$$.

Hi Bunuel,

Would you please have a look at my solution? I've got the same answer yet a different path. Can you point it out what's wrong with my thought? Many thanks.

Question:
What is the smallest value of K, when N+K is the multiple of 35?

My solution process:
step 1. n+k=x(35), I was assuming that if x=1(because the hint is 'the smallest integer value'), then n+k=35, so k=35-n
step 2. how to get n? (the same process as you've stated above) n=31
step 3. k=35-n, and n=31, therefore k=4

Yes, the smallest value of n is 31, adding 4 gives a multiple of 35.
_________________
Intern
Joined: 07 Jun 2015
Posts: 2
Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

### Show Tags

14 Jan 2018, 02:35
Bunuel wrote:
bingzhang wrote:
Bunuel wrote:
SOLUTION

When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35

Positive integer n is divided by 5, the remainder is 1 --> $$n=5q+1$$, where $$q$$ is the quotient --> 1, 6, 11, 16, 21, 26, 31, ...
Positive integer n is divided by 7, the remainder is 3 --> $$n=7p+3$$, where $$p$$ is the quotient --> 3, 10, 17, 24, 31, ....

There is a way to derive general formula for $$n$$ (of a type $$n=mx+r$$, where $$x$$ is divisor and $$r$$ is a remainder) based on above two statements:

Divisor $$x$$ would be the least common multiple of above two divisors 5 and 7, hence $$x=35$$.

Remainder $$r$$ would be the first common integer in above two patterns, hence $$r=31$$.

Therefore general formula based on both statements is $$n=35m+31$$. Thus the smallest positive integer k such that k+n is a multiple of 35 is 4 --> $$n+4=35k+31+4=35(k+1)$$.

Hi Bunuel,

Would you please have a look at my solution? I've got the same answer yet a different path. Can you point it out what's wrong with my thought? Many thanks.

Question:
What is the smallest value of K, when N+K is the multiple of 35?

My solution process:
step 1. n+k=x(35), I was assuming that if x=1(because the hint is 'the smallest integer value'), then n+k=35, so k=35-n
step 2. how to get n? (the same process as you've stated above) n=31
step 3. k=35-n, and n=31, therefore k=4

Yes, the smallest value of n is 31, adding 4 gives a multiple of 35.

thank you! thank you, Bunuel. It's very nice of you.
Re: When positive integer n is divided by 5, the remainder is 1. &nbs [#permalink] 14 Jan 2018, 02:35

Go to page    1   2    Next  [ 22 posts ]

Display posts from previous: Sort by

# 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®.