It is currently 12 Dec 2017, 10:17

### 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

# Events & Promotions

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

# If M and N are positive integers that have remainders of 1

Author Message
TAGS:

### Hide Tags

Manager
Joined: 22 Jul 2009
Posts: 194

Kudos [?]: 554 [0], given: 6

Location: Manchester UK
If M and N are positive integers that have remainders of 1 [#permalink]

### Show Tags

19 Nov 2009, 13:43
6
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

67% (01:19) correct 33% (01:23) wrong based on 455 sessions

### HideShow timer Statistics

If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M+N?

(A) 86
(B) 52
(C) 34
(D) 28
(E) 10
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Apr 2012, 04:25, edited 1 time in total.
Edited the question and added the OA

Kudos [?]: 554 [0], given: 6

Math Expert
Joined: 02 Sep 2009
Posts: 42571

Kudos [?]: 135363 [6], given: 12691

### Show Tags

19 Nov 2009, 13:56
6
KUDOS
Expert's post
5
This post was
BOOKMARKED
sagarsabnis wrote:
If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M+N?
(A) 86
(B) 52
(C) 34
(D) 28
(E) 10

Guys i am going mad for this question according to me the answer should be E but OA is A please explain how come it cannot be 10

M=6p+1, where p is integer >=0, so M can be 1, 7, 13, etc.
N=6q+3. where q is integer >=0, so N can be 3, 9, 15, etc.

M+N=6(p+q)+4, hence M+N is multiple of 6 plus 4 = 10, 16, 22, 28, 34, etc.

Only answer which is not of this type is 86, 86=13*6+2

Hope it's clear.
_________________

Kudos [?]: 135363 [6], given: 12691

Manager
Joined: 22 Jul 2009
Posts: 194

Kudos [?]: 554 [0], given: 6

Location: Manchester UK

### Show Tags

19 Nov 2009, 14:31
but still can you give me the the value of M and N which will add up to 10. If you check the other way round the least numbers are 7 and 9 which adds up to 16 so how come 10 is possible?

Kudos [?]: 554 [0], given: 6

Math Expert
Joined: 02 Sep 2009
Posts: 42571

Kudos [?]: 135363 [2], given: 12691

### Show Tags

19 Nov 2009, 14:43
2
KUDOS
Expert's post
sagarsabnis wrote:
but still can you give me the the value of M and N which will add up to 10. If you check the other way round the least numbers are 7 and 9 which adds up to 16 so how come 10 is possible?

The problem in your reasoning is that you are not considering the values of M and N when, p and q are 0. If you do you'll get the least values: M=1 and N=3. So if you take the value of M when p=1 and value of N when q=0 you'll get:

M=7 and N=3 --> M+N=10.
_________________

Kudos [?]: 135363 [2], given: 12691

Manager
Joined: 22 Jul 2009
Posts: 194

Kudos [?]: 554 [0], given: 6

Location: Manchester UK

### Show Tags

19 Nov 2009, 14:47
Yeah now i do get it...Thanks mate!!!

Kudos [?]: 554 [0], given: 6

Intern
Joined: 17 Apr 2010
Posts: 38

Kudos [?]: 22 [0], given: 25

### Show Tags

01 May 2010, 12:27
If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M+N?

a) 86
b) 52
c) 34
d) 28
e) 10

[Reveal] Spoiler:

This question is fairly easy, but I don't get it, why 10 is not false.
When divided by 6, the smallest numbers that have a remainder of 1 and 3, respectively, are 7 and 9. 7+9=16.
So I don't get it, why 10 is correct.

_________________

Consider giving kudos, if you like my thread!

FYI: Questions are from the workbook of my prep course.

Kudos [?]: 22 [0], given: 25

Manager
Joined: 06 Apr 2010
Posts: 80

Kudos [?]: 45 [0], given: 2

### Show Tags

31 Aug 2010, 23:28
The smallest numbers are 1 and 3:
1 div by 6 = 0 with remainder if 1
3 divided 6 =0 with remainder of 3

so 10 could be 7+3 => 7 gives a remainder of 1 and 3 gives a remainder of 3 when divided by 6
or 1+9 => 9 gives a remainder of 3 and 1 gives a remainder of 1 when divided by 6

Kudos [?]: 45 [0], given: 2

Intern
Joined: 19 Apr 2012
Posts: 25

Kudos [?]: 5 [0], given: 8

Re: If M and N are positive integers [#permalink]

### Show Tags

24 Apr 2012, 02:53
My Approach:

When m and n are divided by 6, their remainders are 1 and 3.

In order to have a remainder of 1, M must be 7,13,19, 25 and so on. We can set up an equation: 6*n + 1 for every number M has to be. The same is with N. Their we have 9,15,21,27 etc. = 6*n+9.

If we sum it up, we have: 16+12n.

Now we can test every solution and the one that doesen't work is solution A.

16 + 12n = 86

12n = 74

n = 6,1666 <- no integer.

Sorry for my english.:/

Kudos [?]: 5 [0], given: 8

Intern
Joined: 27 Feb 2012
Posts: 3

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

Re: If M and N are positive integers that have remainders of 1 [#permalink]

### Show Tags

24 Apr 2012, 04:54
I did it this way
M=6p+1
N=6q+1

We need M+N=6p+1+6q+3=6(p+q)+4

Pick numbers for p & q
Since it is an addition of p & q and the answer to this expression should be an integer (because all the numbers being added are integers),
we just need to choose values so that we get integer multiples of 6 so
p+q=0 ; M+n = 4
p+q=1 ; M+N = 10
P+q=2 ; M+N = 16

and so on, so basically you get something like - 4,10,16,22,28,34,..... all the other options were turning up.

Then I directly tried p+q=12 because it was closer to 86 for the first option, i got 76, then tried with 13 - got 84, 14 got 90.. no 86.

Is the approach correct?

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

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7791

Kudos [?]: 18107 [1], given: 236

Location: Pune, India
Re: If M and N are positive integers that have remainders of 1 [#permalink]

### Show Tags

24 Apr 2012, 09:33
1
KUDOS
Expert's post
nsvarunns wrote:
I did it this way
M=6p+1
N=6q+1

We need M+N=6p+1+6q+3=6(p+q)+4

Your approach was not wrong but a little cumbersome. Instead of picking values for p and q and trying to get to the options, pick the options and find out whether they suit this format.

52 = 6*8 + 4 so p+q = 8. Hence 52 can be the value of M+N
Similarly check for other options. At every step, you get closer to the solution else you could end up waiting for a long time before you get 4 of the 5 options.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 18107 [1], given: 236

Intern
Joined: 27 Feb 2012
Posts: 3

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

Re: If M and N are positive integers that have remainders of 1 [#permalink]

### Show Tags

24 Apr 2012, 18:08
thanks Karishma. Will keep that in mind. It did take a long time for me to get it through.

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

Intern
Joined: 11 Oct 2011
Posts: 5

Kudos [?]: [0], given: 7

Re: If M and N are positive integers that have remainders of 1 [#permalink]

### Show Tags

25 Apr 2012, 10:18
1+3=4
Using six times table to recognize the the nearest integer uner each option i.e.- 6*4 =24.
Whichever answer does not have four as a remainder wins the selection.

A

Posted from my mobile device

Kudos [?]: [0], given: 7

Intern
Joined: 23 May 2012
Posts: 31

Kudos [?]: 43 [0], given: 11

Re: If M and N are positive integers that have remainders of 1 [#permalink]

### Show Tags

19 Oct 2012, 00:22
sagarsabnis wrote:
If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M+N?

(A) 86
(B) 52
(C) 34
(D) 28
(E) 10

M & N are divided by 6 : R =1 & 3 resp

M+ N divided by 6 : R 1+ 3 = 4

So divide the options by 6 .. & R should be 4.

Only ..86 has a remiander of 2 .....instead of 4..

Kudos [?]: 43 [0], given: 11

Manager
Joined: 04 Jan 2013
Posts: 79

Kudos [?]: 19 [0], given: 1

Re: If M and N are positive integers [#permalink]

### Show Tags

10 Jan 2013, 00:55
Bunuel wrote:
shikhar wrote:
19. If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M+N?

(A) 86
(B) 52
(C) 34
(D) 28
(E) 10

@bunuel..why are we considering the integers 1 and 3 in the list of integers when the stem question says that when each of the numbers is divided by six,the remainder is 1 and 3 respectively..when 1 is divided by six am sure we would not get 1 as a remainder neither would we get 3 as a remainder when the integer 3 is divided by 6

Posted from my mobile device

Kudos [?]: 19 [0], given: 1

Math Expert
Joined: 02 Sep 2009
Posts: 42571

Kudos [?]: 135363 [0], given: 12691

Re: If M and N are positive integers [#permalink]

### Show Tags

10 Jan 2013, 03:42
chiccufrazer1 wrote:
Bunuel wrote:
shikhar wrote:
19. If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M+N?

(A) 86
(B) 52
(C) 34
(D) 28
(E) 10

@bunuel..why are we considering the integers 1 and 3 in the list of integers when the stem question says that when each of the numbers is divided by six,the remainder is 1 and 3 respectively..when 1 is divided by six am sure we would not get 1 as a remainder neither would we get 3 as a remainder when the integer 3 is divided by 6

Posted from my mobile device

That's not correct.

• When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller integer.
For example, 7 divided by 11 has the quotient 0 and the remainder 7 since $$7=11*0+7$$

Hence, 1 divided by 6 yields the remainder of 1 and 3 divided by 6 yields the remainder of 3.

For more check Remainders chapter of Math Book: http://gmatclub.com/forum/remainders-144665.html
_________________

Kudos [?]: 135363 [0], given: 12691

Non-Human User
Joined: 09 Sep 2013
Posts: 14897

Kudos [?]: 287 [0], given: 0

Re: If M and N are positive integers that have remainders of 1 [#permalink]

### Show Tags

17 Aug 2014, 03:09
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.
_________________

Kudos [?]: 287 [0], given: 0

Non-Human User
Joined: 09 Sep 2013
Posts: 14897

Kudos [?]: 287 [0], given: 0

Re: If M and N are positive integers that have remainders of 1 [#permalink]

### Show Tags

18 Aug 2015, 18:06
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.
_________________

Kudos [?]: 287 [0], given: 0

Intern
Joined: 15 Feb 2015
Posts: 6

Kudos [?]: [0], given: 10

Re: If M and N are positive integers that have remainders of 1 [#permalink]

### Show Tags

21 Sep 2015, 02:13
Assume M = 6a+1 and N = 6b+3
M+N = 6(a+b)+4
Hence answer should be such that when subtracted by 4, it should be divisible by 6.
86-4 = 82 is not divisible by 6. Hence A is the answer.

Kudos [?]: [0], given: 10

Non-Human User
Joined: 09 Sep 2013
Posts: 14897

Kudos [?]: 287 [0], given: 0

Re: If M and N are positive integers that have remainders of 1 [#permalink]

### Show Tags

26 Sep 2016, 18:24
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.
_________________

Kudos [?]: 287 [0], given: 0

Re: If M and N are positive integers that have remainders of 1   [#permalink] 26 Sep 2016, 18:24
Display posts from previous: Sort by