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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A group of n students can be divided into equal groups of 4 [#permalink]

Show Tags

21 Jan 2012, 14:37

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

65% (01:26) correct
35% (01:26) wrong based on 334 sessions

HideShow timer Statistics

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a) 33 b) 46 c) 49 d) 53 e) 86

Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one.

A group of n students can be divided into equal groups of 4 with 1 student left over --> n=4q+1 --> n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4)

A group of n students can be divided into equal groups of 5 with 3 students left over --> n=5p+3 --> n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5)

Therefor two smallest possible values of n are 13 and 33 --> 13+33=46.

Answer: B.

Else you can derive general formula based on n=4q+1 and n=5p+3.

Divisor will be the least common multiple of above two divisors 4 and 5, hence 20.

Remainder will be the first common integer in above two patterns, hence 13 --> so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) --> 13+33=46.

Hi Bunuel - can the values of q and p be ZERO? I don't think they can be and therefore n cannot be 1 & 3. Am I wrong?

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

For example we are told that when positive integer n is divided by 25, the remainder is 13 --> \(n=25q+13\). Now, the lowest value of \(q\) can be zero and in this case \(n=13\) --> 13 divided by 25 yields the remainder of 13. Generally when divisor (25 in our case) is more than dividend (13 in our case) then the reminder equals to the dividend. For example: 3 divided by 24 yields a reminder of 3 --> \(3=0*24+3\); or: 5 divided by 6 yields a reminder of 5 --> \(5=0*6+5\).

Also note that you shouldn't worry about negative numbers in divisibility questions, as every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

OUR ORIGINAL QUESTION: We are told that "a group of n students can be divided into equal groups of 4 with 1 student left over" --> n=4q+1. Here q also can be zero, which would mean that there is only 1 student and zero groups of 4.

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a)33 b)46 c)49 d)53 e) 86

OA is a and this is how I arrived at.

Let say n = 4x+1 and n = 5y+3 -----> From the question Stem

I get these above values by putting the same values for x and y. Is my concept correct?

I wrote a blog post discussing this concept in detail. I have discussed a couple of questions very similar to this one in the post. You can check it out if you like.

Again since the max student left over are 3 so any no with 3 can be common between the two groups ; 3, 13, 23, 33, 43, 53 etc. using in the formula pf n= 4q+1 = 5r+ 3 ; will get 13 and 33 ; addd then =46
_________________

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a) 33 b) 46 c) 49 d) 53 e) 86

Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one.

A group of n students can be divided into equal groups of 4 with 1 student left over --> n=4q+1 --> n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4)

A group of n students can be divided into equal groups of 5 with 3 students left over --> n=5p+3 --> n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5)

Therefor two smallest possible values of n are 13 and 33 --> 13+33=46.

Answer: B.

Else you can derive general formula based on n=4q+1 and n=5p+3.

Divisor will be the least common multiple of above two divisors 4 and 5, hence 20.

Remainder will be the first common integer in above two patterns, hence 13 --> so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) --> 13+33=46.

Could you explain what you mean by 'divisor'? Where is there a divisor in those two equations....Is there work here that you did in your head, but left out? I'm trying to learn how to do these, so if you could show any work that was done mentally I would appreciate it! Thanks!

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a) 33 b) 46 c) 49 d) 53 e) 86

Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one.

A group of n students can be divided into equal groups of 4 with 1 student left over --> n=4q+1 --> n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4)

A group of n students can be divided into equal groups of 5 with 3 students left over --> n=5p+3 --> n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5)

Therefor two smallest possible values of n are 13 and 33 --> 13+33=46.

Answer: B.

Else you can derive general formula based on n=4q+1 and n=5p+3.

Divisor will be the least common multiple of above two divisors 4 and 5, hence 20.

Remainder will be the first common integer in above two patterns, hence 13 --> so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) --> 13+33=46.

Could you explain what you mean by 'divisor'? Where is there a divisor in those two equations....Is there work here that you did in your head, but left out? I'm trying to learn how to do these, so if you could show any work that was done mentally I would appreciate it! Thanks!

20 is the divisor in n=20m+13. Please follow the links in my post.
_________________

Re: A group of n students can be divided into equal groups of 4 [#permalink]

Show Tags

26 Mar 2016, 03:00

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.
_________________

Re: A group of n students can be divided into equal groups of 4 [#permalink]

Show Tags

16 Apr 2017, 09:55

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.
_________________

Re: A group of n students can be divided into equal groups of 4 [#permalink]

Show Tags

06 Sep 2017, 14:31

1

This post received KUDOS

enigma123 wrote:

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

A. 33 B. 46 C. 49 D. 53 E. 86

assume difference of 1 between quotients (n-1)/4-(n-3)/5=1 n=13=smallest n 13+4*5=33=second smallest n 13+33=46 B

Re: A group of n students can be divided into equal groups of 4 [#permalink]

Show Tags

06 Sep 2017, 15:19

gracie wrote:

enigma123 wrote:

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

A. 33 B. 46 C. 49 D. 53 E. 86

assume difference of 1 between quotients (n-1)/4-(n-3)/5=1 n=13=smallest n 13+4*5=33=second smallest n 13+33=46 B

Re: A group of n students can be divided into equal groups of 4 [#permalink]

Show Tags

06 Sep 2017, 15:24

gracie wrote:

enigma123 wrote:

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

A. 33 B. 46 C. 49 D. 53 E. 86

assume difference of 1 between quotients (n-1)/4-(n-3)/5=1 n=13=smallest n 13+4*5=33=second smallest n 13+33=46 B

also, how can you assume that the difference is 1, it can be 2 or 3 or 4

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...