Last visit was: 24 Apr 2026, 14:28 It is currently 24 Apr 2026, 14:28
Close
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
iNumbv
Joined: 02 May 2012
Last visit: 06 Jan 2016
Posts: 9
Own Kudos:
93
 [15]
Given Kudos: 5
Posts: 9
Kudos: 93
 [15]
Kudos
Add Kudos
15
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
cyberjadugar
Joined: 29 Mar 2012
Last visit: 01 Apr 2026
Posts: 264
Own Kudos:
1,814
 [8]
Given Kudos: 23
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
GMAT 3: 730 Q50 V38
Posts: 264
Kudos: 1,814
 [8]
7
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
General Discussion
User avatar
DonQuixote
Joined: 25 Jun 2012
Last visit: 11 Feb 2013
Posts: 49
Own Kudos:
Given Kudos: 2
Posts: 49
Kudos: 45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
hfbamafan
Joined: 29 Mar 2010
Last visit: 27 Feb 2019
Posts: 90
Own Kudos:
592
 [2]
Given Kudos: 16
Location: United States
Concentration: Finance, International Business
GMAT 1: 590 Q28 V38
GPA: 2.54
WE:Accounting (Hospitality and Tourism)
GMAT 1: 590 Q28 V38
Posts: 90
Kudos: 592
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
I just took a simpler route.

Because there are no more than 7 children # of cookies has to be divisible by 7.

Question does not tell us how many children are going to show up it has to be divisible by 6 as well. And 5....1.
So what I did was take 6*7=42 and the answer has to be divisible by 5, 4, 3 ,2 1 there is only one answer choice that satisfies these constraints which is 420



So
avatar
damstamnd
Joined: 04 Apr 2011
Last visit: 10 Aug 2016
Posts: 7
Own Kudos:
Given Kudos: 3
Schools: Sloan '16
Schools: Sloan '16
Posts: 7
Kudos: 11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I also took the simpler route. Found the answer that was divisible by 1 thru 7.

Question: Why did you multiply 6*7? Did you look for an answer that was also divisible by 42?
avatar
damham17
Joined: 06 Mar 2012
Last visit: 01 Jun 2018
Posts: 32
Own Kudos:
Given Kudos: 32
Posts: 32
Kudos: 225
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I agree with the last 2 responses in terms of how to attack the problem. Do not get caught up with the story of the problem. This type of problem is shown in the OG problem solving book with several variations. It all comes down to answering what number is divisible by the constraint (in this case the numbers 1 through 7).
User avatar
abhimahna
User avatar
Board of Directors
Joined: 18 Jul 2015
Last visit: 06 Jul 2024
Posts: 3,481
Own Kudos:
5,779
 [1]
Given Kudos: 346
Status:Emory Goizueta Alum
Products:
Expert
Expert reply
Posts: 3,481
Kudos: 5,779
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
It is known that no more than 7 children will be attending a party. What is the smallest number of cookies that must be brought to the party so that each child receives the same number of cookies?

A. 35
B. 105
C. 180
D. 210
E. 420

No more than 7 children attending the party means the number of children could be 1,2,3,4,5,6,7.

We need to find a number that is divisible by each of these. Out of the choices given only 420 is divisible by each of these numbers.

Hence , answer is E.
User avatar
rever08
Joined: 21 Jul 2017
Last visit: 13 Jan 2020
Posts: 148
Own Kudos:
115
 [2]
Given Kudos: 143
Location: India
Concentration: Social Entrepreneurship, Leadership
GMAT 1: 660 Q47 V34
GPA: 4
WE:Project Management (Education)
Products:
GMAT 1: 660 Q47 V34
Posts: 148
Kudos: 115
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
The question states that there are no more than 7 children in the party. Thus, the possibility is that there could be 1, 2, 3, 4, 5, 6 or 7 children.
Cookies are to be equally divided among the children attending party.
Thus, the question asks for a number of cookies which can be distributed to any number of children (from 1 to 7).

And therefore the smallest number of cookies would be lcm of (1, 2, 3, 4, 5, 6, 7), i.e., 420.

Answer (E)
--------------------------------------------------------------------------
Hit Kudos if this helped!
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 24 Apr 2026
Posts: 5,986
Own Kudos:
Given Kudos: 163
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
GMAT 1: 690 Q50 V34
Posts: 5,986
Kudos: 5,859
Kudos
Add Kudos
Bookmarks
Bookmark this Post
iNumbv
It is known that no more than 7 children will be attending a party. What is the smallest number of cookies that must be brought to the party so that each child receives the same number of cookies?

A. 35
B. 105
C. 180
D. 210
E. 420

Show SpoilerMY DOUBT
** Can someone please explain to me the intuition behind this answer? I figure it's a very straightforward answer, but what I don't understand is assuming the maximum number of children attend the party which is 7 and among the answers 35 looks to suffice as each child will receive 5 cookies, which makes it the smallest number among the answers. I'm a bit confused can someone point me to the right direction, Thanks. **

Show SpoilerSOLUTION

Since the number of cookies must be divisible by 1, 2, 3, 4, 5, 6, and 7, let's find the least common multiple of the integers 1, 2, 3, 4, 5, 6, and 7. Every integer is divisible by 1. Let's find the prime factorizations of 2, 3, 4, 5, 6, and 7.

So we want to find the least common multiple of these integers, which we will write in a column.

The largest number of times the prime factor 2 appears in any of these integers is 2, in . The largest number of times the prime factor 3 appears in any of these integers is 1. Similarly, the largest number of times the prime factors 5 appears in any of these integers is 1, and the largest number of times the prime factors 7 appears in any of these integers is 1.

So the least common multiple of the first 7 positive integers must contain 2 prime factors of 2, 1 prime factor of 3, 1 prime factor of 5, and 1 prime factor of 7.

The least common multiple of the first 7 positive integers is 2 × 2 × 3 × 5 × 7 = 420.

The smallest number of cookies that must be brought to the party is 420, choice (E).


Given: It is known that no more than 7 children will be attending a party.
Asked: What is the smallest number of cookies that must be brought to the party so that each child receives the same number of cookies?

Prime factorisation of 1,2,3,4,5,6,7
1 = 1
2 = 2
3 = 3
4= 2^2
5 = 5
6 = 2*3
7 = 7
Since we don't know the exact number of children attending the party, we have to bring number of cookies which will be equally divided irrespective of number of children attending the party. Number of children = {1,2,3,4,5,6,7}

2^2*3*5*7 = 420

Now the cookies can be divided equally among children irrespective of number of children attending the party.

IMO E
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,974
Own Kudos:
Posts: 38,974
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109818 posts
Tuck School Moderator
853 posts