Last visit was: 19 Nov 2025, 13:48 It is currently 19 Nov 2025, 13:48
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
505-555 Level|   Min-Max Problems|               
User avatar
jamifahad
User avatar
Joined: 03 Mar 2010
Last visit: 14 Mar 2015
Posts: 257
Own Kudos:
1,676
 [74]
Given Kudos: 22
Schools: Simon '16 (M$)
Schools: Simon '16 (M$)
Posts: 257
Kudos: 1,676
 [74]
Coins are to be put into 7 pockets so that each pocket contains at lea 29 Mar 2011, 11:52
6
Kudos
Add Kudos
67
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

15% (low)

Question Stats:

81% (01:45) correct 19% (02:03) wrong based on 2177 sessions

History

Date
Time
Result
Not Attempted Yet
Coins are to be put into 7 pockets so that each pocket contains at least one coin. At most 3 of the pockets are to contain the same number of coins, and no two of the remaining pockets are to contain an equal number of coins. What is the least possible number of coins needed for the pockets?

(A) 7
(B) 13
(C) 17
(D) 22
(E) 28

Show Spoiler
To determine the least possible number of coins needed, the smallest possible number of coins should be placed in each pocket, subject to the constraints of the problem. Thus, one coin should be put in three of the pockets, 2 coins in the fourth pocket, 3 coins in the fifth, 4 coins in the sixth, and 5 coins in the seventh. The least possible number of coins is therefore 1 + 1 + 1 + 2 + 3 + 4 + 5 = 17, so the best answer is C.
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,351
 [19]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,351
 [19]
Coins are to be put into 7 pockets so that each pocket contains at lea 04 Mar 2012, 18:48
8
Kudos
Add Kudos
10
Bookmarks
Bookmark this Post
eybrj2
Coins are to be put into 7 pockets so that each pocket contains at least one coin. At most 3 of the pockets are to contain the same number of coins, and no two of the remaining pockets are to contain an equal number of coins. What is the least possible number of coins needed for the pockets?

A. 7
B. 13
C. 17
D. 22
E. 28
Show more

Since at most 3 of the pockets can contain the same number of coins, we minimize the number of coins in each of those; thus, let each contain just 1 coin.

Next, we are told that no two of the remaining 4 pockets should contain an equal number of coins. Therefore, they should contain 2, 3, 4, and 5 coins each, which is the minimum possible.

Total: 1+1+1+2+3+4+5=17.

Answer: C.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,000
 [6]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,000
 [6]
Coins are to be put into 7 pockets so that each pocket contains at lea 29 Mar 2011, 18:22
5
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
jamifahad
Coins are to be put into 7 pockets so that each pocket contains at least one coin. At most 3 of the pockets are to contain the same number of coins, and no two of the remaining pockets are to contain an equal number of coins. What is the least possible number of coins needed for the pockets?

(A) 7 (B) 13 (C) 17 (D) 22 (E) 28

I can figure this out by hit and trial. But that's not correct approach. Please help.
Show more

Hit and Trial and Step by step analysis are two different things. My guess is that if you arrived at the correct answer, it was by step by step analysis.
It can be done in many ways. The following is how I would do it:

We need to minimize the number of coins needed.
Each pocket contains at least one coin so give one coin to each pocket - 7 coins used.
At most 3 pockets can have the same number of coins so let three pockets retain just one coin each (to minimize the number of coins) ...
Now we are left with 4 pockets - no two of these should have same number of coins so they should have 2, 3, 4 and 5 coins (to minimize the total number of coins)

Hence, minimum number of coins needed = 1+1+1+2+3+4+5 = 17
General Discussion
User avatar
shrouded1
User avatar
Retired Moderator
Joined: 02 Sep 2010
Last visit: 29 Apr 2018
Posts: 609
Own Kudos:
3,191
 [1]
Given Kudos: 25
Location: London
Products:
My Reviews
Posts: 609
Kudos: 3,191
 [1]
Coins are to be put into 7 pockets so that each pocket contains at lea 29 Mar 2011, 13:59
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Problems like these can be solved by the greedy approach

We start filling the pockets one by one all the time keeping the constraints in mind and see whats the least number of balls we can get away with.

Pocket 1 - Use 1 ball
Pocket 2 - Can still use just 1 ball
Pocket 3 - Can still use just 1 ball
Pocket 4 - Now we need a different number, and the least we can pick is 2
Pocket 5 - Similar to above, min we need to use is 3
Pocket 6 - Again, min we can use is 4
Pocket 7 - Finally min is 5

So net number is 1+1+1+2+3+4+5 = 17
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,351
 [2]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,351
 [2]
Coins are to be put into 7 pockets so that each pocket contains at lea 14 Jun 2013, 05:26
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Min/Max Problems to practice: search.php?search_id=tag&tag_id=42
All PS Min/Max Problems to practice: search.php?search_id=tag&tag_id=63
User avatar
MathRevolution
User avatar
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Last visit: 27 Sep 2022
Posts: 10,070
Own Kudos:
19,392
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Expert
Expert reply
GMAT 1: 760 Q51 V42
Posts: 10,070
Kudos: 19,392
Coins are to be put into 7 pockets so that each pocket contains at lea 02 Dec 2020, 10:01
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Total 7 pockets with 1 coin each: 7 * 1 = 7 coins.

At most '3' can have the same number and we have to find the minimum possible number of total coins: '1' being smallest, '3' pockets will have '1' coin each

Total coins are still '7' as we have assigned '1' to each pocket.

No two of the remaining pockets are to contain an equal number of coins: So, the remaining '4' pockets will have a different number of coins with minimum values: 2 + 3 + 4 + 5.

'4' pockets already had 4 * 1 = 4 coins and now their total is 2 + 3 + 4 + 5 = 14 coins. So, we have included 14 - 4 = 10 coins.

Overall total: 7 coins [1 each in each pocket] + 10 [new included] = 17 coins.

Answer C
User avatar
Tejas183
User avatar
Joined: 23 Oct 2018
Last visit: 02 Apr 2025
Posts: 50
Own Kudos:
10
 [1]
Given Kudos: 96
Location: India
Schools: Tuck '25 (D) Darden '25 (A) Cornell '25 (WL)
GRE 1: Q169 V167
GPA: 3
Products:
My Reviews
Schools: Tuck '25 (D) Darden '25 (A) Cornell '25 (WL)
GRE 1: Q169 V167
Posts: 50
Kudos: 10
 [1]
Coins are to be put into 7 pockets so that each pocket contains at lea 18 Dec 2020, 12:07
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jamifahad
Coins are to be put into 7 pockets so that each pocket contains at least one coin. At most 3 of the pockets are to contain the same number of coins, and no two of the remaining pockets are to contain an equal number of coins. What is the least possible number of coins needed for the pockets?

(A) 7
(B) 13
(C) 17
(D) 22
(E) 28

Show Spoiler
To determine the least possible number of coins needed, the smallest possible number of coins should be placed in each pocket, subject to the constraints of the problem. Thus, one coin should be put in three of the pockets, 2 coins in the fourth pocket, 3 coins in the fifth, 4 coins in the sixth, and 5 coins in the seventh. The least possible number of coins is therefore 1 + 1 + 1 + 2 + 3 + 4 + 5 = 17, so the best answer is C.
Show more




No two of the remaining pockets..... That means after first three pockets - can;t we repeat the pocket of 1

Eg: 1 1 1 | 4 3 2 1
User avatar
adgarg
User avatar
Joined: 02 Feb 2023
Last visit: 21 Dec 2024
Posts: 28
Own Kudos:
3
Given Kudos: 31
Location: India
Posts: 28
Kudos: 3
Coins are to be put into 7 pockets so that each pocket contains at lea 19 Aug 2023, 11:12
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
eybrj2
Coins are to be put into 7 pockets so that each pocket contains at least one coin. At most 3 of the pockets are to contain the same number of coins, and no two of the remaining pockets are to contain an equal number of coins. What is the least possible number of coins needed for the pockets?

A. 7
B. 13
C. 17
D. 22
E. 28

Since at most 3 of the pockets are to contain the same number of coins then minimize # of coins in each, so let each contain just 1 coin;

Next, we are told that no two of the remaining 4 pockets should contain an equal number of coins, so they should contain 2, 3, 4, and 5 coins each (also minimum possible);

Total: 1+1+1+2+3+4+5=17.

Answer: C.
Show more

I have the same question as the other person, the question said, No two of the remaining pockets..... That means after first three pockets - can;t we repeat the pocket of 1

Eg: 1 1 1 | 4 3 2 1
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,351
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,351
Coins are to be put into 7 pockets so that each pocket contains at lea 21 Aug 2023, 00:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
adgarg
Bunuel
eybrj2
Coins are to be put into 7 pockets so that each pocket contains at least one coin. At most 3 of the pockets are to contain the same number of coins, and no two of the remaining pockets are to contain an equal number of coins. What is the least possible number of coins needed for the pockets?

A. 7
B. 13
C. 17
D. 22
E. 28

Since at most 3 of the pockets are to contain the same number of coins then minimize # of coins in each, so let each contain just 1 coin;

Next, we are told that no two of the remaining 4 pockets should contain an equal number of coins, so they should contain 2, 3, 4, and 5 coins each (also minimum possible);

Total: 1+1+1+2+3+4+5=17.

Answer: C.

I have the same question as the other person, the question said, No two of the remaining pockets..... That means after first three pockets - can;t we repeat the pocket of 1

Eg: 1 1 1 | 4 3 2 1
Show more

No. The question clearly states that "at most 3 of the pockets are to contain the same number of coins", so only three pockets can have the same number of coins. So, after the initial three pockets with 1 coin each, the next pockets cannot have just 1 coin. The correct minimum distribution is 1, 1, 1, 2, 3, 4, 5, totaling 17 coins.
User avatar
MMishra012
User avatar
Joined: 09 Oct 2024
Last visit: 19 Nov 2025
Posts: 3
Given Kudos: 7
Location: India
Posts: 3
Kudos: 0
Coins are to be put into 7 pockets so that each pocket contains at lea 12 Nov 2025, 03:37
Kudos
Add Kudos
Bookmarks
Bookmark this Post
You can fill 7 pockets like this : X,X,X,X+1,X+2,X+3,X+4
- 3 X for 3 same
- No two four can be same so the minimum values in four other pockets keeping them different from first three pockets would be X+1, next being X+2 etc. This way 3 are same & 4 are unique. Sum of this is 7X +10. Now to minimise this, keep X as minimum 1. You get 17.
jamifahad
Coins are to be put into 7 pockets so that each pocket contains at least one coin. At most 3 of the pockets are to contain the same number of coins, and no two of the remaining pockets are to contain an equal number of coins. What is the least possible number of coins needed for the pockets?

(A) 7
(B) 13
(C) 17
(D) 22
(E) 28

Show Spoiler
To determine the least possible number of coins needed, the smallest possible number of coins should be placed in each pocket, subject to the constraints of the problem. Thus, one coin should be put in three of the pockets, 2 coins in the fourth pocket, 3 coins in the fifth, 4 coins in the sixth, and 5 coins in the seventh. The least possible number of coins is therefore 1 + 1 + 1 + 2 + 3 + 4 + 5 = 17, so the best answer is C.
Show more
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts