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Re: Coins are to be put into 7 pockets so that each pocket contains at lea [#permalink]
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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
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Re: Coins are to be put into 7 pockets so that each pocket contains at lea [#permalink]
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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
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Re: Coins are to be put into 7 pockets so that each pocket contains at lea [#permalink]
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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
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Re: Coins are to be put into 7 pockets so that each pocket contains at lea [#permalink]
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jamifahad wrote:
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

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.





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
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Re: Coins are to be put into 7 pockets so that each pocket contains at lea [#permalink]
Bunuel wrote:
eybrj2 wrote:
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
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Re: Coins are to be put into 7 pockets so that each pocket contains at lea [#permalink]
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adgarg wrote:
Bunuel wrote:
eybrj2 wrote:
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


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.
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Re: Coins are to be put into 7 pockets so that each pocket contains at lea [#permalink]
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