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Coins are to be put into 7 pockets so that each pocket contains at lea

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

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New post 29 Mar 2011, 10:52
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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

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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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New post 04 Mar 2012, 17:48
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.
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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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New post 29 Mar 2011, 12:59
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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New post 29 Mar 2011, 17:22
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

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


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
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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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New post 14 Jun 2013, 04:26
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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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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