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29 Mar 2011, 11:52
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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 2378
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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
(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.
04 Mar 2012, 18:48
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eybrj2
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.
29 Mar 2011, 18:22
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jamifahad
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
