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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Karishma
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