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

Question

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.

Bunuel · Accepted Answer

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.

KarishmaB · Answer

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

shrouded1 · Answer

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

Bunuel · Answer

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MathRevolution · Answer

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   + 10   = 17 coins.

Answer C

Tejas183 · Answer

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

adgarg · Answer

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

Bunuel · Answer

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.

MMishra012 · Answer

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.

SRIVISHUDDHA22 · Answer

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

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