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Que: Few coins are put into 7 boxes such that each box contains at least two coins. At the most '3' boxes can contain

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Que: Few coins are put into 7 boxes such that each box contains at least two coins. At the most '3' boxes can contain

by Max@Math Revolution » Wed Dec 02, 2020 11:59 pm

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Que: Few coins are put into 7 boxes such that each box contains at least two coins. At the most '3' boxes can contain the same number of coins, and the remaining boxes cannot contain an equal number of coins. What is the minimum possible number of coins in the 7 boxes?

(A) 18
(B) 20
(C) 24
(D) 27
(E) 30
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Re: Que: Few coins are put into 7 boxes such that each box contains at least two coins. At the most '3' boxes can contai

by Max@Math Revolution » Sat Dec 05, 2020 10:53 pm
Solution: Since each box contains at least two coins, we have 7 * 2 = 14 coins minimum.

We know that at the most 3 boxes can have the same number of coins. Since we need to minimize the total number of coins, we must have as many boxes having the same number (minimum possible number, i.e. 2 coins) of coins as possible. Thus, for each of the 3 boxes containing an equal number of coins, we have ‘2’ coins. Thus, number of coins in the 3 boxes = 2 × 3 = 6.

Since each of the remaining 4 boxes have a different number of coins, let us put in 3, 4, 5, and 6 coins in those boxes. Thus, the total number of coins = 6 + (3 + 4 + 5 + 6) = 24.

Therefore, C is the correct answer.

Answer C
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