Question

Anush has fifteen coins and five boxes labeled A, B, C, D and E. He drops coins into all the boxes such that, each box has at least one coin and no two boxes have the same number of coins. Later, he found that B contains more coins than E and C contains the least number of coins when compared to A and D. If B contains twice the number of coins in E, then which of the following is necessarily true?

• A. C has an even number of coins
• B. C has an odd number of coins
• C. A has an odd number of coins
• D. A has an even number of coins
• E. D has more coins than A

Answer 1

The Correct Option is B

To solve this problem, let's break it down step-by-step:

1. We have five boxes labeled A, B, C, D, and E containing a total of 15 coins. 
2. Each box must contain at least one coin, and no two boxes can have the same number of coins.
3. B contains twice the number of coins as E, and B has more coins than E. Also, C has the fewest coins compared to A and D.

Given these conditions, let's denote the number of coins in each box as follows:

• B = 2E, meaning B has twice the coins of E.
• C is the least populated box among A, D, and itself.

Assuming the minimum values satisfying these conditions:

• Let E = 1 (minimum number of coins required because each box must contain at least one coin).
• Then B = 2 × 1 = 2.
• Since C has the fewest coins among A, C, and D, we try C = 1, but C must be unique as E also has 1. Hence C can be 2 or more considering it has the least among A, D but different from every box.

We continue with C = 2 because it's the next unique smallest possible value:

• C = 2, resulting in B needing a unique higher amount: B already is 2, so we need another setup.
• We need a reconsideration to ensure there is no duplication due to these constraints: If we use C = 3 for next available smallest:
• Given B already is 2, which uniquely distinguishes between working.

Thus:

• E = 1
• B = 2 (thus B = 2×E holds)
• C = 3

Remaining Coins Calculation:

• Total taken: 1 + 2 + 3 = 6
• Remaining = 15 - 6 = 9
• Assigning to A and D, since they need to be higher with different values for distinct uniqueness:
• A possible valid assignment: A = 4, D = 5 or vice versa. Check: D = 5 from remaining 11 post remainder for value A = 4, or again vice versa.

Conclusion:

• Every box has different coins satisfying all conditions.
• Since C = 3 (being unique and lesser, satisfying from among other choices).

Observation counts C as an odd value, thus confidently supports the correct option:

The necessarily true statement: C has an odd number of coins.