Question

On a cardboard there are 32 squares. 1 coin is kept on the first box, 2 coins on the second box and so on. Collect all the coins and add some more coins to them. These coins are equally distributed on all the squares of the cardboard. What is the minimum number of coins that are added?

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

The coins double on each box: 1, 2, 4, 8, and so on. The total collected is one short of a power of 2 that is divisible by 32, so check how many coins are needed to reach that next power of 2.

Answer 1

The Correct Option is B

Step 1: Understand the coin pattern on the cardboard.
The first box gets 1 coin, the second box gets 2 coins, and following that pattern, each next box gets double the coins of the one before it. This is the classic doubling pattern, so the coins on the boxes are \(1, 2, 4, 8, \ldots\), that is, \(2^0, 2^1, 2^2, \ldots\), all the way up to the 32nd box, which gets \(2^{31}\) coins.

Step 2: Add up all the coins already on the board.
The sum of a doubling sequence starting at 1 and going up to \(2^{31}\) is \(2^{32} - 1\).
\[ \text{Total coins} = 2^{31} + 2^{30} + \ldots + 2^1 + 2^0 = 2^{32} - 1 \]

Step 3: Check how far this total is from a multiple of 32.
We want to add just enough coins so the new total can be split equally among all 32 boxes, which means the new total must be exactly divisible by 32.
Since \(32 = 2^5\), and \(2^{32}\) is clearly divisible by \(2^5\), it has far more factors of 2 than that, so \(2^{32}\) itself leaves a remainder of 0 when divided by 32.
But our total is \(2^{32} - 1\), one less than that, so it leaves a remainder of \(32 - 1 = 31\) when divided by 32.

Step 4: Find the minimum coins to add.
To move from a remainder of 31 up to the next full multiple of 32, we need to add just 1 more coin.
\[ (2^{32} - 1) + 1 = 2^{32}, \text{ which is exactly divisible by 32.} \]

Final Answer:
The minimum number of coins that need to be added is 1. \[ \boxed{1} \]