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
• E. 4

Hint:

Coins double each box: total = \(2^{32}-1\). Since \(2^{32}\) is divisible by 32, \(2^{32}-1\) leaves remainder 31 mod 32, so just 1 more coin is needed.

Answer 1

The Correct Option is B

Step 1: Read the coin pattern correctly.
"1 coin on the first box, 2 coins on the second box and so on" doubles each time (this is the classic doubling pattern), so box \(n\) gets \(2^{n-1}\) coins: 1, 2, 4, 8, ..., \(2^{31}\) on the 32 boxes.

Step 2: Find the total coins collected.
Total \(= 2^0 + 2^1 + 2^2 + \ldots + 2^{31} = 2^{32} - 1\) (sum of a geometric series).

Step 3: Check divisibility by 32.
For equal distribution over 32 squares, the total (after adding extra coins) must be divisible by \(32 = 2^5\).
Since \(32 > 5\), \(2^{32}\) is exactly divisible by \(2^5\), so \(2^{32} \equiv 0 \pmod{32}\).
Therefore \(2^{32} - 1 \equiv -1 \equiv 31 \pmod{32}\).

Step 4: Find the minimum coins to add.
The total currently leaves remainder 31 when divided by 32. To reach the very next multiple of 32, we need to add \(32 - 31 = 1\) coin.

Step 5: Conclusion.
Minimum coins added = 1.