Question:medium

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?

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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.
Updated On: Jul 21, 2026
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The Correct Option is B

Solution and Explanation

The coins double from one box to the next: 1, 2, 4, 8, and so on, up to the 32nd box. Rather than working with the huge number \(2^{32} - 1\) directly, we can spot the pattern using smaller cases and then apply it to 32 boxes.

Let's check what happens with just a few boxes at a time, always asking whether the coins collected so far are divisible by the number of boxes:

  1. 1 box: total coins = 1. This is already divisible by 1, so 0 extra coins are needed.
  2. 2 boxes: total coins = 1 + 2 = 3. Dividing by 2 leaves a remainder of 1, so we need 1 more coin to reach 4, which splits evenly as 2 coins per box.
  3. 4 boxes: total coins = 1 + 2 + 4 + 8 = 15. Dividing by 4 leaves a remainder of 3, so we need 1 more coin, reaching 16, which splits evenly as 4 coins per box.

Every time the number of boxes is itself a power of 2, say \(2^k\), the coins collected total \(2^{2k} - 1\), which is one short of the perfect power \(2^{2k}\), and \(2^{2k}\) is always divisible by \(2^k\), since there are more than enough factors of 2 to go around. So the shortfall is always just 1 coin.

Here we have 32 boxes, and 32 is \(2^5\). Following the same pattern, the total coins collected is \(2^{32} - 1\), which is 1 short of \(2^{32}\), and \(2^{32}\) divides evenly by 32.

Let's summarize:

  • The coins on the boxes double each time, giving a total of \(2^{32} - 1\) coins across all 32 boxes.
  • Testing smaller cases, 2 boxes and 4 boxes, shows the shortfall to the next multiple is always exactly 1 coin, whenever the box count is a power of 2.
  • 32 boxes follow the same rule, so only 1 extra coin is needed.

So the minimum number of coins added is 1.

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