Question:hard

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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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.
Updated On: Jul 20, 2026
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The Correct Option is B

Solution and Explanation

Instead of computing the full sum $2^{32}-1$, work modulo 32 directly and notice most terms vanish.

Total coins $= 2^0+2^1+2^2+\cdots+2^{31}$.

Every term from $2^5$ onward is a multiple of $2^5 = 32$, so modulo 32 those terms all contribute $0$. Only the first five terms matter:
$2^0+2^1+2^2+2^3+2^4 = 1+2+4+8+16 = 31$

So the total number of coins $\equiv 31 \pmod{32}$, meaning the total is 1 short of the next multiple of 32.

Hence the minimum number of coins that must be added so the total splits evenly across the 32 squares is:

\[\boxed{1}\]
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