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Number Theory Level 3

Our hero Batman likes to play with coins. He has $n$ number of coins all of which are white on one side and black on the other. He puts all the coins on a straight line with white side up. Now he starts to flip the coins. First he flips every coins placed at position of multiple of $1$ , then at multiple of $2$ and so on till the multiple of $n$ . He wants you to guess the last position of the coin whose black side is up if $n$ is given to be 1000000007 .

The answer is 999950884.

1 solution

Purushottam Abhisheikh
Apr 19, 2016

Answer of this question is $\left( \left\lfloor \sqrt { n } \right\rfloor \right) ^{ 2 }$ . Now let me explain why.

A coin at $i^{th}$ place is flipped each time when flipping is done for the multiple of divisors of $i$ .

$\because$ Starting color is white and Ending color is black.

$\therefore$ No. of flipping should be odd.

Which means number of divisors of $i$ should be odd so that coin at $i^{th}$ position is black. Therefore, i is a perfect square and the last perfect square number is our solution which is $\left( \left\lfloor \sqrt { n } \right\rfloor \right) ^{ 2 }$ .

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The answer is 999950884.

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