Question

There are 1100 coins with one side as heads and other side as tails and there are 1100 people First person goes and places all the coins to show heads Second person goes and places every second coin (2, 4, 6, …) to show tails The third person goes to every third coin (3, 6, 9, ...) and, if it is heads, then places it to show tails, and if it is tails, then places it to show Heads The fourth person does this to every fourth coin (4, 8, 12, ...), and so on. After all the 1100 people take their turn how many coins will be showing heads?

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The number of coins which will be showing heads after 1100 iterations is found to be 33.

Step-by-step explanation:

In the process explained to us, there are $n$ people and the same $n$ coins present with $n$ having a value that ranges between 1 to 1100.

Now, if the $n^{th}$ person goes and flips the coin, the trick will be to spot if a coin is being visited by people even or an odd number of times. Pay attention to the coin number.

Only perfect squares have their factors in odd numbers and all the other numbers have even number of factors.

For all those numbered coins which have an odd number of their factors will be facing heads upwards and for all those coins which have even no. of their factors will be facing tails upwards.

So, coins numbered 1, 4, 9, 16, 25…. will have heads and numbers  3, 5, 6, 7, 10, 11, … will have tails facing upwards.

The number of perfect squares till 1100 is 33. Thus, the number of coins with heads upwards will be 33.

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