How many coins are still heads up?

$ \begingroup $ A utilitarian proficiency in solving such problems is to scale things down — bring on a smaller font, to detect patterns .
For case, suppose we have lone $ 6 $ coins. Put them in a row, and total them from $ 1 $ to $ 6 $ : $ $ \ { 1, 2, 3, 4, 5, 6\ }. $ $
now look at any single coin, say coin $ 4 $. How many times does it get flipped when we go through this procedure ? Well, it gets flipped once when going through the multiples of $ 2 $ : $ \ { 2, 4, 6\ } $. It does not get flipped when going through the multiples of $ 3 $, since $ 4 $ is not a multiple of $ 3 $. And then it does get flipped once again when we go through the multiples of $ 4 $. So it get flipped a total of $ 2 $ times.

How many times does $ 5 $ get flipped ? Well, lone once, when we go through multiples of $ 5 $ .
And how many times does $ 6 $ get flipped ? Three times.

distinctly, this situation has something to do with the divisors of the number on the coin. The divisors of $ 4 $ are $ 1, 2, 4 $. And the divisors of $ 5 $ are $ 1, 5 $, and the divisors of $ 6 $ are $ 1, 2, 3, 6 $. So it looks like the numeral of times a given coin is flipped is one less than its number of divisors. The reason why it is one less, is because we do n’t start the action by flipping every coin — we start with the multiples of $ 2 $ .
Let ‘s work out a larger example. Say we are concerned in mint number $ 100 $. The divisors of $ 100 $ are : $ $ 1, 2, 4, 5, 10, 20, 25, 50, 100. $ $ There are $ 9 $ divisors, so we suspect that this coin would be flipped $ 8 $ times. Can you check that this is true ?

so if there are $ 1000 $ coins, the number of coins that are heads up is peer to the count of coins that have an odd total of divisors ( including $ 1 $, and the number itself ). Can you go through a list of the first $ 1000 $ such numbers and see which ones have an leftover number of divisors ? I will start you out with a partial derivative number :
$ $ \begin { align } { c|c } n & \sigma_0 ( normality ) \\ \hline 1 & 1 \\ 2 & 2 \\ 3 & 2 \\ 4 & 3 \\ 5 & 2 \\ 6 & 4 \\ 7 & 2 \\ 8 & 4 \\ 9 & 3 \\ 10 & 4 \\ 11 & 2 \\ 12 & 6 \\ 13 & 2 \\ 14 & 4 \\ 15 & 4 \\ 16 & 5 \\ \end { array } $ $
The moment column with heading $ \sigma_0 ( n ) $ represents the act of divisors of $ north $. Which of these are curious ? Do you see a model ? If you do, how would you go about proving this model ?

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Category : Finance