Hats in 2015

You may have seen an easier problem with only two or three hat colors possible. Nevertheless, the strategy here is still the same. First, the people set each hat color to a distinct residue $\pmod{2015}$ i.e. color 1 is $0 \pmod{2015}$ , color 2 is $1 \pmod{2015}$ , and so forth.

When the person in the back is asked to guess her own hat color, instead she adds up the values of the colors of all the hats in front of her and says the hat color that corresponds with her result. For example, if the first person had a hat with color 5 and the rest had hats with color 2, she would calculate $2013(1) + 4 \pmod{2015}$ , which is $2 \pmod{2015}$ . The hat color corresponding to $2 \pmod{2015}$ is color 3, so she would say that color, whatever it is. It is not guaranteed that she will guess her own hat color correctly, though.

Continuing with our example, the person that is second to last hears the last person say "color 3" and immediately knows that the sum of the $2013$ hat in front of him plus his own is congruent to $2 \pmod{2015}$ . He sees $2012$ hats of color 2 and $1$ hat of color 5, so he calculates the sum of the hats in front of him as $2012(1) + 4 \equiv 1 \pmod{2015}$ . Then, he would know that the color of the hat on his head was congruent to $2 - 1 \equiv 1 \pmod{2015}$ , which corresponds to color 2. So, he says that color out loud, and he would get to keep his hat.

This continues up to the first person in line. Therefore, at a minimum, the first 2014 people get to keep their hats and the last person does not, so the answer is $\boxed{2014}$ .

Everyone tells the person in front of them their colour of their hat. That guarantees 2014 hats. The first person says nothing. The last person guessing from 2015 colours is highly unlikely to get it correct.

e.g. If the order of a line up is ABCDEFG, G tells F the colour of F's hat. F tells E the colour of E's hat. E tells D the colour of D's hat and so on. A has no clue about the colours of the hats and G guesses out of the many possibilities of hat colours.

Since there's no regulations mentioned except standing in a line and guessing the hat's color correctly to keep it, a strategy where person A would tell the B's hat color (assume that B is standing in front of A) is possible and so on, this strategy ensures 2014 hats to be kept. Therefore a = 2014 or possibly 2015 if the first person guessed the right color.

The first person to guess could install an observant behind him/her to ensure that 2015 hats will be kept.