, 2025

Throughout the solution, we will use diagrams like
the one below to represent a way to colour the flower. Each circle
corresponds to a petal, and the capital letters $R$, $O$, $Y$, and $B$ represent the colours red, orange, yellow, and blue, respectively. For example, the diagram representing the petals coloured red, blue, yellow, blue, red, orange starting at the top petal in clockwise order is [[IMAGE0]] For now, we will count the number of ways to colour the petals according to the rules with the additional assumption that the top petal is red. Using this assumption, we will consider cases based on how many times the colour red is used. Before reading on, you should convince yourself that it is not possible to colour four or more petals red without causing two neighbouring petals to be coloured red. Case 1: Red is used three times. Because two neighbouring petals must have different colours, there is only one choice of which three petals are coloured red [[IMAGE1]] The rest of the petals can be coloured in any way using the other three colours, so there are $3\times3\times3=27$ ways to colour the petals in this case. Case 2: Red is used two times. There are three possibilities of which two petals are coloured red. [[IMAGE2]] [[IMAGE3]] [[IMAGE4]] Consider the second of these configurations. The two petals that neighbour the top red petal can each be coloured any of the three remaining colours. Once these two (independent) choices are made, the remaining two petals can be independently coloured using one of exactly two possible colours. Therefore, the colouring of the middle configuration can be completed in $3\times3\times2\times2=36$ different ways. In the first configuration, the petal between the two red petals can be coloured in any of $3$ ways, and this choice is independent of the way the other three petals are coloured. The other three petals can either be coloured three different colours ($x$, $y$, and $z$), or they can be coloured using two different colours $(u$ and $v$), as shown below: [[IMAGE5]] [[IMAGE6]] If three colours are used, then there are $6$ ways to choose them since we can choose $x$ in any of $3$ ways, then choose $y$ in any of two ways, and then $z$ is forced. If two colours are used, then there are $3$ ways to choose $u$ and then $2$ ways to choose $v$, for a total of $6$ ways. Thus, there are $6+6=12$ ways to colour the three petals (other than the one between the two red petals). Since there are $3$ ways to colour the petal between the red petals, there are $3\times12=36$ ways to complete the colouring for this configuration. The third configuration can be coloured in $36$ ways as well, so there are $36+36+36=108$ ways to colour the petals in this case. Case 3: Red is used one time. We can imagine colouring the petals one at a time in the clockwise direction starting from the red petal. There will be $3$ choices for the first petal since the only restriction is that we cannot use red. The next petal has $2$ options since it cannot be the same colour as the previous petal, but it also cannot be red. Continuing in this way, there are $2$ options for the $4$th petal, and $2$ options for the $5$th petal. The final petal also has two options since its two neighbours have two different colours. This is because we have not used red other than to colour the top petal. Therefore, there are $3\times2\times2\times2\times2=48$ ways to colour the petals in this case. From the three cases, we get a total of $27+108+48=183$ ways to colour the petals so that the top petal is red. There are $4$ choices for the colour of the top petal and each will lead to a count of $183$. Therefore, the number of ways to colour the petals according the the given rules is $4\times183=732$. The last two digits of $732$ make the integer $32$, which is the answer to
the question.