10. Jacopo has 6 colors (including white) available to color all integers. However, he wants to respect the following conditions: $n$ and $n+5$ must have the same color for every integer $n$, and if $a b$ is white, then at least one of $a$ and $b$ must be white. In how many ways can Jacopo color the integers?
(A) 156
(B) 656
(C) 3181
(D) 3906
(E) 3936
10. The answer is $\mathbf{( E )}$. The condition that $n$ and $n+5$ have the same color is equivalent to restricting the problem to the 5 congruence classes modulo 5 (which we will continue to call $0,1,2,3,4$). Now let's study the second condition; write down all possible ways to obtain a number as the product of two others: $0 \equiv 0 \cdot k$ with $k \in\{0,1,2,3,4\}, 1 \equiv 1 \cdot 1 \equiv 2 \cdot 3 \equiv 4 \cdot 4 ; 2 \equiv 1 \cdot 2 \equiv 3 \cdot 4$; $3 \equiv 1 \cdot 3 \equiv 2 \cdot 4 ; 4 \equiv 1 \cdot 4 \equiv 2 \cdot 2 \equiv 3 \cdot 3$. Each congruence implies that if the product is white, then one of the two factors is white. We see how the color of 0 does not affect the colors of the other classes. Instead, if 1 is white, then 4 is also white; if 4 is white, then both 2 and 3 are white; if 2 (or 3) is white, then either 3 (or 2) is white or 4 is white, but 4 white implies that both 2 and 3 are white. In summary, if 1 is white, then 2, 3, and 4 are also white. If 4 is white but 1 is not, then 2 and 3 are white. Finally, if 2 is white but neither 1 nor 4 are, then 3 is also white and vice versa. Therefore, we have four cases for the classes in $\{1,2,3,4\}$ colored white: $\{1,2,3,4\} \circ\{2,3,4\} \circ\{2,3\} \circ \emptyset$. Remembering that 0 can be colored independently, we get $6 \cdot\left(1+5+5^{2}+5^{4}\right)=3936$ different colorings.