# WARNING: The text below provides a guidance for a Project Euler problem.

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### A long row of dice

Striping the problem of its dicey narrative, the task boils down to finding the value $f(10^{36})$ of the following function $f$:

where $\sigma_0(n)$ is the divisor function, that is a function that counts the number of positive divisors. We now turn to characterise numbers $n$ for which it is the case that $\sigma_0(n) \equiv 1 \text{ mod } 6$. For this, recall the standard formula (when we know the prime factorization of $n = \prod_{i=1}^r p_i^{\alpha_i}$):

Assuming that $\prod_{i=1}^r (\alpha_i + 1) \equiv 1 \text{ mod } 6$, we immediately see that none of the $(\alpha_i + 1)$ can be 0, 2, 3, or 4 (by simply looking at this equation modulo 2, 3) modulo 6. Each $(\alpha_i + 1)$ thus gives remainder 1 or 5 = -1, and each $\alpha_i$ gives remainder 0 or 4 modulo 6. Moreover, the number of $(\alpha_i + 1)$’s giving remainder -1 must be even. Going back to the prime factorization of number $n$, we may rewrite it as:

and pulling together the sixth powers:

Here now comes the crucial observation: to construct an integer $n$ satisfying the above relation, the first chunk $(q_1 \cdots q_{2u})$ can be taken to be any integer $B$ for which the Möbius function $\mu(B) = 1$ (this comes straight from the definition of the Möbius function). Afterwards the first chunk, $( p_1^{k_1} \cdots p_m^{k_m} q_1^{l_1} \cdots q_{2u}^{l_{2u}} )$, can be set to be an arbitrary integer! For a fixed value of $B$, there will be hence $\left\lfloor \sqrt{\frac N {B^4}} \right\rfloor$ possible values for $A$ among $1, 2, \dots, N$. This yields a feasible formula for $f(N)$:

The following piece of compact Julia code gives answer in a (somehow) reasonable time. The bottleneck here is the unnecessary computation of all of the values of the Möbius function.