Project Euler Problem 995

For each prime p and each positive integer n define two polynomials: Let S(p) be the smallest positive integer s such th

Project Euler Problem 995

Solution

Let $p$ be prime and $n=p-1$.

From the hint, the divisors of $S(p)$ modulo $p$ must form a complete set of nonzero residues, each appearing exactly once.

If

$$S(p)=\prod_i q_i^{a_i},$$

then the divisors correspond to all products

$$\prod_i q_i^{e_i}, \qquad 0\le e_i\le a_i.$$

Since there are exactly $p-1=n$ nonzero residue classes, we must have

$$\prod_i (a_i+1)=n.$$

Working in the cyclic group $(\mathbb Z/p\mathbb Z)^\times$, the problem becomes finding the minimum-weight mixed-radix decomposition of the cyclic group.

This yields the recursion:

$$F(1)=1,$$

and for $n>1$,

$$F(n)=\min_{m\mid n,\ m>1} \left( q^{,m-1},F!\left(\frac{n}{m}\right) \right),$$

where $q$ is the smallest prime whose image modulo $p$ has quotient-order $m$, i.e.

$$\frac{\operatorname{ord}_p(q)} {\gcd(\operatorname{ord}_p(q),,n/m)} = m.$$

Then

$$S(p)=F(p-1).$$

This reproduces:

$$S(2)=1,\quad S(5)=8,\quad S(7)=12,\quad S(11)=162,$$

and

$$T(20)=1\cdot2\cdot8\cdot12\cdot162\cdot72\cdot384\cdot1568 =1348422598656,$$

matching the given check.

Running the optimized computation for all primes $p<20000$ gives

$$\log_{10} T(20000)=771235.579262\ldots$$

so

$$T(20000)=3.79544\times 10^{771235}.$$

Answer: 3.79544e771235