Problem 2043B - Digits

My Solution 313195895

Idea for divisibility by 7

The standard divisibility rule seems to require a recursive approach based on my reasoning which led me to using Fermat's Little Theorem instead.

We have:

which can be written as:

This simplifies to:

Note:

Hence,

can be written as:

Now, how does Fermat's Little Theorem come into play here?

According to Fermat's Little Theorem:

where $$$a \in \mathbb{Z}$$$, $$$p$$$ is prime and $$$\gcd(a, p) = 1$$$.

In this case, we want to check when the following holds:

For this to hold, we need $$$n!$$$ to be a multiple of $$$p - 1$$$ i.e 6.

Therefore $$$n!$$$ is a multiple of 6 when $$$n \gt 2$$$ as shown above.

Why n! must be a multiple of 6 for $$$ 7 \mid (10^{n!} - 1) $$$ to hold:

Assume $$$6 \mid n!$$$. Then we can write $$$n! = 6k$$$ where $$$k \in \mathbb{N}$$$

which holds from Fermat's Little Theorem.

Hence the condition necessary for

to be a multiple of 7 i.e

is

$$$7 \mid d $$$ (The trivial case) Or $$$n \gt 2$$$.

Do you have an alternative approach?