Finding consecutive naturals that all fail to have inverses modulo $70$

Any number that is coprime to a modulus will have an inverse, so we need to find $5$ consecutive numbers that share a factor with $70$.

$70$ has three primes factors: $2,5,7$. Of any $5$ consecutive numbers, two or three will be even, but at most one will be divisible by $5$ or $7$. So we need three even numbers with an odd multiple of $5$ and an odd multiple of $7$ in the second and fourth positions. Since odd multiples of $5$ are all $\equiv 5\bmod 10$, it's apparent this means we need to look for cases where $7k \equiv \{3,7\} \bmod 10$. There are two such cases below $70$: $k=1$ and $k=9$ (giving $7$ and $63$), with the two options of $5$ consecutive numbers:

$$\{4,5,6,7,8\} \text{ and } \{62,63,64,65,66\}$$

For those comfortable with negative values in modular arithmetic, the second set is the negation of the first, that is, $\{62,63,64,65,66\} \equiv \{-8,-7,-6,-5,-4\} \bmod {70}$ .

Note that $x \in \mathbb{N}$ has an inverse modulo $n$ if and only if $\text{gcd}(x,n) = 1$. Looking for the prime decomposition of $70$, we see that $$70 = 2 \cdot 5 \cdot 7.$$ Now clearly $4, 5, 6, 7, 8$ don't have greatest common divisor $1$ with $70$ and therefore no inverse modulo $70$.

The numbers $[4,5,6,7,8]$ satisfy the required property.