Digit sequence that is not prime in any base Is there a sequence of base-$b$ digits of length greater than one with all digits $\ne 0$ that does not represent a prime number in any base?
Example: $12_{10}=12$ is not prime, but $12_3=5$ is.
 A: The smallest example (unless you are in base 2) is $22=2 \cdot 11$
A: If $d_nd_{n-1}\ldots d_0$ is a string of digits, it is convenient to consider the polynomial $f(x)=d_nx^n+d_{n-1}x^{n-1}+\ldots+d_0$. Then $(d_nd_{n-1}\ldots d_0)_b=f(b)$. The OP's question is related to the question "Which integer polynomials take only composite values?".
The answers given so far describe two different phenomena (both illustrated in Mark Bennet's answer).

*

*If $f(x)$ factors as a polynomial, then $f(b)$ is composite for all sufficiently large $b$. For example, $1111_b=101_b\cdot 11_b$ is composite for all $b$, which is equivalent to the polynomial factorization $x^3+x^2+x+1=(x^2+1)(x+1)$.

*Sometimes there is a particular prime $p$ that divides $f(x)$ for all integers $x$. For instance, $2$ always divides $112_b$ (equivalently, $2\big|\,f(b)$, where $f(x)=x^2+x+2$).

The Bunyakovsky conjecture would imply that these are the only reasons a polynomial takes only composite values.

Conjecture: Let $f(x)\in\mathbb{Z}[x]$ be non-constant with positive leading coefficient. Suppose:

*

*$f(x)$ is irreducible, and


*there is no integer $d>1$ such that $d\,\big|\,f(b)$ for all integers $b$.
Then $f(b)$ is prime for infinitely many positive integers $b$.

There are deterministic algorithms for checking whether a given polynomial satisfies the hypotheses of this conjecture.
A: $112$ is always divisible by $2$ for bases above base $2$
$1111=11\times 101$ in any base.
A: $121$. Suppose it is prime in base $b$. Then $b^2+2b+1=(b+1)(b+1)$ is prime.
A: Extending Jorge's example, 
$$\sum_{k=0}^{n}{\binom{n}{k}b^k}=(b+1)^n$$
is not prime in any base large enough to have the binomial coefficients as digits.
A: How about $121$.  We have $1b^2+2b+1=(b+1)^2$.
Similarly for $1331$ we have $1b^3+3b^2+3b+1=(b+1)^3$.
