Any even number $2n$ can be written as 22 in base $n-1$. In particular $1994=22996$.

We have to show that we cannot write $1993=$ aaa ... $a_{b}$. If the number has $n$ digits, then 1993 $=a(1+b+\ldots+b^{n-1})=a(b^n-1)/(b-1)$. But 1993 is prime, so $a$ must be 1. Hence $b^{n-1}+\ldots+b -1992=0$. So $b$ must divide $1992=2^3 \cdot 3 \cdot 83$. We cannot have $n=2$, for then $b=1992$ and we require $b>2$. But $83^2=6889>1993$, so $b$ must divide 24. Hence $b=2,3,4,6,8,12$, or 24. But we can easily check that none of these work:

```
$1+2+2^2+\ldots+2^9=1023,1+\ldots+2^{10}=2047$.
$1+3+\ldots+3^6=1093,1+\ldots+3^7=3280$
$1+4+\ldots+4^5=1365,1+\ldots+4^6=5461$
```

```
$1+6+\ldots+6^4=1555,1+\ldots+6^5=9331$
$1+8+8^2+8^3=585,1+\ldots+8^4=4681$
$1+12+12^2+12^3=1885,1+\ldots+12^4=22621$
$1+24+24^2=601,1+\ldots+24^3=14425$
```