1. If $[a \cdot b] - 1$ is a nilless number, then it follows from $[[a \cdot b] - 1] = 1$ that $[a \cdot b] = 2$. This case was covered above in the solution for klas 4 and below. Now assume that $[a \cdot b] - 1$ is not nilless. The difference is unequal to 1, so $[a \cdot b] - 1$ is equal to 10, 100, 1000, etcetera, and hence $[a \cdot b]$ is equal to 11, 101, 1001, etcetera. But since $[a \cdot b]$ does not contain zeros, we only have the option $[a \cdot b] = 11$. Hence, $a \cdot b$ is a number consisting of two ones and some zeros. Since $a, b < 100$ we have $a \cdot b < 10000$, so $a \cdot b$ consists of at most four digits. We look at all the possibilities and find a nilless factorisation.

* $a \cdot b = 11$ and $a \cdot b = 101$ are not possible, because those are prime numbers and $a > 1$. In the following we disregard factorisations with $a = 1$.
* $a \cdot b = 110 = 2 \cdot 5 \cdot 11$ gives solutions $(a, b) = (2, 55)$ and $(a, b) = (5, 22)$. The option $(a, b) = (10, 11)$ is not possible because 10 is not nilless.
* $a \cdot b = 1001 = 7 \cdot 11 \cdot 13$ gives solutions $(a, b) = (11, 91)$ and $(a, b) = (13, 77)$. The option $(a, b) = (7, 143)$ is not possible because it does not have $b < 100$.
* $a \cdot b = 1010 = 2 \cdot 5 \cdot 101$ also has three factorisations: $2 \cdot 505$, $5 \cdot 202$ and $10 \cdot 101$. None of them is nilless.
* $a \cdot b = 1100 = 2 \cdot 2 \cdot 5 \cdot 5 \cdot 11$ gives only the solution $(a, b) = (25, 44)$. We already saw that $a$ and $b$ both do not have a factor 2 and a factor 5, so the only other option is $(a, b) = (4, 275)$ and this is a contradiction with $b < 100$.

In total, we find seven solutions: $(4, 5)$, $(8, 25)$, $(2, 55)$, $(5, 22)$, $(11, 91)$, $(13, 77)$, and $(25, 44)$.