Dutch Mathematical Olympiad

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 $(25,44)$.