
1. A number is called nillless if it is integer and positive and contains no zeros. You can make a positive integer nillless by simply omitting the zeros. We denote this with square brackets, for example $[2050]=25$ and $[13]=13$. When we multiply, add, and subtract we indicate with square brackets when we omit the zeros. For example, $[[4 \cdot 5]+7]=[[20]+7]=[2+7]=[9]=9$ and $[[5+5]+9]=[[10]+9]=[1+9]=[10]=1$. The following is known about the two numbers $a$ and $b$ :

- $a$ and $b$ are nilless,
- $1<a<b<100$,
- $[[a \cdot b]-1]=1$.

Which pairs $(a, b)$ satisfy these three requirements?

## 1. Version for klas $5 \&$ klas 6

If $[a \cdot b]-1$ is a nillless 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 $10,100,1000$, et cetera, and hence $[a \cdot b]$ is equal to 11,101, 1001, et cetera. 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 $a, b1$. 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 nillless.
- $a \cdot b=1100=2 \cdot 2 \cdot 5 \cdot 5 \cdot 11$ 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 $(25,44)$.
