
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 $[13]=13$. The following is known about the two numbers $a$ and $b$ :

- $a$ and $b$ are nillless,
- $1<a<b<100$
- $[a \cdot b]=2$.

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

## 1. Version for klas 4 and below

If $[a \cdot b]=2$, then $a \cdot b$ equals $2,20,200$, or 2000 . Bigger is not possible, because $a, b<100$ so $a \cdot b<10000$. We will look at the prime decomposition of these four possibilities to see what options there are for $a$ and $b$.

Since $a$ and $b$ are both greater than $1, a \cdot b=2$ is not possible. For $a \cdot b=20=2 \cdot 2 \cdot 5$, we find the possibility $(a, b)=(4,5)$. Since $a$ and $b$ are nilless, $(a, b)=(2,10)$ and $(a, b)=(1,20)$ are impossible. Then we look at $a \cdot b=200=2 \cdot 2 \cdot 2 \cdot 5 \cdot 5$. Due to their nillessness both $a$ and $b$ cannot have a factor 2 and a factor 5 at the same time. Numbers that do have those, are divisible by 10 and hence end in a 0 . So the only solution with $a \cdot b=200$ is $(a, b)=(8,25)$. Finally, we consider $a \cdot b=2000=2 \cdot 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5$. The only nilless factorisation is $(a, b)=(16,125)$, but that contradicts $b<100$.

So the only two solutions are $(a, b)=(4,5)$ and $(a, b)=(8,25)$.

