1. A number is called non-null if it is whole and positive and contains no zeros. You can nullify a positive whole number by simply removing the zeros. We denote this with square brackets, for example $[2050]=25$ and $[13]=13$. For two numbers $a$ and $b$, the following is known:

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

Which pairs $(a, b)$ satisfy these three conditions?
## 1. Version for Class 4 and Lower

If $[a \cdot b]=2$, then $a \cdot b$ is equal to 2, 20, 200, or 2000. Larger is not possible because $a, b<100$ so $a \cdot b<10000$. We will examine the prime factorization of these four possibilities to see which 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 $(a, b)=(4,5)$. Because $a$ and $b$ are non-trivial, $(a, b)=(2,10)$ and $(a, b)=(1,20)$ are not possible. Then we look at $a \cdot b=200=2 \cdot 2 \cdot 2 \cdot 5 \cdot 5$. Due to their non-triviality, $a$ and $b$ do not both contain a factor of 2 and a factor of 5. Numbers for which this is the case are divisible by 10 and thus end in a 0. The only solution with $a \cdot b=200$ is therefore $(a, b)=(8,25)$. Finally, we look at $a \cdot b=2000=2 \cdot 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5$. The only non-trivial factorization is $(a, b)=(16,125)$, but this contradicts $b<100$.

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