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 multiplication, addition, and subtraction, we use square brackets to indicate when we are nullifying. For example, $[[4 \cdot 5]+7]=[[20]+7]=[2+7]=[9]=9$ and $[[5+5]+9]=[[10]+9]=[1+9]=[10]=1$. For two numbers $a$ and $b$, the following is known:

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

Which pairs $(a, b)$ satisfy these three conditions?
## 1. Version for Grade 5 & Grade 6

If $[a \cdot b]-1$ is a non-trivial number, then from $[[a \cdot b]-1]=1$ it follows that $[a \cdot b]=2$. This case has been discussed in the solution for Grade 4 and below. Now assume that $[a \cdot b]-1$ is not non-trivial. The nullification is equal to 1, so $[a \cdot b]-1$ is equal to $10, 100, 1000$, etc., thus $[a \cdot b]$ is equal to 11, 101, 1001, etc. But since $[a \cdot b]$ contains no zeros, we only have the option $[a \cdot b]=11$. Therefore, $a \cdot b$ is a number consisting of two ones and some zeros. Since $a, b > 1$. In the following, we will exclude factorizations with $a=1$.
- $a \cdot b=110=2 \cdot 5 \cdot 11$ gives solutions $(a, b)=(2,55)$ and $(a, b)=(5,22)$. The possibility $(a, b)=(10,11)$ is invalid because 10 is not non-trivial.
- $a \cdot b=1001=7 \cdot 11 \cdot 13$ gives solutions $(a, b)=(11,91)$ and $(a, b)=(13,77)$. The possibility $(a, b)=(7,143)$ is invalid because $b<100$ does not hold.
- $a \cdot b=1010=2 \cdot 5 \cdot 101$ has three factorizations: $2 \cdot 505, 5 \cdot 202$, and $10 \cdot 101$. None of these are non-trivial.
- $a \cdot b=1100=2 \cdot 2 \cdot 5 \cdot 5 \cdot 11$ gives the only solution $(a, b)=(25,44)$. We already saw that $a$ and $b$ cannot both contain a factor of 2 and a factor of 5, so the only other possibility is $(a, b)=(4,275)$, which contradicts $b<100$.

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