Find the smallest number written only with the digits 0 and 1, which is divisible without remainder by the product of the six smallest natural numbers.

(Bednářová)
First, we determine the product of the six smallest natural numbers: $1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6=2^{4} \cdot 3^{2}$. $\cdot 5=10 \cdot 2^{3} \cdot 3^{2}$. The sought number $n$ can thus be written as $n=10 \cdot a$, where the number $a$ is the smallest number written only with the digits 0 and 1 that is divisible by the number $2^{3} \cdot 3^{2}=8 \cdot 3^{2}$. According to the rule of divisibility by 8, the last three digits of the number $a$ must be divisible by 8, but this triplet must also be composed only of the digits 0 and 1. Therefore, only the triplet 000 is possible. This means that the number $a=1000 \cdot b$, where $b$ is the smallest natural number written only with 0 and 1 that is divisible by the number $3^{2}=9$, because $n=10 \cdot a=10000 \cdot b$ and the number 3 does not divide 10000. According to the rules of divisibility by 9, the digit sum of the sought number $b$ must be divisible by 9, which means that the number $b$ consists of only nine digits 1. The sought number $n$ is therefore equal to 111111110000.