6. Find all natural numbers that are not divisible by ten and which, in their decimal notation, have two zeros next to each other, and after striking out these two zeros, the original number is reduced by 89 times.
SOLUTION. a) First, assume that the zeros are in the third and second positions from the right. The sought number $x$ then has the form $x=1000a+b$, where $a$ is a natural number (as in the following cases) and $b$ is a non-zero digit. The condition from the problem $1000a+b=89(10a+b)$ can be rearranged to $5a=4b$, from which it follows that $b$ is a multiple of five. Only $b=5$ and $a=4$ satisfy this, so $x=4005$.

b) If the sought number $x$ has zeros in the fourth and third positions from the right, then $x=10000a+b$, where $b$ is a two-digit number. The condition from the problem $10000a+b=89(100a+b)$ can be rearranged to $25a=2b$, from which it follows that $b$ is an odd multiple of 25 (we remind that $x$, and thus $b$, is not divisible by ten). Hence, $b=25, a=2$ or $b=75, a=6$, so $x \in \{20025, 60075\}$.

c) If the sought number $x$ has zeros in the fifth and fourth positions from the right, then $x=100000a+b$, where $b$ is a three-digit number. The condition from the problem $100000a+b=89(1000a+b)$ can be rearranged to $125a=b$, from which it follows that $b$ is an odd multiple of 125. Only $b=125$ and $a=1, b=375$ and $a=3, b=625$ and $a=5, b=875$ and $a=7$ satisfy this, so $x \in \{100125, 300375, 500625, 700875\}$.

d) From the previous cases, we see that for the sought number $x$ of the form $x=10^{n+2}a+b$, where $b$ is an $n$-digit number, we get the condition $10^{n+2}a+b=89(10^{n}a+b)$ or $11 \cdot 10^{n}a=88b$, from which for $n \geq 4$ we get the condition $125 \cdot 10^{n-3}a=b$, according to which $b$ is a multiple of ten. Therefore, no further $x$ that satisfies the problem exists.

Conclusion. The sought numbers are 4005, $20025, 60075, 100125, 300375, 500625$, and 700875.

## EXERCISES FOR PRACTICE: