# Problem 8.5 (7 points)

Find the smallest natural number that is divisible by $48^{2}$ and contains only the digits 0 and 1.
Solution:

$48^{2}=2^{8} \cdot 3^{2}$. For a number to be divisible by $2^{8}$, it must end with at least 8 zeros according to the divisibility rule, because otherwise, with fewer zeros ($n \leq 7$), it would have the form $1 \ldots .1 \cdot 10^{\text {n }}$ and would only be divisible by the n-th power of two, but we need divisibility by the 8th power of two. For a number to be divisible by $3^{2}=9$, it must have a sum of digits that is a multiple of 9, meaning the number must contain a multiple of 9 ones, i.e., at least 9. Therefore, our number must have at least 9 ones and end with at least 8 zeros, and the smallest such number is 11111111100000000 (9 ones and 8 zeros), which is divisible by both $2^{8}$ and $3^{2}$, and thus, due to their mutual simplicity, by their product $2^{8} \cdot 3^{2}=48^{2}$.

| Criteria | Points |
| :--- | :---: |
| Complete solution | 7 |
| Necessity of 8 zeros derived | 3 |
| Divisibility by 9 shown | 3 |
| Incorrect solution | 0 |

Answer: 11111111100000000 (9 ones and 8 zeros)