A ten-based number is converted into base seven, eight, and nine number systems. If the resulting three sequences of digits are considered as if they were in the decimal system, and then added together, the sum approximates a number that is less than 25,000 by half a percent. What could the original number be?
Let the representation of the number $n$ in the septenary (base-7) numeral system (the sequence of digits) be denoted by $\alpha_{n}$, and the value of this digit sequence in the base-$k$ numeral system by $\alpha_{n}^{k}$. Similarly, for the octal (base-8) and nonary (base-9) numeral systems, we introduce the notations $\beta_{n}, \gamma_{n}$, and $\beta_{n}^{\frac{k}{k}}, \gamma_{n}^{k}$. It is clear that

$$
n=\alpha_{n}^{\frac{7}{n}}=\beta_{n}^{\frac{8}{n}}=\gamma_{n}^{\frac{9}{n}}
$$

Furthermore, we need to determine those numbers $n$ for which

$$
24875<\alpha_{n}^{10}+\beta_{n}^{10}+\gamma_{n}^{10}<25125
$$

Notice that if we increase the value of $n$, then $\alpha_{n}^{\frac{7}{n}}, \beta_{n}^{\frac{8}{n}}, \gamma_{n}^{\frac{9}{n}}$ also increase, and with them, $\alpha_{n}^{\frac{10}{n}}, \beta_{n}^{\frac{10}{n}}$, and $\gamma_{n}^{10}$ also increase. Thus, the numbers satisfying (1) are consecutive integers, so it is sufficient to find the smallest $n$ for which $\alpha_{n}^{10}+\beta_{n}^{\frac{10}{10}+\gamma_{n}^{10}}$ is already greater than 24875, and the largest for which the sum is still less than 25125.

We will show that $\beta_{n}$ cannot be a five-digit number. Suppose $\beta_{n}^{10} \geqq 10000$, i.e., $n=\beta_{n}^{\frac{8}{n}} \geqq 10000^{\frac{8}{2}}=4096$. Then $\alpha_{n}^{\frac{7}{n}} \geqq 4096=14641^{7}$ and $\gamma_{n}^{\frac{9}{n}} \geqq 4096=5551^{\frac{9}{10}}$, so $\alpha_{n}^{\frac{10}{n}}+\beta_{n}^{\frac{10}{n}}+\gamma_{n}^{\frac{10}{n}} \geqq 14641+10000+5551=30192$. Thus, $\beta_{n}$ can have at most four digits, i.e., $\beta_{n}^{\frac{10}{0}} \leqq 7777$. Moreover, the first digit of $\beta$ cannot be 7. Similarly, based on $3584=13310^{7}=7000^{8}=4822^{9}$, in this case

$$
\alpha_{n}^{\frac{10}{n}}+\beta_{n}^{\frac{10}{10}}+\gamma_{n}^{\frac{10}{n}} \geqq 13310+7000+4822=25132
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_39ee572aff9363df9339g-1.jpg?height=57&width=1198&top_left_y=1018&top_left_x=152)

$$
\alpha_{3583}^{10}+\beta_{3583}^{10}+\gamma_{3583}^{10}=13306+6777+4821=24904
$$

With this, we have found the largest of the sought numbers, 3583. If we now decrease 3583 by at most 6, $\alpha_{n}^{\frac{10}{n}}$ and $\beta_{n}^{\frac{10}{n}}$ will decrease by at most 6, and $\gamma_{n}^{\frac{10}{n}}$ by at most 7, so their sum will still be more than 24875. However,

![](https://cdn.mathpix.com/cropped/2024_05_02_39ee572aff9363df9339g-1.jpg?height=79&width=1800&top_left_y=1283&top_left_x=152)
the smallest is 3577.

In summary, the initial number could have been 3577, 3578, 3579, 3580, 3581, 3582, or 3583.