A natural number raised to the sixth power, with its digits arranged in ascending order, is as follows:

$$
0,2,3,4,4,7,8,8,9
$$

What is this number?
If there exists a number $x$ that meets the condition, then it is a two-digit number because its sixth power is greater than the 7-digit number $10^{6}$ but less than the 13-digit number $100^{6}$. Tighter bounds for $x^{6}$ are given by the smallest and largest numbers that can be written with the given digits. Since 0 cannot be the first digit, we get

$$
203447889 \leq x^{6} \leq 988744320
$$

Slightly increasing the upper bound,

$$
x^{6}(600)^{3}=$ 216000 000. Further, $24^{6}=\left((3 \cdot 8)^{3}\right)^{2}=(27 \cdot 512)^{2}=13824^{2}<14000^{2}=196000000$ is already smaller than the lower bound. Thus, the number in question can only fall between 24 and 32 (not including the bounds)①

Notice that the sum of the digits of $x^{6}$ is divisible by 3, so $x^{6}$ itself is divisible by 3. This is only possible if $x$ itself is divisible by 3 (the sixth power of a number not divisible by 3 is not divisible by 3). Among the natural numbers greater than 24 and less than 32, only 27 and 30 are divisible by 3, but 30 cannot be the solution because its sixth power ends in 6 zeros. However, $27^{6}=387420$ 489, the digits match the given ones, so $x=27$.

Remarks. 1. The solution can also be selected based on the last digit of the number. If the last digit of $x$ is respectively

$$
0,1,2,3,4,5,6,7,8,9
$$

then the last digit of $x^{6}$ is respectively

$$
0,1,4,9,6,5,6,9,4,1
$$

Since the digits 1, 5, and 6 do not appear among the given digits, $x$ can only be 27, 28, or 30 among the numbers already calculated. 30 is not a solution for the already mentioned reason, $28^{6}$ does not have the given digits, but 27 does.

2. Several contestants concluded from the divisibility of $x^{6}$ by 9 that $x$ must be divisible by 9, and marked 27 as the only possible solution among the calculated bounds. This conclusion is false, because for example, $3^{6}$ is divisible by 9, but the base: 3 is not.[^0]

[^0]:    ${ }^{1} \mathrm{The}$ inequality pair obtained for $x^{6}$ can most quickly yield bounds for $x$ using logarithms. Since we are looking for a two-digit bound, it is enough to keep four (or even three) significant digits from the 9-digit numbers, of course rounding the lower bound down and the upper bound up in the process.