Problem 1162

Let's find such a number. Since the number must be greater than 5, and numbers greater than 5 in base 6 are two-digit numbers, the number we are looking for must be at least two digits. The number in base 10, when its digits are reversed, gives the number in base 6, so the number in base 6 can have at most as many digits as in base 10. However, the largest five-digit number in base 6, $6^5-1=7775$, is only a four-digit number in base 10, so it is certain that numbers in base 10 with at least five digits will have more digits in base 6 than in base 10. Therefore, the numbers in the problem can have at most four digits.

Since in base 6 only the digits 0 to 5 are used, the digits in our number can only be these.

Finally, the first digit in the base 10 representation of the number cannot be zero. Let's examine separately the numbers that can be written with two digits, three digits, and four digits.

I. For a two-digit number, let it be $A B$. This is in base 10, and its base 6 representation is $B A$, that is,

$$\begin{gathered} 10 A + B = 6 B + A \\ 9 A = 5 B \end{gathered}$$

so $B$ must be divisible by 9. But since $B \leq 5$, $B$ can only be 0, in which case $A$ would also be 0, which we have excluded. Therefore, there is no suitable two-digit number.

II. For a three-digit number, $A B C_{10} = C B A_{6}$, that is,

$$\begin{gathered} 100 A + 10 B + C = 36 C + 6 B + A \\ 99 A + 4 B = 35 C \leq 175 \end{gathered}$$

since $C \leq 5$. Thus, the equality can only hold for $A \neq 0$ when $A=1$:

$$99 + 4 B = 35 C$$

$B$ is a number between 0 and 5, so $35 C$ must be between 99 and $99 + 20 = 119$. Only $35 \cdot 3 = 105$ fits; thus, $C=3$ and

$$4 B = 105 - 99 = 6$$

from which $B$ would not be an integer. Therefore, there is no three-digit number.

III. For a four-digit number, $A B C D_{10} = D C B A_{6}$, that is,

$$\begin{aligned} & 1000 A + 100 B + 10 C + D = 216 D + 36 C + 6 B + A \\ & 999 A + 94 B = 26 C + 215 D \leq 26 \cdot 5 + 215 \cdot 5 = 1205 \end{aligned}$$

thus the equality can only hold for $A=1$:

$$999 + 94 B = 26 C + 215 D$$

Since $999 > 990 = 26 \cdot 5 + 215 \cdot 4$, and $D \leq 5$, only $D=5$ can hold:

$$94 B = 26 C + 76 < 26 \cdot 5 + 76 = 206$$

so $B=0, 1, 2$ are the only possibilities.

For $B=0$, $0 = 26 C + 76$, which is not possible.

For $B=1$, $26 C = 94 - 76 = 18$, which is not possible.

For $B=2$, $26 C = 188 - 76 = 112$, which is not a multiple of 26.

Thus, there is no four-digit number.

As mentioned at the beginning of the solution, the base 6 representation of numbers with more than four digits will have more than four digits, so we have completed the solution.