We have written an even number in binary. By omitting the trailing $0-\mathrm{s}$, we get the ternary representation of the same number. Determine the number!

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The sought number, in its binary form, has the second-to-last digit with a place value of 2. If we remove the 0 at the end of the number, the place value of this digit becomes 1, which means the transformation reduces the place value by 1 at this position.

Let's examine what happens to the place values of the other digits. The third digit from the end originally has a place value of 4. After the transformation, this digit will be the second-to-last digit of a ternary number, so its new place value is 3, which is again 1 less than the original place value. Since both place values decrease, the sought number must have other non-zero digits as well, because the transformation does not change the value of the number.

The fourth digit from the end in binary has a place value of 8, and its new place value will be 9, which means the place value increases by 1. Thus, we get two valid numbers depending on which of the previous 1-unit decreases balances this 1-unit increase. These numbers are 1010 and 1100 in binary, which are 10 and 12 in decimal, and their ternary forms are 101 and 110.

There are no other solutions, since the digits of binary numbers cannot be greater than 1, so the combined decrease in place values seen so far can be at most 2, while the place values of the subsequent digits increase by more than that. Indeed, the next digit's original place value is 16, and its new place value is 27, with an increase of 11. Further on, the difference continues to grow, as the original place value only doubles, while the new place value triples.

Thus, the sought number in decimal form is either 10 or 12.

Note. With a bit of imagination, 0 can also be considered a solution, if we accept that the absence of any digits after removal is represented by the symbol for zero. Since the transformation has a similar effect on negative numbers, the numbers -10 and -12 can also be included in the solutions. Many found the original two solutions, but several papers had to be marked as incomplete despite this. For a complete solution, it is generally not enough to list all the final results; it must also be proven that no other cases can yield valid results. In this case, papers that made no mention of numbers greater than 12 were considered incomplete. Those who only provided the two sought numbers without any justification received no points at all. (U. G.)