We are informed that there is a two-digit (natural) number, the square and cube of which contain all ten digits. With the school (four-digit) function table at hand, how many cubing operations must we perform to provide a number that exhibits the mentioned property?
For writing the square and cube of a two-digit number, at most ten digits are needed, and exactly ten digits appear in the square and cube of numbers ranging from 47 to 99 (see, e.g., problem 1126.1). From this, it follows, on the one hand, that the number with the mentioned property can only be found between 47 and 99, and on the other hand, that each digit appears exactly once in its square and cube.

We also saw in the cited problem that the square of the number to be examined is a four-digit number, and its cube is a six-digit number. Thus, we can precisely extract the square from our function table, and round the cube to four significant digits. Generally, we can be certain that the first three digits of this rounded value match the first three digits of the number's cube, because the fourth digit could be either retained or rounded up. Moreover, if we find the fourth digit of the rounded cube to be 0, the mentioned double possibility also applies to the third digit. In every case, if for any number to be examined, we find two matching digits among the four digits of the square and the first three (or fewer) digits of the cube, this number can be discarded, as it cannot have the desired property.

We can also consider the sixth (last) digit of the cube as known, because the cube of any number ending in $1,2,3, \ldots, 9,0$ will end in the same digit as the cube of $1,2,3, \ldots, 9,10$, i.e., 1, or $8,7,4,5,6,3,2,9,0$. Similarly, from the last digit of the base, we can determine the last digit of the square without a table, which is respectively $1,4,9,6,5,6,9,4,1,0$. Therefore, numbers ending in 1, 5, 6, and 0 can be immediately discarded, as the last digit of their square and cube is the same.

In the square and cube of the remaining numbers, except for two cases, we always find a repeating digit, which is shown in the following table. The letters $N$ and $K$ indicate that the repetition can already be determined in the square or the cube, respectively.

| Units digit: |  | 2 | 3 | 4 | 7 | 8 | 9 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Tens place | 4 | - | - | - | $N 2$ | $K 1$ | $K 1$ |
| value | 5 | 0 | 8 | 1 | 3 | $N 3$ |  |
| digit: | 6 | $N 4$ | $N 9$ | $K 2$ | $N 4$ | $N 4$ |  |
|  | 7 | $K 3$ | 3 | 4 | $N 9$ | $K 4$ | 4 |
|  | 8 | $K 5$ | $N 8$ | 5 | 6 | $N 7$ | 7 |
|  | 9 | $N 4$ | 8 | $N 8$ | $N 9$ | 9 | 9 |

Further examination is required only for 59 and 69. However, $59^{2}=3481$ and $59^{3} \approx 205,4 \cdot 10^{3}$, so the digit after the decimal point in the cube, whether rounded up or not, will always match one of the first two digits of the square. Therefore, only the number 69 can have the mentioned property. And since the announcement states that such a number exists, we can state that the number 69 has this property. For this, we did not need to calculate the exact cube of any number.

Verification: $69^{2}=4761, 69^{3}=328509$, and indeed, all ten digits appear exactly once in these numbers.

Pavel Dezső Zoltán (Budapest, Petőfi S. g. IV. o. t.)[^0]


[^0]:    ${ }^{1}$ See K. M. L. 24 (1962/5) 207.