Problem 1066

The second number, except for the others, gives a composite number in any number system whose base is 3 or greater - in other words, in which the digits $0,1,2$ are used, or there are three different digits $A, B, C$ to replace them. Indeed, these numbers can be written in a product form:

$$\begin{aligned} & 12002110=1200211 \cdot 10, \quad 121212=12 \cdot 10101 \\ & 102102=102 \cdot 1001, \quad \overline{A B C A B C}=\overline{A B C} \cdot 1001 \end{aligned}$$

and neither of the factors written can be 1.

The second number is also a composite number in any $b$ (integer) base number system because it is an even number. Indeed, by rearrangement, summing up the powers of $b$ with equal digits, it can be written as:

$$2\left(b^{9}+b^{8}+b^{4}+1\right)+b\left(b^{6}+b^{4}+b^{2}+1\right)$$

(the exponents are understood in the decimal system), and here the value of the second parenthesis is even for odd $b$ as well, because the powers of an odd number are odd, and the sum of 4 odd numbers is even. With this, we have proven the statement.

Kálmán András (Budapest, Petőfi S. g. II. o. t.)

Remark. The evenness of the second number can also be seen when restricted to the base 3, as in any $b$ base number system, the divisibility rule for the number one less than the base is the same as the divisibility rule for 9 in the decimal system. Indeed, in determining the latter, we only used the fact that for any positive integer exponent $b^{n}-1$ is divisible by $b-1$, so $b^{n}=(b-1) D+1$, where $D$ is an integer, and thus, for example, the number

$$\begin{gathered} \overline{A B C}=A b^{2}+B b+C=A \cdot[(b-1) D+1]+B \cdot[(b-1)+1]+C= \\ =(b-1)(A \cdot D+B)+(A+B+C) \end{gathered}$$

the remainder when divided by $b-1$ is the same as the remainder when the sum of the digits $A, B, C$ is divided by $b-1$. In the second number to be examined, the sum of the digits is 12 (understood in the decimal system), which is even, so the number itself is even.