\section*{Exercise 4 - 251014}

 Represent the number 1985

a) in the 2-adic positional system (binary system)

b) in the 3-adic positional system!

c) How can one recognize from the representations in these positional systems that the number is odd?

Note: By the representation of a number in the \(m\)-adic positional system, we mean the one that uses the base \(m\) and the digits \(0,1, \ldots, m-1\).
a) It holds that

\[
1985=1 \cdot 2^{10}+1 \cdot 2^{9}+1 \cdot 2^{8}+1 \cdot 2^{7}+1 \cdot 2^{6}+0 \cdot 2^{5}+0 \cdot 2^{4}+0 \cdot 2^{3}+0 \cdot 2^{2}+0 \cdot 2^{1}+1 \cdot 2^{0}
\]

Thus, the number 1985 has the representation \([11111000001]_{2}\) in the binary system.

b) It holds that

\[
1985=2 \cdot 3^{6}+2 \cdot 3^{5}+0 \cdot 3^{4}+1 \cdot 3^{3}+1 \cdot 3^{2}+1 \cdot 3^{1}+2 \cdot 3^{0}
\]

Thus, the number 1985 has the representation [2201112] \(]_{3}\) in the ternary system.

c) Since all powers of 2 except \(2^{0}=1\) are divisible by 2, a natural number is divisible by 2 if and only if the term \(1 \cdot 2^{0}\) appears in its binary representation, i.e., it is odd if and only if the representation ends in 1.

The number 3 is odd; consequently, all its powers are also odd. Therefore, a natural number is odd if and only if the (only) odd digit 1 appears an odd number of times in its ternary representation.

Equivalently, one can also use the criterion that the "cross sum" of the number in its ternary representation is odd. (The "nines rule" known for the decimal system thus becomes the "twos rule" in the ternary system.)

Solutions of the 1st Round 1985 adopted from [5]

\subsection*{7.27.2 2nd Round 1985, Class 10}