Think of a Number/Examples/Bachet/1

Classic Problem

 * A person chooses secretly a number, and trebles it, telling you whether the product is odd or even.


 * If it is even, he takes half of it,
 * or if it is odd, he adds one and then takes one half.
 * Next he multiplies the result by $3$,
 * and tells you how many times $9$ will divide into the answer, ignoring the remainder.


 * The number he chose is -- what?

Solution
Let $n$ be the number given at the end.

If $n$ is stated as being even, then the number originally chosen was $2 n$.

If $n$ is stated as being odd, then the number originally chosen was $2 n + 1$.

Proof
Let $x$ be the number chosen.

Let $x$ be either $2 n$ or $2 n + 1$, depending on whether it is odd or even.

If $x$ is even, the successive operations do the following:
 * $2 n \to 6 n \to 3 n \to 9 n \to n$

If $x$ is odd, the successive operations do the following:
 * $2 n + 1 \to 6 n + 3 \to 6 n + 4 \to 3 n + 2 = 9 n + 6 \to n$