Premise: So, my father came up with a game where you would look at a 4-digit license plate make 2 pairs of numbers that were made from 2 of the digits from the license plate and were Identical Odd Numbers.

Some example games may be: \begin{align} \fbox{ΑΒΕ 5409} & \rightarrow 5+4=9 \\ & \rightarrow 9+0=9 \\ & \rightarrow 9,9 \\ \end{align}

\begin{align} \fbox{ΗΙΚ 3162} & \rightarrow 3\times1=3 \\ & \rightarrow \frac{6}{2}=3 \\ & \rightarrow 3,3 \\ \end{align}

By the way, all types of basic elementary operations (addition, subtraction, etc...) are allowed.

Question: Given any 4-digit number, how can I achieve a successful game?

Is this even possible?

• You'll really need to specify 'all types of operations' if you want some kind of intelligent answer to this. – Bram28 Nov 10 '17 at 21:05
• If you don't specify what you mean by "all types of operations" you'll get a smart aleck answer like us defining a new operation $\heartsuit$ such that $a\heartsuit b = 1$ regardless the values of $a$ and $b$ in which case you have $5\heartsuit 4 = 9\heartsuit 0 = 1$. Using this newly defined operation trivially allows you to win the game every time. – JMoravitz Nov 10 '17 at 21:15
• If you are specifically wanting odd results., and you limit yourself to adding, subtracting, multiplying and dividing, and only a single operation, then the license plate with numbers $8444$, the only possible results that could be made using the eight and one of the fours is even, and the game could not be won. Once you allow more than these four operations or allow iterated operations, then you run into the problem of me defining some stupid operation or sequence of operations that lets me reach $1$ every time like my earlier comment. – JMoravitz Nov 10 '17 at 22:27
• Another problem number: $0000$. If all digits are nonzero however and repeated application of our four basic operations as well as allowing each digit to appear multiple times and the use of parenthesis, we can always win. As an example, we know that it is possible to represent $\gcd(a,b)$ using repeated applications of addition and subtraction. Then, we can also use $\gcd(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)})$ which will trivially equal one. (this doens't work if $a$ or $b$ is equal to zero) – JMoravitz Nov 10 '17 at 22:45