Show that every integer $a>0$ can be shown in the form: $$a=\sum_{i=1}^{k}2^{x_i}\cdot 3^{y_i}$$ where $0\le x_1< x_2< \dots < x_k$ and $0\le y_k < y_{k-1} < \dots < y_1$ are integers.
I thought of showing the number a in ternary, but then the condition for the exponents won't hold.
You can use strong induction. $1=2^03^0$ is the base case. Now assume all numbers up through $m$ can be represented. If $m+1$ is even, take the representation of $\frac 12(m+1)$ and increase all the $x$'s by $1$. If $m+1$ is an odd multiple of $3$, take the representation of $\frac 13(m+1)$ and increase all the $y$'s by $1$. If $m+1$ is coprime to $6$, let $3^p$ be the largest power of $3$ less than $m+1$ and $2^q$ the largest power of $2$ less than $m+1-3^p$ that you can subtract and leave a multiple of $3$. Now $m+1-3^p-2^q$ is a multiple of $6$, so we can represent $m+1$ as $2^03^p+6\cdot\frac 16(m+1-3^p-2^q)+2^q3^0$ To handle the middle term we increase all the $x$'s and $y$'s by $1$ in the representation of $\frac 16(m+1-3^p-2^q)$.
As an example, let us represent $12343$. It is coprime to $6$. The largest power of $3$ in it is $3^8=6561$, leaving $5782$, which is $1 \pmod 3$. The largest power of $2$ in it that is $1 \pmod 3$ is $2^{12}=4096,5782-4096=1686=6\cdot 281$ so our representation so far is $3^82^0+6(281)+2^{12}3^0$ Then $281=3^5+38=3^5+6+2^5$, so $12343=3^82^0+3^62^1+3^22^2+3^12^6+2^{12}3^0$
Added: there is a simpler algorithm. Given $a$, if it is even divide by $2$ as many times as possible until you have an odd value. Now subtract the largest power of $3$ you can. The result will be even, so divide by $2$ again as many times as possible. Stop when you hit a power of $2$. The example of $12343$ goes $12343=3^8+5782=3^8+2\cdot 2891=3^8+2(3^7+704)=3^8+2(3^7+2^6\cdot 11)=3^8+2\cdot3^7+2^7(3^2+2)=3^82^0+2\cdot 3^7+2^73^2+2^83^0$ This shows the representation is not unique. We can show that the powers of $3$ will be strictly decreasing. If $3^n$ is the largest power of $3$ less than $n$, the number at the next step is $\frac 12(n-3^n) \lt \frac 12(3^{n+1}-3^n)=3^n$