<s>[INST] Putnam 2005 A1 Solution 
Show that every positive integer is a sum of one or more numbers of the form $2^r3^s,$ where $r$ and $s$ are nonnegative integers and no summand divides another.
  (For example, $23=9+8+6.)$

Suppose for $k = \{1, 2, 3... ,n-1\}$ this equality holds. 
Let $n$ be an even integer. Since $\frac{n}{2} \le n -1$ if $n \ge 2$ then,
$$\frac{n}{2} = \sum 2^x 3^y \implies n = \sum 2^{x+1} 3^{y}$$
Proof hold for $P(n)$ from $P(n-1)$. 
With odds, it gets trickier. 
Suppose $n$ is a odd integer, NOT PRIME. So:
$$n = \{1, 9, 15, ...\}$$
Since, $\frac{n}{3} \le n-1 $ if $n \ge 3$ It follows,
$$\frac{n}{3} = \sum 2^x e^y \implies n = \sum 2^x 3^{y+1}$$
Suppose $n$ is PRIME. 
$$n = \{3, 5, 7, 11, 13, 17, ...  \}$$
There exists $t$ such that: 
$$n < 3^{t-1} \implies \log_{3}(n) < t-1 = \sum 2^x 3^{y}$$
By strong induction,
$$\log_{3}(n) = \sum 2^x 3^y$$
Is there any thing I can do now ?I am stuck!
 [/INST] Your start is right, but odd numbers that are not prime are not all divisible by $3$.
Rather, you need to first show that if $n$ is divisible by $3$ or $2$, then you can proceed inductively. That is what you have (sort of) done already.
If $n$ is not divisible by $3$ or $2$, pick the largest, $k$, such that $n-2^k$ is positive and divisible by $3$. This will alway (why?) be either $k=\lfloor\log_2 n\rfloor$ or $k=\lfloor \log_2 n\rfloor-1$, which means that $4\cdot 2^k > n$.
Then proceed by induction using $\frac{n-2^k}{3}$. Since $\frac{n-2^k}{3}<2^k$, none of the terms are multiples of $2^k$. Multiply those terms by $3$ and then include $2^k$.
Lemma: Prove that if $n,m$ are both not divisible by $3$, then one of $n-m$ or $n-2m$ is divisible by $3$.
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