# Breaking down numbers to $2$'s and $3$'s [closed]

How can I prove that all numbers greater than 3 could be written as a sum of $$2$$'s and $$3$$'s only?

• Hint: can there be a smallest number that can't be written as sum of 2s and 3s? Commented Feb 16, 2021 at 12:42
• Can you see that once you have two consecutive numbers which can be written in this way, all the rest can? Commented Feb 16, 2021 at 12:43
• Every even greater than $3$ can be written $2+2+\dots+2$. Every odd greater than $3$ can be written $3+2+2+\dots+2$. These can be phrased as straightforward induction proofs. Commented Feb 16, 2021 at 12:44
• Of course... there are in fact many different ways a particular number can be written as a sum of $2$'s or $3$'s, but that doesn't matter to us. As a fun challenge though, you might try revisiting this problem later once you have learned about recurrence relations to see if you can come up with an expression for the number of ways a number $n$ can be written as a sum of $2$'s or $3$'s oeis.org/A182097 Commented Feb 16, 2021 at 12:49

Let $$n$$ be a number, $$n>3$$. There are only two possibilities, either $$n$$ is even or odd. If $$n$$ is even, then $$\exists\;k$$, s.t. $$2k=n$$. Then it is clear that, $$n$$ can be written as sum of $$k$$ twos. On the other hand, if $$n$$ is odd, then $$\exists\;l$$, s.t. $$2l+1=n\Rightarrow 2(l-1)+3=n$$. Then it is clear that, $$n$$ can be written as sum of $$l-1$$ twos, and one $$3$$.
Note that numbers $$1$$ and $$3$$ can't be written as sum of $$2$$ and $$3$$. That is why the condition greater than $$3$$ is needed.