1. ## STRONG induction

i have this question that i couldnt solve its about STRONG INDUCTION i need ur help pls.

Prove by strong induction that any natural number n greater than 7 can
be written as a sum of 3's and 5's, being understood that no 3's or no 5's is
also accepted, as in 12 = 3 +3 + 3 + 3.

thank you

2. ??? 9 can't. Do you mean sum or difference of 3s and 5s?

3. Originally Posted by Debsta
??? 9 can't. Do you mean sum or difference of 3s and 5s?
Why not?

9 = 3+3+3 is an accepted form, according to his post.

Base case: n=8. 8=5+3 so we are good to go.

Assume that $n \in \mathbb{N}$ can be written as a sum of 3's and 5's. We want to prove that we can also express n+1 as such a sum.

Lemma: The implementation of n as such a sum has either at least three 3s or one 5 in it.
Proof: If n has a 5 in its sum, we are done. Assume it doesn't, then n = 3m for some positive integer m. But we only take n > 7, therefore $m\geq 3$ and the lemma is correct.

According to the inductive assumption, we know that $n = 3k + 5m$ where $k\geq 3 \ \text{or} \ m \geq 1$.

Substituting that into n+1, we get: $n+1 = 3k + 5m + 1$. Now, if:

(I) $k <3$:
We know that there is at least one 5 in the sum. This gives us that: $n+1 = 3k + 5(m-1) + 5 + 1 = 3k + 5(m-1) + 6 = 3(k+2) + 5(m-1)$ and we are done.

(II) $m < 1$:
Then, $n = 3m: \ m \in \mathbb{N}, \ m\geq 3$. This gives us that: $n+1 = 3m+1 = 3(m-3)+3\cdot3 + 1 = 3(m-3)+10 = 3(m-3) + 2\cdot 5$ and then n+1 is also a sum of 3,5s. Therefore it always is, and we are done.