prove that for $n\geq 8$ exist $p,q\epsilon\mathbb{N}$ so that $n=5p +3q$

Question

I ve been asked to prove that for $n\geq 8$ exist $p,q\in \mathbb{N}_0$ so that $n=5p +3q$ anyone who could help and explain too?

You only need to verify that the claim holds for $n=8,9,10$. — Batominovski, Nov 20 '17 at 10:02
See https://math.stackexchange.com/questions/2527496/divisible-by-multiple-numbers-induction — lab bhattacharjee, Nov 20 '17 at 10:11
In particular see Robjohn's answer for an elementary derivation of the more general result. — Jyrki Lahtonen, Nov 20 '17 at 11:04

score 2 · Accepted Answer · answered Nov 20 '17 at 10:03

Base of induction $n=8,9,10$: \begin{eqnarray*} 8 &=5\cdot 1+3\cdot 1\\ 9 &=5\cdot 0+3\cdot 1\\ 10 &=5\cdot 2+3\cdot 0 \end{eqnarray*}

Induction step $n\to n+3$. By induction hypothetis there are $p,q$ such that $n=5p+3q$. Then:

$$ n+3 = 5p+3q+3 = 5p'+3q'$$ where $q' = q+1$ and $p'=p$.

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