Prove by induction that any amount of postage of at least 24 cents can be made up with only 4-cent and 9-cent stamps.


closed as off-topic by qbert, mrp, Juniven, Chris Godsil, Namaste May 5 '17 at 13:01

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  • $\begingroup$ I believe you can at least get the base case! $\endgroup$ – qbert May 5 '17 at 0:42

We have a proposition $$P_n:\exists x,y\in\mathbb{N}\quad\text{such that}\quad n=4x+9y,\quad n\geq24$$

First we must show its true for the base case $n=24:$

$24=4\cdot 6\implies$ $P_n$ holds

Assume $P_n$ is true, now we must prove $P_{n+1}$ is true.

$(n+1) = 4x+9y+1\quad$ There are some cases to consider.

$(1): x\geq 2\implies n+1= 4(x-2)+9y+8+1=4(x-2)+9(y+1)$

$(2):x=1\implies y\geq 3\implies n+1=4+9y+1 =5+9(y-3)+27=4\cdot8+9(y-3)$

$(3):x=0\implies y\geq 4 \implies n+1=9y+1= 9(y-4)+36=9(y-4)+4\cdot 9$

In all cases we have shown that $n+1$ can be written in the required form, hence our statement $P_n$ holds $\forall n\in\mathbb{N_{n\geq 24}}$






28=24+4, 29=25+4, 30=26+4, ...


Statement to be proven: $\forall n\in \Bbb N, \exists \{k,\ell\} : n+23 = 4k+9\ell$

Equivalent statement: $$\forall m\in \Bbb N, \left\{ \begin{array}{c} \exists \{k_0,\ell_0\} : 4m+20+0 = 4k_0+9\ell_0\\ \exists \{k_1,\ell_1\} : 4m+21+1 = 4k_1+9\ell_1\\ \exists \{k_2,\ell_2\} : 4m+22+2 = 4k_2+9\ell_2\\ \exists \{k_3,\ell_3\} : 4m+23+3 = 4k_3+9\ell_3 \end{array} \right.$$

We will prove the equivalent statement (ES) by induction on $m$.

Basis: when $m=1$, ES(1) reads $$ \left\{ \begin{array}{c} \exists \{k_0,\ell_0\} : 24 = 4k_0+9\ell_0\\ \exists \{k_1,\ell_1\} : 25 = 4k_1+9\ell_1\\ \exists \{k_2,\ell_2\} : 26 = 4k_2+9\ell_2\\ \exists \{k_3,\ell_3\} : 27 = 4k_3+9\ell_3 \end{array} \right.$$ with all the $k_i$ and $\ell_i$ non-negative integers, which is demonstrated by $$\{ k_0 = k_1=k_2=k_3 = 6, \ell_0 = 0, \ell_1 = 1, \ell_2 = 2, \ell_3 = 3\}$$

Induction: assume ES is true for some $m$ using some set of $k^{(m)}_i,\ell^{(m)}_i$. . Then ES$(m+1)$ is shown to be true by using

$$\begin{array}{cccc} k_0 = k^{(m)}_0 +1& k_1 = k^{(m)}_1+1&k_2 = k^{(m)}_2+1&k_3 = k^{(m)}_3+1\\ \ell_0 = \ell^{(m)}_0&\ell_1 = \ell^{(m)}_1&\ell_2 = \ell^{(m)}_2&\ell_3 = \ell^{(m)}_3& \end{array}$$

Thus induction is established, and (ES) is true for all natural $m$, so the original statement is true for all integer $m\geq 24$.


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