I've read a few articles but its not doing the trick.

can you show step by step how a postage of 12 cents or more can be formed using 4 cents and 5 cents stamps

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- May 5th 2010, 07:16 AMstatman101Induction for postage stamps
I've read a few articles but its not doing the trick.

can you show step by step how a postage of 12 cents or more can be formed using 4 cents and 5 cents stamps - May 5th 2010, 10:14 AMundefined
Ah, you're referring to the fact that the Frobenius number of {4,5} is 4*5-(4+5) = 11.

The easiest demonstration I can think of involves congruence with a modulus, but you may not be familiar with the notation.

We can immediately obtain all numbers congruent to 0 (mod 4). These are simply the multiples of 4.

Starting with 5, we can obtain all numbers congruent to 1 (mod 4). These are numbers of the form 4k+1. Just start with 5 and continue adding 4 to it: 5, 9, 13, 17, ...

Starting with 10 we can obtain all numbers congruent to 2 (mod 4).

Starting with 15 we can obtain all numbers congruent to 3 (mod 4).

So certainly, starting with 15 we can obtain all integers.

14 we got via 5+5+4. 13 we got via 5+4+4. 12 is 4+4+4. So that accounts for all integers greater than or equal to 12. - May 5th 2010, 11:40 AMnovice
If you are interested in learning math, you need a habit of tinkering by do small experiments such as this:

$\displaystyle 4\cdot3+5\cdot0=12$

$\displaystyle 4\cdot2+5\cdot1=13$

$\displaystyle 4\cdot1+5\cdot2=14$

$\displaystyle 4\cdot0+5\cdot3=15$

$\displaystyle 4\cdot4+5\cdot0=16$

$\displaystyle 4\cdot3+5\cdot1=17$

$\displaystyle 4\cdot2+5\cdot2=18$

$\displaystyle 4\cdot1+5\cdot3=19$

$\displaystyle 4\cdot0+5\cdot4=20$

$\displaystyle 4\cdot4+5\cdot1=21$

$\displaystyle 4\cdot3+5\cdot2=22$

$\displaystyle \cdot$

$\displaystyle \cdot$

$\displaystyle \cdot$

These are: $\displaystyle (4q+r)+(5s+t)$, where $\displaystyle q,r,s,t \in \mathbb{N}$, and $\displaystyle 0\leq r < 4$ and $\displaystyle 0\leq t < 5$

Now you can prove it by the Strong Principle of Induction. - May 5th 2010, 02:01 PMArchie Meade
If some value y is a multiple of fours and fives,

then

$\displaystyle y=4n+5m$

$\displaystyle y+1=4n+1+5m=(n-1)4+4+1+5m=(n-1)4+5+5m=(n-1)4+(m+1)5$

$\displaystyle y+2=(n-2)4+4+4+2+5m=(n-2)4+(2)5+5m=(n-2)4+(m+2)5$

$\displaystyle y+3=(n-3)4+4+4+4+3+5m=(n-3)4+15+5m=(n-3)4+(m+3)5$

y+4 is a sum of fours and fives if y is.

y+5 is a sum of fours and fives if y is or if y+1 is.

y+6 is a sum of fives and fours if y+2 or y+1 is.

y+7 is a sum of fours and fives if y+3 or y+2 is.

The pattern repeats. - May 5th 2010, 03:48 PMnovice
Let's have a little fun of writing the proof by the Strong Principle of Mathematical Induction, which says:

*For each positive integer*n,*let*$\displaystyle P(n)$*be a statement If*

$\displaystyle (1) P(1) $*is true and*

$\displaystyle (2) \forall k \in \mathbb{N}, P(1)\wedge P(2) \wedge P(3) \wedge...\Rightarrow P(k+1)$*is true*.

Proof:

For the__basis step__, we want to prove the a postage of 12 cents is made of some numbers of 4 cents and 5 cents stamps. So

We say there exist nonnegative integers $\displaystyle a$ and $\displaystyle b $ such that $\displaystyle 12 = 4a+5b$. In fact we have integers $\displaystyle a=3$ and $\displaystyle b=0$. So this part of proof is done. We move the next step, the Inductive Step:

Here we assume the there exist integers $\displaystyle a$ and $\displaystyle b$ such the $\displaystyle i=4a+5b$, where $\displaystyle 0 \leq i <k$, where $\displaystyle k>12$. Now we consider the integer $\displaystyle k+1$. We show that there are nonnegative integers such that $\displaystyle k+1=4a+5b$. Since $\displaystyle k>12$, we we have $\displaystyle 13=4\cdot2 + 5 \cdot 1$, then we have $\displaystyle 14=4\cdot1 + 5 \cdot 2$, and we have $\displaystyle 15=4\cdot0+5\cdot3$. Hence we assume $\displaystyle k+1\geq 16. $ It follows that $\displaystyle 12 \leq (k+1) - 4 < k$. By the induction hypothesis, there are nonnegative integers $\displaystyle a$ and $\displaystyle b$ such that

$\displaystyle (k+1)-4=4a+5b$, and so

$\displaystyle (k+1)=(4a+4)+5b$ or equivalently, $\displaystyle (k+1)=4(a+1)+5b$, where $\displaystyle (a+1)$ is and integer.

This concludes the proof. - May 5th 2010, 04:30 PMArchie Meade
Or...

**P(k)**

$\displaystyle y+k=(n-k)4+(m+k)5$

**P(k+1)**

$\displaystyle (n-k)4+(m+k)5+1$ is a combination of integer multiples of 4 and 5 if y+k is

**Proof**

$\displaystyle (n-k)4+(m+k)5+1=(n-k)4+(m+k)5+5-4$

$\displaystyle =[n-(k+1)]4+[m+(k+1)]5$

bingo - May 5th 2010, 05:11 PMnovice
- May 5th 2010, 05:22 PMArchie Meade
Hi novice,

sorry about that,

it's a follow on from my earlier post above,

i guess they look a bit disjointed,

i might write out another post. - May 5th 2010, 05:44 PMnovice
While you are at it, I have a question about the line I marked in red.

It seems that the line in question happened to work out nicely. I am wondering whether that was only an accident.

In fact the stamps can also be made of some 3 cents and 5 cents stamps. In this particular case, the line in red might even become false. - May 5th 2010, 05:59 PMundefined
Since there are so many formal proofs popping up here, I might as well formalise my earlier post.

Define "k is attainable" as: there exist nonnegative integers m and n such that k = 4m + 5n.

Base cases:

12 is attainable, (m, n) = (3, 0).

13 is attainable, (m, n) = (2, 1).

14 is attainable, (m, n) = (1, 2).

15 is attainable, (m, n) = (0, 3).

Induction step:

Let k be an integer such that k-4 is attainable. Then k is attainable.

Proof:

There exist nonnegative integers m and n such that k-4 = 4m + 5n.

Then k = 4(m+1) + 5n.

Therefore, k is attainable.

One approach to tie together the induction step with the base cases is to use strong induction, but why not just partition the set of nonnegative integers by congruence (mod 4)?

S0 = {0, 4, 8, ...}

S1 = {1, 5, 9, ...}

S2 = {2, 6, 10, ...}

S3 = {3, 7, 11, ...}

It can be seen that the intersection of the sets is the empty set, and the union of the sets is the nonnegative integers.

So we treat each subset the same way we would treat the nonnegative integers and apply weak induction four times.

Therefore, all integers greater than or equal to 12 are attainable.

This concludes the proof. - May 5th 2010, 06:27 PMnovice
- May 6th 2010, 02:57 AMArchie Meade
$\displaystyle y=4n+5m$

$\displaystyle y+1=(n-1)4+(m+1)5$

$\displaystyle y+2=(n-2)4+(m+2)5$

$\displaystyle y+3=(n-3)4+(m+3)5$

The base case has y=12 and k=0.

The general case is P(k) and we use it to establish that it causes P(k+1) to be valid.

**P(k)**

$\displaystyle y+k=(n-k)4+(m+k)5$ is a combination of multiples of 4 and 5

**P(k+1)**

$\displaystyle y+k+1=\left(n-(k+1)4\right)+\left(m+(k+1)5\right)$ is also a combination of multiples of 4 and 5

**Proof**

$\displaystyle y+k+1=\left(n-(k+1)\right)4+\left(m+(k+1)\right)5$

$\displaystyle =\left((n-k)4-4\right)+\left((m+k)5+5\right)$

$\displaystyle =(n-k)4+(m+k)5+5-4$

which is definately a combination of multiples of 4 and 5 if y+k is.