Induction for postage stamps

**statman101** · May 5th 2010, 07:16 AM

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

**undefined** · May 5th 2010, 10:14 AM

Originally Posted by statman101

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

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.

**novice** · May 5th 2010, 11:40 AM

Originally Posted by statman101

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

If you are interested in learning math, you need a habit of tinkering by do small experiments such as this:

$4\cdot3+5\cdot0=12$
$4\cdot2+5\cdot1=13$
$4\cdot1+5\cdot2=14$
$4\cdot0+5\cdot3=15$
$4\cdot4+5\cdot0=16$
$4\cdot3+5\cdot1=17$
$4\cdot2+5\cdot2=18$
$4\cdot1+5\cdot3=19$
$4\cdot0+5\cdot4=20$
$4\cdot4+5\cdot1=21$
$4\cdot3+5\cdot2=22$
$\cdot$
$\cdot$
$\cdot$

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

Now you can prove it by the Strong Principle of Induction.

**Archie Meade** · May 5th 2010, 02:01 PM

Originally Posted by statman101

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

If some value y is a multiple of fours and fives,
then

$y=4n+5m$

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

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

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

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.

**novice** · May 5th 2010, 03:48 PM

Originally Posted by statman101

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

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 $P(n)$ be a statement If

$(1) P(1)$ is true and
$(2) \forall k \in \mathbb{N}, P(1)\wedge P(2) \wedge P(3) \wedge...\Rightarrow P(k+1)$ 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 $a$ and $b$ such that $12 = 4a+5b$ . In fact we have integers $a=3$ and $b=0$ . So this part of proof is done. We move the next step, the Inductive Step:

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

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

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

This concludes the proof.

**Archie Meade** · May 5th 2010, 04:30 PM

Or...

P(k)

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

P(k+1)

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

Proof

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

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

bingo

**novice** · May 5th 2010, 05:11 PM

This part here is interesting:

Originally Posted by Archie Meade

P(k+1)

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

Proof

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

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

bingo

Why do you have y+k on the left?

Originally Posted by Archie Meade

Or...

P(k)

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

I think the y+k is a little too muddy.

**Archie Meade** · May 5th 2010, 05:22 PM

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.

**novice** · May 5th 2010, 05:44 PM

Originally Posted by Archie Meade

Proof

$(n-k)4+(m+k)5+1=(n-k)4+(m+k)5+5-4$ <---This here?

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

bingo

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.

**undefined** · May 5th 2010, 05:59 PM

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.

**novice** · May 5th 2010, 06:27 PM

Originally Posted by undefined

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) = (3, 0).

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.

Beautiful proof using equivalence classes. I am quite sure there many more ways with Strong induction. Thanks for sharing.

**Archie Meade** · May 6th 2010, 02:57 AM

$y=4n+5m$

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

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

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

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)

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

P(k+1)

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

Proof

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

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

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

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

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