7 Cents and 11 cents Stamps Mathematical Induction

Question

Assume you can only use 7-cent and 11-cent stamps.
a) Determine which amounts of postage can be formed by the given stamps.
b) Prove your answer using the principle of mathematical induction.
c) Prove your answer using strong induction.

By doing a) i found out that all the numbers after 59 can be created using a combination of 7 cants and 11 cents stamps
In part b , i assumed that the $$ n=7k+11l$$ where k is the amount of 7 cents stamps and l is the amount of 11 cents stamps. But how do i proceed? And in the strong induction step what is our inductive hypothesis?

Answer 1

Elementary pedestrian proof (although, I like generating series). - The lowest postage is 7 - Then 11 - After you must be able to decrease $l$ to feed $k$ (and conversely) a rapid examination of the cases leads you to the following amounts [14,18,21,22,25,28,29,32,33,35,36,39,40,42 ... 60,61] - the last one $61$ allows the recurrence using

1. ) $1=2\times 11-3\times 7$
2. ) $1=8\times 7- 5\times 11$

then for your recurrence use alternatively the first and the second expression. From the number $61=3\times 11+4\times 7$ the recurrence works because, if $n=a\times 7+b\times 11\geq 61$, you cannot have $a<8,b<2$ both ($60=7\times 7+1\times 11$) and then you can use (1) or (2).

Answer 2

To answer a) you can use generating functions:

$\left(1+x^7+x^{14}+x^{21}+x^{28}+x^{35}+x^{42}+x^{49}+x^{56}\right) \left(1+x^{11}+x^{22}+x^{33}+x^{44}+x^{55}\right)$

$= 1+x^7+x^{11}+x^{14}+x^{18}+x^{21}+x^{22}+x^{25}+x^{28}+x^{29}+x^{32}+\ x^{33}+x^{35}+x^{36}+x^{39}+x^{40}+x^{42}+x^{43}+x^{44}+x^{46}+x^{47}+\ x^{49}+x^{50}+x^{51}+x^{53}+x^{54}+x^{55}+x^{56}+x^{57}+x^{58}$

By looking at the powers we can see those numbers that can be represented. For instance, we see x^25 so we know that 25 is representable. We already know by the Frobenius numbers that 59 is the largest number that cannot be represented ( see my comment above ).

Answer 3

Not a complete answer, but might help:

• $A_{60} = \{7,7,7,7,7,7,7,11\}$
• $A_{61} = \{7,7,7,7,11,11,11\}$
• $A_{62} = \{7,11,11,11,11,11\}$
• $A_{63} = \{7,7,7,7,7,7,7,7,7\}$
• $A_{64} = \{7,7,7,7,7,7,11,11\}$
• $A_{65} = \{7,7,7,11,11,11,11\}$
• $A_{66} = \{11,11,11,11,11,11\}$
• $A_{n} = \{7\} \cup A_{n-7}$ inductive step

Answer 4

Consider $n=7k+11 t$, where $k\geq 3$ or $t\geq 5$. Otherwise, where $k<3$ and $t<5$, we have $n<58$ which is contradiction. To proceed, if $k\geq 3$, then we have $n+1 =7 (k-3)+11(t+2)$. If $t\geq 5$, then we have $n+1 =7(k+8)+11(t-5)$.

Answer 5

2x11 - 3x7 = 1 and 8x7 -5x11=1

So if you can get n, you can get n+1 by either adding 2 11s and removing 3 7s. This assumes you have at least 3 7s. If you don't you can add 8 7s and remove 5 11s (and then you'll have more than 3) which assumes you have at least 5 11s. If you never get less than 3 11s you can switch back and forth with these techniques.

For n=60 we can do 7 7s and 1 11. Then we add 2 11s and subtract 3 7s to get 61. Do it again to get 62. We will have 5 11s and 1 seven. We then have to switch method but then then at at least 63 if we ever get down to 3 11s we will have at least 4 7s so we can add two 11s and subtract 3 to get the next.