Question

1.Inspired by the Harry Potter books, where the valucs of the types of coins that are available are prime multiples of one another (17 Sickles in a Gallcon, and 29 Knuts in a Sickle), the dictator of a small nation decided that only three types of coins would have legal tender, all based on a unit called thc GSK Coins worth 5 GSK Coins worth 7 GSK. Coins worth 11 GSK a)Prove that there is no way to buy an item worth 13 GKS using exact change. b) Prove that it is possible to find a collection of coins that add up to r GSK for every 214. Hint: use mathematical induction. Clcarly state the base casc(s), your induction hypothcsis, and your proofs for the base case(s) and induction step

Answer 1

a) Suppose that it is possible to buy an item worth GSK using exact change. Then we will have integers $n_5,n_7,n_{11}\geq 0$ such that $13=5n_5+7n_7+11n_{11}$ . Note that $0\leq n_{11}\leq 1$ , and if then we should have integers $n_5,n_7\geq 0$ such that $13=5n_5+7n_7+11~\Rightarrow~5n_5+7n_7=2$ ; but this is impossible because if $5n_5+7n_7\geq 5$ unless $n_5=n_7=0$ , and if $n_5=n_7=0$ then $5n_5+7n_7=0$ . Therefore, $n_{11}=0$ .

Thus, $13=5n_5+7n_7$ . Again, $0\leq n_{7}\leq 1$ (because $n_{7}\geq 2$ implies $13=5n_5+7n_7\geq 14$ , a contradiction). If $n_7=1$ then $13=5n_5+7n_7~\Rightarrow~5n_5+7=13~\Rightarrow~5n_5=6~\Rightarrow~n_5=1.2$ which is absurd because $n_5,n_7,n_{11}\geq 0$ are integers. If $n_7=0$ then $13=5n_5+7n_7~\Rightarrow~5n_5=13~\Rightarrow~n_5=2.6$ which is absurd because $n_5,n_7,n_{11}\geq 0$ are integers.

Therefore, no integers $n_5,n_7,n_{11}\geq 0$ exists such that $13=5n_5+7n_7+11n_{11}$ . This proves the statement.

b) For integers let be the statement that there are integers $n_5,n_7,n_{11}\geq 0$ such that

$x=5n_5+7n_7+11n_{11}$

We need to prove that is true for all integers $x\geq 14$ . We do so via mathematical induction.

Base case: If $x=14$ then

$14=5(0)+7(2)+11(0)$

Thus, is true.

Induction hypothesis: Suppose is true for some $x\geq 14$ . Then there are integers $n_5,n_7,n_{11}\geq 0$ such that

$x=5n_5+7n_7+11n_{11}$

We consider the following scenario:

i) If $n_5\geq 2$ then we get $x+1=5(n_5-2)+7n_7+11(n_{11}+1)$ .

ii) If $n_7\geq 2$ then we get $x+1=5(n_5+3)+7(n_7-2)+11n_{11}$ .

iii) If $n_{11}\geq 1$ then we get .

iv) If $n_5<2,~n_7<2,~n_{11}<1$ then $x=5n_5+7n_7+11n_{11}\leq 5+7=12$ which is impossible since $x\geq 14$ . But then, we have proved that either i) or ii), or iii) must be true, in which case is true.

Thus, by induction, is true for all $x\geq 14$ .