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Use mathematical induction to show that for all integers, any price equal or greater than $8$, can be paid for by $3c$ and $5c$ coins.

Harsh Kumar
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user3602479
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3 Answers3

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$n = p*5+q*3$
$n+1 = (p-1)*5+5+1+q*3 = (p-1)*5+(q+2)*3$

Carsten Kühn
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  • This doesn't work as you reduce the number of 5-cent coins used at every step, so eventually you will use a negative number of them. – Especially Lime Feb 03 '17 at 13:11
  • i don't think so: It's just the inductive step, the base case n=5+3 q=1 and p=1is trivial! Or, to put it the other way: if the result is proven for n, then it follows for n+1. (And i'm not talking about "steps" or reducing!) – Carsten Kühn Feb 03 '17 at 13:20
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    No, what you've proven is that for any $n$ there are integers $p,q$ such that $n=5p+3q$. The problem is that they need to be non-negative to translate to a valid way to pay $n$ cents with 3- and 5-cent coins. If you follow your induction for two steps it gives $p=-1$, $q=5$ for $n=10$, which isn't valid. – Especially Lime Feb 03 '17 at 13:26
  • Sorry to insist: The line $n= p5+q3$ "expands" to: Let n be a number, that can be written as the sum a positiv multiple of 5 and a positiv multiple of 3. The next line expands to: if we have the base statement, then the claim holds for n+1. And again, induction is not to be understood as a construction. Regarding your example: The result for 10 holds because it holds for 9 because it holds for 8! p and q may vary from "step" to "step" to put it in the "constructive parlor" – Carsten Kühn Feb 03 '17 at 13:43
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    Just because $n=5p+3q$ is the sum of a positive multiple of 5 and a positive multiple of 3, it doesn't follow that $n+1=5(p-1)+3(q+2)$ is the sum of a positive multiple of 5 and a positive multiple of 3, because $p-1$ might not be positive. – Especially Lime Feb 03 '17 at 13:51
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This is my first time posting on this website but this is what I think the answer is.What we are trying to prove is that any amount of money can be obtained using 3 cents and 5 centers for number n>=8.

Base step: 8 cents can be made using one 3 cents and one 5 cents therefore its is true.

Inductive step: For all K which is greater then 8 there must a combination of 3 cents and 5 cents used.

First case: if there is 5 cent coin used. Then we have to replace the 5 cent coin with two 3 cent coins, then that will be (k+1) Example: k=8 we have a 5 cent and a 3 cent. For k+1=9 we replace that five cent coin with 2 3 cents so we have 3+3+3=9

Other case: if we have a 5 cent coin used then we know that 3 cent coins replaced it. So we should replace three 3 cents coins with two 5 cents to make up (k+1). Example: k=9 we have 3 3 cent coins. For k+1=10 we replace the 3 3 cents coins with 2 five cent coins. 5+5=10.

I hope this is the answer that you are looking for.

yuvaly
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    Nice try, but I think you need to show more clearly that there will be $3$ 3-cent coins available in the "other case". – Joffan Feb 06 '17 at 16:56
  • Since $k\geq8$, if there is no 5-cents, there need to be at least 3 3-cents. 2 3-cents equals to 6. – Alain Remillard Feb 06 '17 at 17:09
  • For k=8 we had one five cents and one 3 cents. So for k+1 we take out the five cents and add 2 3 cents. Originally we had 5+3 but then we would have (3+3)+3 – yuvaly Feb 06 '17 at 17:37
  • Yes, but you need to demonstrate that for all larger numbers you will either have a 5-cent coin or 3 3-cent coins. I think treating $8$ as a special case and arguing for $k\ge 9$ would help you here. – Joffan Feb 06 '17 at 18:05
  • If there is a 5c coin in $n$, replacing it by 2 × 3c gives $n + 1$; if there is no 5c coin, as $n \ge 8$ there are at least 3 × 3c, replace by 2 × 5c to get $n +1$. – vonbrand Aug 15 '19 at 16:50
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We define a recursive function $\mu = (\mu_1,\mu_2): \{8,9,10,\dots\} \to \Bbb N \times \Bbb N$ as follows:

$\quad \mu(8) \,\;= (1,1)$
$\quad \mu(9) \,\;= (3,0)$
$\quad \mu(10) = (0,2)$

$\quad$ If $\mu(k)$, $\mu(k+1)$, and $\mu(k+2)$ are defined then set $\mu(k+3) = \bigr(\mu_1(k)+1,\, \mu_2(k)\bigr)$

It is easy to prove by induction that

$\tag 1 \mu_1(k) \cdot 3 + \mu_2(k) \cdot 5 = k \quad \text{ for all } k \ge 8$

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