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With only 15 and 21 dollar bills...

a)What is the cheapest item you could buy in this country? (What would you give the store clerk and what change would the store clerk give you?)

I know that the gcd of 15 and 21 is 3 and thus you can buy a 3 dollar item by giving the store clerk 3 15 dollar bills for 2 21 dollar bills. How do I prove that this is the cheapest item i.e. you cannot buy a 1 or 2 dollar item?

b)Suppose the store clerk does not have any change, what is the value of the items that you could buy? (Describe all such numbers)

I figured out that I can make any number that is a multiple of 3 and is 72 or greater. The numbers I can make below 72 are combinations of some number of 15s, 21, or a combination:

15s: 15, 30, 45, 60

21s: 21, 42, 63

15+21 (36): 36, 72

How would I describe the numbers and explain my answer?

JMoravitz
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user510631
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  • with algebra the equation $k\cdot 21 + l\cdot 15 = m$ you can rewrite $3(7k+5l) = m$, therefore $m$ must have at least one factor 3 in it's prime factor representation. – mathreadler Dec 10 '17 at 20:36
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    For (b), see coin problem (that link assumes $\gcd=1$, so you need to adapt it to your $\gcd=3$ case). – Arthur Dec 10 '17 at 20:36
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    The number of dollars that changes hands in any transaction will always be a multiple of three. Change or no change. So your idea of using $\gcd$ is a good one. – Jyrki Lahtonen Dec 10 '17 at 20:37
  • For part b I like to refer the askers to this explanation by robjohn. You are interested in the numbers $7k+5\ell$ as pointed out also by mathreadler. So use $a=7$, $b=5$ in that argument, and multiply everything by three in the end. As Arthur pointed out this is an instance of the coin problem. If you want to go to the deep end google for more material on numerical semigroups. – Jyrki Lahtonen Dec 10 '17 at 20:38
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    You asked this same question several times over the past week. What was unsatisfactory about the comments and answers you got previously? What more would you want from us without holding your hand? The comments you receive here will likely be the the same as before unless you explain why you didn't understand earlier commenters or explain why you feel the earlier comments were insufficient. Rather than opening a new question to continue the conversation about the problem, you could have and should have responded to the comments in the older problem. – JMoravitz Dec 10 '17 at 20:55
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    Further, you tagged this with "combinatorics" and "permutations." By a rather liberal interpretation, I suppose you could call this combinatorics, but by no definition or interpretation should the term "permutations" be able to describe this problem or any aspect of it. I'm changing the tags for you to something more appropriate. – JMoravitz Dec 10 '17 at 20:58
  • I agree with JMoravtiz, this is a repetitive question with many satisfactory answers, what else do you need? – The Math Guy Dec 10 '17 at 21:23

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