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I came across this problem in an examination...

Mtrpia, a small country, has the following coins in circulation:

1 paisa, 2 paise, 5 paise, 10 paise, 20 paise, 50 paise, and 1 rupee.

Suppose it is known that you can pay A paise with B coins. Prove that you can pay B rupees with A coins. [Assume that there are infinitely many coins of each type.]

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    Isn't the only way to pay $B$ rupees with $B$ coins? or am I missing something? – 2017-02-15
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    oh, apparently $100$ paise make a rupee. – 2017-02-15
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    yes correct 1 rupee=100 paise – 2017-02-15

2 Answers 2

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Let $x_1$, $x_2$, $x_5$, $x_{10}$, $x_{20}$, $x_{50}$ and $x_{100}$ be the ammounts of 1, 2, 5, 10, 20, 50 and 100 paises when we give $A$ paises with $B$ coins. Then $$A=x_1+2x_2+5x_5+10x_{10}+20x_{20}+50x_{50}+100x_{100}$$ and $$B=x_1+x_2+x_5+x_{10}+x_{20}+x_{50}+x_{100}.$$

Then $$100B=100x_1+100x_2+100x_5+100x_{10}+100x_{20}+100x_{50}+100x_{100}=\\=100(x_1)+50(2x_2)+20(5x_5)+10(10x_{10})+5(20x_{20})+2(50x_{50})+(100x_{100}).$$

So we can give $100x_{100}$ 1 paise coins, $50x_{50}$ 2 paise coins and so on.

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Suppose that you have $B$ coins with value equal to $A$ paise.

We change each of these $B$ coins as follows: If the coins is worth $x$ paise change it for $x$ coins worth $100/x$ paise. This gives us $x$ coins with total value equal to $1$ rupee.

So in total we get $A$ coins whose total value is $B$.