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A student wishes to travel to the USA and changes amounts of Euro and Swiss Franc in Dollar (in whole numbers!). The exchange rates are \$ 1.35 = € 1.00 and \$ 1.12 = CHF 1.00. If the student receives \$ 66.49 in total, then how much money of each currency has been changed?

So I don't really understand this task. The student has got \$ 66.49.

This seems problematic "how much money of each currency has been changed?".

Am I asked to convert \$ 66.49 into € and CHF?

If so then: $$\frac{\$66.49}{\$1.35}=€49.25$$

and

$$\frac{\$66.49}{\$1.12}=50.37 \text{ CHF}$$

But wait, it was mentioned "in whole numbers"... So what to do? Or the task is just bad? I think it cannot be as easily done as I tried here, no way : /

Edit: I heard there is a recommendation using extended euclidean algorithm but I don't see the use of it here at all.

Matt Samuel
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cnmesr
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    You are not asked to convert $ 66.49 into € and CHF. You are told that the amount of currency of each type (Euros and Francs) exchanged is a whole number. I suspect that this is going to lead towards a diophantine equation. – Shraddheya Shendre Nov 17 '16 at 19:31
  • @nbro integers. – cnmesr Nov 17 '16 at 19:33
  • Something's amiss. I followed the interpretation I mentioned in my first comment, but the resulting diophantine equation does not have a solution. @cnmesr - As to your recent edit, diophantine equations are solved using Extended Euclidean Algorithm. – Shraddheya Shendre Nov 17 '16 at 19:39
  • Maybe you can tell me how the equations would look then I'd like to see if solvable or not, assuming I get that far.. ^.^ – cnmesr Nov 17 '16 at 19:41
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    Assume that $e$ Euros and $f$ Francs were exchanged. Then $$1.35e + 1.12f = 66.49$$ Multiply throughout by 100 and solve to get the integral answers to your equation. – Shraddheya Shendre Nov 17 '16 at 19:51
  • Thanks a lot I will do that :-) But still, I would never get the idea of doing it if this was an exam.. : / – cnmesr Nov 17 '16 at 20:21

2 Answers2

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Scaling by $\,100\,$ we need to solve $\ 112x + 135 y = 6649\ $ for $\,x,y\in\Bbb N.\,$ Using Gauss's algorithm

$\quad {\rm mod}\ 112\!:\,\ y \equiv \dfrac{6649}{135}\equiv \dfrac{41}{23}\,\overset{\times 5}\equiv\, \dfrac{205}{115}\equiv \dfrac{93}3\equiv 31,\ $ so $\,\ x = \dfrac{6649-135\cdot 31}{112} = 22$

Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.

Community
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Bill Dubuque
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The problem is asking you to find an integer amount of Euros and an integer amount of Swiss Franc which when converted gives you USD 66.49, so you need to solve

$1.35e + 1.12f = 66.49$

where $e$ and $f$ are integers, which, as someone else has already said in the comments, is a diophantine equation.

Now, I don't know how to properly solve a diophantine equation, but I have found a solution.

You know that the final digit is 9, and since multiplying 1.35 will only ever get you final digits 0 or 5 and multiplying 1.12 will never get you final digit 9, it must be that $e$ is odd and $f$ will get you a 4.

So, first I tried to find an odd integer $a$ such that

$1.35 a + 2.24 = n + 0.49$

where n in any integer. I literally got a calculator and attempted values and found $a = 15$ and $n = 22$. So now we have 22.49 and we need 44 still.

Since 44 is an integer you know you have to multiply 1.12 by a multiple of 5 and 1.35 by an even number. $1.12 \times 5 = 5.6$ and $1.35 \times 4 = 5.4 $ so that

$1.12 \times 5 + 1.35 \times 4 = 11$

If you multiply that by 4 you have the missing 44 and finally

$ 1.35 \times 15 + 1.12 \times 2 + 4 \times (1.12 \times 5 + 1.35 \times 4) = 31 \times 1.35 + 22 \times 1.12 = 66.49$

Thus,

$e = 31$

$f = 22$

P. Ewald
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