Tourist Trouble


On the last day of his trip, a tourist from Paupa New Guinea wanted to buy a $7 vase from my shop. However, he ran out of Australian dollars and only had Paupa New Guinean coins--both circular and triangular--with him. The exchange rate on that day was that 7 circular coins were worth $6 and 7 triangular coins were worth $11.

Would he be able to pay for the vase without requiring change from me?

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1 solution

Each circular coin is worth 6 7 \dfrac{6}{7} Australian dollars and each triangular coin is worth 11 7 \dfrac{11}{7} Australian dollars. So to be able to pay without change being made, there must exist non-negative integers m , n m,n such that

6 7 m + 11 7 n = 7 6 m + 11 n = 49 \dfrac{6}{7}m + \dfrac{11}{7}n = 7 \Longrightarrow 6m + 11n = 49 .

But by the Chicken McNugget Theorem we know that the greatest integer that cannot be written in this form is m n m n = 6 × 11 6 11 = 49 mn - m - n = 6 \times 11 - 6 - 11 = 49 , and so the answer is No \boxed{\text{No}} .

Were you surprised to see the Chicken McNugget theorem pop up in this setting?

To create the problem, i needed ( a 1 ) ( b 1 ) = c d + 1 (a-1)(b-1) = cd + 1 , and I decided to set c = d c = d , which made c = d = 7 , c d + 1 = 50 c = d =7 , cd+1 = 50 a good choice for the values.

Chung Kevin - 4 years, 2 months ago

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Yes, I was pleasantly surprised. Nicely crafted problem. :)

Brian Charlesworth - 4 years, 2 months ago

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