AMC 10 Daily Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 13 Medium

A gold coin is worth x\% more than a silver coin. The silver coin is worth y\% less than the gold coin. Both x and y are positive integers. How many possible values for x are there?

  • A.

    0

  • B.

    3

  • C.

    6

  • D.

    9

  • E.

    12

Answer:E

Let the values of a gold coin and a silver coin be g and s dollars, respectively. Since a gold coin is worth x \% more than a silver coin, g=\frac{100+x}{100} \times s. Hence \frac{g}{s}=\frac{100+x}{100}. Since a silver coin is worth y \% less than a gold coin, s=\frac{100-y}{100} \times g. Hence \frac{g}{s}=\frac{100}{100-y}. Therefore \frac{100+x}{100}=\frac{100}{100-y}. Hence (100+x)(100-y)=100 \times 100=10000. Because x is a positive integer, it follows that 100+x is a factor of 10000 with 100+x>0. Therefore we need to count the factors of 10000 which are greater than 100. There are 12 such factors.

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