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Algebra Difficulty 4.8 AIME Prove it South Africa

Let ff be a function satisfying
f(xy)=f(x)y f(xy) = \frac{f(x)}{y}
for all positive real numbers xx and yy. If f(500)=3f(500) = 3, what is the value of f(100)f(100)?

Solution

We are given 3=f(500)=f(100×5)=f(100)53 = f(500) = f(100 \times 5) = \frac{f(100)}{5}, so f(100)=3×5=15f(100) = 3 \times 5 = 15.

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