Maths Olympiad Prep

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Problem 557

Algebra Difficulty 2.6 Find the answer CEMC Fermat

If x+2y=30x + 2y = 30, what is the value of x5+2y3+2y5+x3\frac{x}{5} + \frac{2y}{3} + \frac{2y}{5} + \frac{x}{3}?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Since x+2y=30x + 2y = 30, then x5+2y3+2y5+x3=x5+2y5+x3+2y3=15x+15(2y)+13x+13(2y)=15(x+2y)+13(x+2y)=15(30)+13(30)=6+10=16\frac{x}{5} + \frac{2y}{3} + \frac{2y}{5} + \frac{x}{3} = \frac{x}{5} + \frac{2y}{5} + \frac{x}{3} + \frac{2y}{3} = \frac{1}{5}x + \frac{1}{5}(2y) + \frac{1}{3}x + \frac{1}{3}(2y) = \frac{1}{5}(x + 2y) + \frac{1}{3}(x + 2y) = \frac{1}{5}(30) + \frac{1}{3}(30) = 6 + 10 = 16

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