Maths Olympiad Prep

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, 2018

Algebra Difficulty 4.8 AIME Find the answer United States

Problem:

Randall proposes a new temperature system called Felsius temperature with the following conversion between Felsius {}E\{ \}^{\circ} E, Celsius {}C\{ \}^{\circ} C, and Fahrenheit {}F\{ \}^{\circ} F:
{}E=7×{}C5+16=7×{}F809. \{ \}^{\circ} E = \frac{7 \times \{ \}^{\circ} C}{5} + 16 = \frac{7 \times \{ \}^{\circ} F - 80}{9}.
For example, 0C=16E0^{\circ} C = 16^{\circ} E. Let x,y,zx, y, z be real numbers such that xC=xEx^{\circ} C = x^{\circ} E, yE=yFy^{\circ} E = y^{\circ} F, zC=zFz^{\circ} C = z^{\circ} F. Find x+y+zx + y + z.

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

Solution

Solution:

Notice that (5k)C=(7k+16)E=(9k+32)F(5 k)^{\circ} C = (7 k + 16)^{\circ} E = (9 k + 32)^{\circ} F, so Felsius is an exact average of Celsius and Fahrenheit at the same temperature. Therefore we conclude that x=y=zx = y = z, and it is not difficult to compute that they are all equal to 40-40.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.