Five-minute 80951

Karel has an average grade of exactly 1.12 from five-minute episodes. Prove that at least 22 of them have one.

Correct answer:

z = 22*1+3*2
z₂ = 23*1+1*2+1*3

μ = 1.12 n = 22 μ_{2} = \frac{n \cdot 1 + 2}{n + 1} = \frac{2 2 \cdot 1 + 2}{2 2 + 1} = \frac{2 4}{2 3} = 1 \frac{1}{2 3} ≐ 1.0435 μ_{3} = \frac{n \cdot 1 + 3}{n + 1} = \frac{2 2 \cdot 1 + 3}{2 2 + 1} = \frac{2 5}{2 3} = 1 \frac{2}{2 3} ≐ 1.087 μ_{4} = \frac{n \cdot 1 + 4}{n + 1} = \frac{2 2 \cdot 1 + 4}{2 2 + 1} = \frac{2 6}{2 3} = 1 \frac{3}{2 3} ≐ 1.1304 μ_{5} = \frac{n \cdot 1 + 5}{n + 1} = \frac{2 2 \cdot 1 + 5}{2 2 + 1} = \frac{2 7}{2 3} = 1 \frac{4}{2 3} ≐ 1.1739 μ_{4} > μ; μ_{5} > μ μ \cdot (n + x) = n \cdot 1 + x \cdot 2 1.12 \cdot (22 + x) = 22 \cdot 1 + x \cdot 2 0.88 x = 2.64 x = \frac{2 . 6 4}{0 . 8 8} = 3 x = 3 z = 22 \cdot 1 + 3 \cdot 2

1.12 = \frac{1 \cdot a + 2 \cdot b + 3 \cdot c}{a + b + c} 112 \cdot (a + b + c) = 100 \cdot (1 \cdot a + 2 \cdot b + 3 \cdot c) 112 a + 112 b + 112 c = 100 a + 200 b + 300 c 12 a = 88 b + 188 c 3 a = 22 b + 47 c; a > 0; b \geq 0; c \geq 0 a_{1} = 22, b_{1} = 3, c_{1} = 0 a_{2} = 23, b_{2} = 1, c_{2} = 1 a_{3} = 44, b_{3} = 6, c_{3} = 0 a_{4} = 45, b_{4} = 4, c_{4} = 1 a_{5} = 46, b_{5} = 2, c_{5} = 2 a_{6} = 47, b_{6} = 0, c_{6} = 3 a_{7} = 66, b_{7} = 9, c_{7} = 0 a_{8} = 67, b_{8} = 7, c_{8} = 1 a_{9} = 68, b_{9} = 5, c_{9} = 2 a_{10} = 69, b_{10} = 3, c_{10} = 3 a_{11} = 70, b_{11} = 1, c_{11} = 4 a_{12} = 88, b_{12} = 12, c_{12} = 0 a_{13} = 89, b_{13} = 10, c_{13} = 1 a_{14} = 90, b_{14} = 8, c_{14} = 2 a_{15} = 91, b_{15} = 6, c_{15} = 3 a_{16} = 92, b_{16} = 4, c_{16} = 4 a_{17} = 93, b_{17} = 2, c_{17} = 5 a_{18} = 94, b_{18} = 0, c_{18} = 6 a_{19} = 110, b_{19} = 15, c_{19} = 0 \dots z_{2} = 23 \cdot 1 + 1 \cdot 2 + 1 \cdot 3

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