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Balti Tee 2019 · Ülesanne 5

Algebra

The 2m2m numbers

1⋅2, 2⋅3, 3⋅4, …, 2m(2m+1)1\cdot2,\ 2\cdot3,\ 3\cdot4,\ \ldots,\ 2m(2m+1)

are written on a blackboard, where m≥2m\ge2 is an integer. A move consists of choosing three numbers a,b,ca,b,c, erasing them from the board and writing the single number

abcab+bc+ca.\frac{abc}{ab+bc+ca}.

After m−1m-1 such moves, only two numbers will remain on the blackboard. Supposing one of these is 43\frac43, show that the other is larger than 4.

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Denoting the new number

g=abcab+bc+ca,g = \frac{abc}{ab + bc + ca},

its reciprocal is

1g=ab+bc+caabc=1a+1b+1c.\frac{1}{g} = \frac{ab + bc + ca}{abc} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.

The sum of all reciprocals is therefore invariantly

∑k=12m1k(k+1)=∑k=12m(1k−1k+1)=1−12m+1=2m2m+1.\sum_{k=1}^{2m} \frac{1}{k(k+1)} = \sum_{k=1}^{2m} \left( \frac{1}{k} - \frac{1}{k+1} \right) = 1 - \frac{1}{2m+1} = \frac{2m}{2m+1}.

The other remaining number xx can then be calculated from the equation

1x+34=2m2m+1,\frac{1}{x} + \frac{3}{4} = \frac{2m}{2m+1},

leading to

x=4(2m+1)2m−3=4+162m−3>4.x = \frac{4(2m+1)}{2m-3} = 4 + \frac{16}{2m-3} > 4.

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