Addition wasn't this hard, the last time I checked

Number Theory Level 5

111. 1 2 + 222. 2 3 + 333. 3 4 + . . . + 888. 8 9 = k 10 111.1_{2} + 222.2_{3} + 333.3_{4} +...+ 888.8_{9} = k_{10}

Find ⌊ k ⌋ \lfloor k \rfloor .

Note: The subscripts denote the base of the number.


The answer is 2022.

Efren Medallo by Efren Medallo , 26, Philippines

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1 solution

Efren Medallo
Jun 20, 2015

This is basically

∑ k = 2 9 ( k − 1 ) ( k 2 + k + 1 + 1 / k ) \sum_{k=2}^{9} (k-1)(k^2 + k + 1 + 1/k)

Simplifying that, we get

∑ k = 2 9 ( k 3 − 1 k ) \sum_{k=2}^{9} (k^3 - \frac{1}{k})

I manually plugged in the values for 1 k 2 \frac{1}{k^2} and combining these two summations, we arrive at approx 2022.17 2022.17 . The equivalent fraction of this number is too lengthy.

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