Robot ant likes even numbers

Algebra Level pending

A robot-ant fell down at the bottom of a well which is H cm H\text{ cm} deep, and it wants to get out.

  • In its first trial, it climbed H 2 cm \frac{H}{2}\text{ cm} up and fell H 3 cm \frac{H}{3}\text{ cm} down.
  • In the second trial, it climbed H 4 cm \frac{H}{4}\text{ cm} up and fell H 5 cm , \frac{H}{5}\text{ cm}, and so on.
  • After 1 0 6 10^{6} trials, the robot-ant is now at a distance of 100.034 cm 100.034\text{ cm} from the bottom of the well.

What is the value of ⌈ H ⌉ ? \lceil H \rceil ?


The answer is 326.

Ossama Ismail by Ossama Ismail , 63, Egypt

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1 solution

David Vreken
Mar 29, 2018

From the description, we have:

H 2 − H 3 + H 4 − H 5 + ⋯ + H 2 ⋅ 1 0 6 − H 2 ⋅ 1 0 6 + 1 = 100.34 \frac{H}{2} - \frac{H}{3} + \frac{H}{4} - \frac{H}{5} + \dots + \frac{H}{2 \cdot 10^6} - \frac{H}{2 \cdot 10^6 + 1} = 100.34

Factoring H H , we have:

H ( 1 2 − 1 3 + 1 4 − 1 5 + ⋯ + 1 2 ⋅ 1 0 6 − 1 2 ⋅ 1 0 6 + 1 ) = 100.34 H(\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots + \frac{1}{2 \cdot 10^6} - \frac{1}{2 \cdot 10^6 + 1}) = 100.34

Since 1 2 − 1 3 + 1 4 − 1 5 + ⋯ + 1 2 ⋅ 1 0 6 − 1 2 ⋅ 1 0 6 + 1 ≈ < 1 2 − 1 3 + 1 4 − 1 5 + ⋯ = 1 − ln ⁡ 2 \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots + \frac{1}{2 \cdot 10^6} - \frac{1}{2 \cdot 10^6 + 1} \approx< \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots = 1 - \ln 2 we have

H ( 1 − ln ⁡ 2 ) ≈ > 100.34 H(1 - \ln 2) \approx> 100.34 and so H ≈ < 325.9999 H \approx< 325.9999 and ⌈ H ⌉ = 326 \lceil H \rceil = \boxed{326} .

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