HMMT November

Algebra Difficulty 5.7 AIME, harder Prove it United States

Problem:

The very hungry caterpillar lives on the number line. For each non-zero integer $i$ , a fruit sits on the point with coordinate $i$ . The caterpillar moves back and forth; whenever he reaches a point with food, he eats the food, increasing his weight by one pound, and turns around. The caterpillar moves at a speed of $2^{-w}$ units per day, where $w$ is his weight. If the caterpillar starts off at the origin, weighing zero pounds, and initially moves in the positive $x$ direction, after how many days will he weigh 10 pounds?

Solution

Solution:

On the $n$ th straight path, the caterpillar travels $n$ units before hitting food and his weight is $n-1$ . Then his speed is $2^{1-n}$ . Then right before he turns around for the $n$ th time, he has traveled a total time of $\sum_{i=1}^{n} \frac{i}{2^{1-i}}=\frac{1}{2} \sum_{i=1}^{n} i \cdot 2^{i}$ . We want to know how many days the caterpillar moves before his weight is 10, so we want to take $n=10$ so that his last straight path was taken at weight 9. Hence we want to evaluate $S=\frac{1}{2} \sum_{i=1}^{10} i \cdot 2^{i}$ . Note that $2S=\frac{1}{2} \sum_{i=2}^{11}(i-1) \cdot 2^{i}$ , so $S=2S-S=\frac{1}{2}\left(11 \cdot 2^{11}-\sum_{i=1}^{10} 2^{i}\right)=\frac{1}{2}\left(10 \cdot 2^{11}-2^{11}+2\right)=9217$ .

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