# 2008 iTest Problems/Problem 10

## Problem

Tony has an old sticky toy spider that very slowly goes down a wall after being stuck to the wall. In fact, left untouched, the toy spider crawls down at a rate of one inch for every two hours it's left stuck to the wall. One morning, at around $9$ o' clock, Tony sticks the spider to the wall in the living room three feet above the floor. Over the next few mornings, Tony moves the spider up three feet from the point where he finds it. If the wall in the living room is $18$ feet high, after how many days (days after the first day Tony places the spider on the wall) will Tony run out of room to place the spider three feet higher?

## Solution

If the toy spider moves down one inch every two hours, then the spider moves down one foot every 24 hours. This means that the daily net gain of the elevation is $2$ feet.

After one day, the spider is 5 feet above the ground. After two days, the spider is 7 feet above the ground. The elevation of the spider (at 9:00) can be modeled as $y = 2x+3$, where $x$ is the number of days elapsed (whole number) and $y$ is the elevation.

After $7$ days, the spider is $17$ feet high, so after $\boxed{8}$ days, Tony can not move the spider three feet higher since the spider at that time is only two feet from the ceiling.