2004 AIME II Problems/Problem 15
Problem
A long thin strip of paper is units in length, unit in width, and is divided into unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a by strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a by strip of quadruple thickness. This process is repeated more times. After the last fold, the strip has become a stack of unit squares. How many of these squares lie below the square that was originally the nd square counting from the left?
Solution 1
Number the squares . In this case , but we will consider more generally to find an inductive solution. Call the number of squares below the square after the final fold in a strip of length .
Now, consider the strip of length . The problem asks for . We can derive some useful recurrences for as follows: Consider the first fold. Each square is now paired with the square . Now, imagine that we relabel these pairs with the indices - then the value of the pairs correspond with the values - specifically, double, and maybe (if the member of the pair that you're looking for is the top one at the final step).
So, after the first fold on the strip of length , the square is on top of the square. We can then write
(We add one because is the odd member of the pair, and it will be on top. This is more easily visually demonstrated than proven.) We can repeat this recurrence, adding one every time we pair an odd to an even (but ignoring the pairing if our current square is the smaller of the two):
We can easily calculate from a diagram. Plugging back in,
Solution 2
More brute force / thinking about the question logically. If the number doesn't change position, then the number of squares below it does not change. Otherwise, it changes position ( for the -th fold). We just take the number of squares under it before we folded and now these are above the square.
Initially it is in position with squares below it.
Solution 3
We can keep track of the position of the square labeled 942 in each step. We use an coordinate system, so originally the 942 square is in the position . In general, suppose that we've folded the strip into an array squares wide and squares tall (so we've made folds). Then if a square occupies the location , we find that after the next fold, it will be in the location described by the procedure Therefore, we can keep track of the square's location in the following table. Therefore, at the end of the process, the square labeled 942 will be in the position , i.e., it will be above squares.
See also
2004 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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