Difference between revisions of "2004 AIME II Problems/Problem 15"
m (→Changed so that it says solution 1 and 2) |
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s_{45, 6} &= 37 \\ | s_{45, 6} &= 37 \\ | ||
s_{82, 9} &= 296 \\ | s_{82, 9} &= 296 \\ | ||
− | s_{941, 10} &= \boxed{593}</cmath> | + | s_{941, 10} &= \boxed{593}\end{align*}</cmath> |
== Solution 2 == | == Solution 2 == |
Revision as of 19:54, 13 March 2015
Contents
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. We can find the number of squares above the number instead. If the number doesn't change position, then we add the number of squares we just folded. Otherwise, we just take the number of squares under it before we folded and now these are above the number.
First its in position with spaces over it. We flip once, since is to the right it gets flipped onto itself, going from position to . Now its in position 83, still has spaces over it.
Flip again, it's still in position 83 but now, since it didn't move position, we add the thickness of the fold we just flipped, which is 2. So now there are spaces over it.
Flip again, its to the left of the fold again, so we add squares to get .
Flip again, it goes from in position, and there was square below our number before we flipped, so now that one number is above it.
Flip again, it goes , and now there were squares below it so now they are above it.
Flip again, it goes , and there were squares below it and now they are above it.
Flip again, it goes and there were squares below it and now they are above it.
Flip again, it stays in position, so we add to get .
Flip again, it goes , and there were squares below it and now they are above it.
Flip again, it goes , and there were squares below it and now they are above it.
Since the question asks for how many squares were below our number, our answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.