Problem 29 · 2012 Math Kangaroo
Hard
Spatial & Visual Reasoning
paper-cutting
A piece of string is folded as shown in the diagram by folding it in the middle, then folding it in the middle again and finally folding it in the middle once more. Then this folded piece of string is cut so that several pieces emerge. Amongst the resulting pieces there are some with length 4 m and some with length 9 m. Which of the following lengths cannot be the total length of the original piece of string? (In the picture, “Schnitt” marks where the cut is made.)
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Answer: C — 72 m
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Hint 1 of 3
Folding in half three times stacks the string into eight equal layers before the single cut.
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Hint 2 of 3
Call the two parts the cut makes in one folded layer \(a\) and \(b\); then the whole string has length \(8(a+b)\).
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Hint 3 of 3
Work out the lengths of the pieces in terms of \(a\) and \(b\), then test each total to see whether both a 4 m and a 9 m piece can appear.
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Approach: lengths of the unfolded pieces
- Three folds make 8 stacked layers, so the folded packet has length \(s=a+b\) where the cut lands distance \(a\) from the folded edge, and the whole string is \(8s=8(a+b)\).
- Unfolding, the cut points split the string into pieces of just three lengths: \(a\) (the two ends), \(2a\), and \(2b\); so both a 4 m piece and a 9 m piece must appear among \(a,2a,2b\).
- A total of 52, 68 or 88 m can be split this way with a 4 m and a 9 m piece, but for a total of 72 m we get \(s=9\), and then \(a,2a,2b\) can never give both 4 and 9 at once.
- So the impossible total is 72 m, answer C.
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