Problem 29 · 2009 Math Kangaroo
Stretch
Counting & Probability
taxicab-distancecareful-counting
A kangaroo is sitting at the origin of a Cartesian coordinate system. With each bounce it can jump one unit in the horizontal or the vertical direction. How many points are there where the kangaroo could be after 10 jumps?
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Answer: A — 121
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Hint 1 of 2
After 10 unit jumps, a reachable point has |x| + |y| ≤ 10, and its parity matches 10 (even).
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Hint 2 of 2
Count all lattice points inside that diamond with even taxicab distance.
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Approach: count even-parity points within taxicab distance 10
- Each jump changes \(x\) or \(y\) by 1, so after 10 jumps \(|x| + |y|\) is at most 10 and has the same parity as 10, i.e. even.
- Every such point is actually reachable, so count the lattice points with \(|x|+|y|\) even and \(\le 10\).
- By distance: 1 point at distance 0, then \(4d\) points at each even distance \(d = 2,4,6,8,10\), giving \(1 + 4(2+4+6+8+10) = 1 + 120 = 121\).
- So there are 121 points.
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