Problem 15 · 2012 Math Kangaroo
Medium
Logic & Word Problems
work-backwardsum-constraint
A flea stands on the floor and wants to climb the 10 steps. Each time it can either jump 3 steps up or jump 4 steps down. What is the smallest number of jumps it must make?
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Answer: E — 8
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Hint 1 of 2
Each up-jump moves him 3 steps, so just adding up-jumps lands on 3, 6, 9, 12, ... — never exactly 10.
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Hint 2 of 2
To fix that miss, mix in a down-jump of 4 and count how many jumps that takes.
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Approach: skip-count the up-jumps, then patch with a down-jump
- Only going up, the flea lands on 3, 6, 9, 12, 15, 18, ... — he hops right over 10 and never lands on it.
- So he must overshoot and then drop down 4: from 12 a down-jump lands on 8, and from 18 a down-jump lands on 14, still not 10.
- Try going higher: 6 up-jumps reach step 18, then 2 down-jumps of 4 take him 18 → 14 → 10, landing exactly on 10.
- That is 6 + 2 = 8 jumps, and nothing shorter ever lands on 10, so the fewest jumps is 8.
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