Problem 8 · 2013 Math Kangaroo
Medium
Logic & Word Problems
sum-constraintcasework
Each of six lone heroes has captured some wanted people. In total they captured 20 wanted people: the first hero caught one, the second two, the third three. The fourth hero caught more than any other hero. What is the smallest number the fourth hero could have caught so that this is possible?
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Answer: B — 6
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Hint 1 of 2
The first three heroes account for 1+2+3 = 6, leaving the rest of the 20.
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Hint 2 of 2
Make the fifth and sixth heroes as large as possible while staying below the fourth.
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Approach: minimise the largest under a sum constraint
- Heroes 1,2,3 capture 1+2+3 = 6, so heroes 4,5,6 share 14.
- Hero 4 must beat heroes 5 and 6, so 14 − h4 ≤ 2(h4 − 1), giving h4 ≥ 6.
- h4 = 6 works (e.g. 6,5,3), so the smallest is 6 = B.
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