Problem 25 · 2021 Math Kangaroo
Stretch
Logic & Word Problems
Counting & Probability
sum-constraintcomplementary-counting
A box contains only green, red, blue and yellow counters. There is always at least one green counter amongst any 27 counters chosen from the box; always at least one red counter amongst any 25 counters chosen; always at least one blue amongst any 22 counters chosen and always at least one yellow amongst any 17 counters chosen. What is the largest number of counters that could be in the box?
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Answer: B — 29
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Hint 1 of 2
'Any 27 chosen contain a green' means you can never pick 27 with no green — so the non-green counters number at most 26.
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Hint 2 of 2
Write the same kind of bound for each colour and add them up.
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Approach: bound the non-colour counts and add
- Non-green ≤ 26, non-red ≤ 24, non-blue ≤ 21, non-yellow ≤ 16.
- Each counter is 'non' for three of the four colours, so summing: 3T ≤ 26+24+21+16 = 87.
- Thus T ≤ 29, and this total is achievable.
- So the answer is B.
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