Problem 23 · 2024 Math Kangaroo
Stretch
Logic & Word Problems
sum-constraintcasework
We have 6 cards and there is one number written on each side of each card. The pairs of numbers on the cards are (5, 12), (3, 11), (0, 16), (7, 8), (4, 14) and (9, 10). The cards can be placed in the empty squares of the calculation \(\square+\square+\square-\square-\square-\square\) in any order with any side up. What is the smallest possible result of the calculation?
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Answer: D — −26
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Hint 1 of 3
The calculation adds three cards and subtracts three, and for any card you may show either side.
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Hint 2 of 3
A subtracted card should show its big number and an added card its small number, so decide which three cards to subtract.
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Hint 3 of 3
Moving a card from the added group to the subtracted group lowers the total by (its small side + its big side), so subtract the three cards with the largest two-number totals.
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Approach: subtract the three cards with the biggest totals, showing their large sides
- Three cards are added and three are subtracted; clearly each added card should show its small number and each subtracted card its big number.
- Switching a card from 'added' to 'subtracted' changes the total by \(-(\text{small}+\text{big})\), so the three subtracted cards should be the ones with the largest pair-totals: \((9,10)=19\), \((4,14)=18\), \((5,12)=17\).
- Subtracting their big sides gives \(10+14+12=36\); the remaining cards add their small sides \(0+3+7=10\).
- The smallest result is \(10-36=\) −26 (answer D).
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