Problem 22 · 2021 Math Kangaroo
Stretch
Number Theory
sum-constraintcasework
The numbers 1, 2, 7, 9, 10, 15 and 19 are written down on a blackboard. Two players alternately delete one number each until only one number remains on the blackboard. The sum of the numbers deleted by one of the players is twice the sum of the numbers deleted by the other player. What is the number that remains?
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Answer: B — 9
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Hint 1 of 2
The whole list sums to 63; if one player's deletions are twice the other's, what must the remaining number satisfy?
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Hint 2 of 2
Check which candidate value can actually be split into the required two groups of three.
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Approach: use the total and a divisibility test, then verify a split
- Total is 1+2+7+9+10+15+19 = 63. If deletions are k and 2k, then 63 − r = 3k, so r is a multiple of 3.
- The multiples of 3 in the list are 9 and 15; only r = 9 allows a valid split: {1,2,15}=18 and {7,10,19}=36 = 2·18.
- So the remaining number is 9.
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