Problem 18 · 2013 Math Kangaroo
Hard
Algebra & Patterns
sum-constraintcasework
Five consecutive positive integers have the following property: the sum of three of the numbers is as big as the sum of the other two. How many sets of 5 such numbers are there?
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Answer: C — 2
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Hint 1 of 2
Let the five numbers be n, n+1, n+2, n+3, n+4 and write the three-equal-two condition as an equation.
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Hint 2 of 2
The three chosen must total half of all five, which is only possible for small n.
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Approach: split the five and require equal halves
- The five numbers total 5n + 10; splitting into a 3-group equal to a 2-group needs each half to be (5n + 10) / 2.
- Testing small starts: {2,3,4,5,6} works (2 + 3 + 5 = 4 + 6 = 10) and {4,5,6,7,8} works (4 + 5 + 6 = 7 + 8 = 15).
- For larger n the two largest can no longer balance the three smallest, so there are exactly 2 such sets.
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