Problem 17 · 2013 AMC 8
Easy
Algebra & Patterns
consecutive-integers
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
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Answer: B — 338.
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Hint 1 of 2
Consecutive numbers are balanced around their middle, so their sum is just (average) × (count). Divide 2013 by 6 to land right in the center of the six numbers — then the largest is a short hop away.
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Hint 2 of 2
For evenly spaced numbers, sum = average × count, and the average sits dead center. Find the center first; the endpoints follow.
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Approach: divide to find the center, then step to the largest
- Six consecutive integers average to sum ÷ 6 = 2013 ÷ 6 = 335.5. With 6 numbers the center falls between the 3rd and 4th, so those two are 335 and 336.
- Counting outward, the six are 333, 334, 335, 336, 337, 338, so the largest is 338.
- You'll see this again: "sum = average × count" cracks every consecutive-integer problem — find the middle, then walk to whichever term you need.
Another way — algebra with a variable:
- Let the smallest be x. The sum is x + (x+1) + … + (x+5) = 6x + 15.
- Set 6x + 15 = 2013, so 6x = 1998 and x = 333. Largest = 333 + 5 = 338.
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