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2003 AMC 10B Problem 3

Problem 3 of 25EasierAlgebra

The sum of 55 consecutive even integers is 44 less than the sum of the first 88 consecutive odd counting numbers. What is the smallest of the even integers?

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Solution

The first 88 odd counting numbers sum to 1+3+⋯+15=64.1+3+\cdots+15=64. Letting nn be the smallest even integer, n+(n+2)+(n+4)+(n+6)+(n+8)=5n+20=60, \begin{gathered} n+(n+2)+(n+4) \\ {}+(n+6)+(n+8) \\ = 5n+20 = 60, \end{gathered} so n=8.n=8. Thus, the correct answer is B.

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Concepts: summation · linear equation

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.