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2018 AMC 8

Problem 5

Problem 5 · 2018 AMC 8 Easy
Arithmetic & Operations groupingarithmetic-series

What is the value of

1 + 3 + 5 + … + 2017 + 2019 − 2 − 4 − 6 − … − 2016 − 2018 ?
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Answer: E — 1010.
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Hint 1 of 2
Don't add hundreds of numbers. Notice the odds and evens almost interleave: pair each even with the odd just above it (3 with 2, 5 with 4…) and every pair collapses to the same tiny number.
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Hint 2 of 2
The technique is pairing for a constant difference: line up the two lists so neighbors differ by 1, count the pairs, and the leftover term is what's sticking out at the start.
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Approach: pair adjacent odd/even terms
  1. Regroup as 1 + (3 − 2) + (5 − 4) + … + (2019 − 2018). The 1 at the front has no even partner; every other odd pairs with the even just below it, and each pair equals 1.
  2. The evens run 2, 4, …, 2018, which is 1009 numbers, so there are 1009 pairs — each contributing 1.
  3. Total: 1 + 1009 = 1010. Sanity check: there are 1010 odds and 1009 evens, so one extra positive term survives — a positive answer near 1000, ruling out the negative choices instantly.
  4. You'll see it again: when two long sequences are subtracted term-by-term, pairing turns the whole thing into "(how many pairs) × (the per-pair value)."
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